RE: Ajax unable to perform markup update
Thanks namesake :) I just put the formPanel in a webmarkupcontainer and updated the container. And it works fine. Setting the markup id didn't work for me due to unknown reasons. Anyway, you helped a lot! Thanks! :) Regards, Martin -Original Message- From: Martin Makundi [mailto:martin.maku...@koodaripalvelut.com] Sent: Monday, January 18, 2010 9:55 AM To: users@wicket.apache.org Subject: Re: Ajax unable to perform markup update Or even better.. ajax-update the parent panel not the child panel. ** Martin 2010/1/18 Martin Makundi martin.maku...@koodaripalvelut.com: Hi! If the markupid number changes then ofcourse.. you should maybe call : newPanel.setMarkupid(oldPanel.getMarkupId()); when ajax updating ** Martin 2010/1/18 Martin Asenov mase...@velti.com: Hello, everyone! Although I've managed to handle such errors so far, I am unable to deal with this one... I've got an abstract class MyFormPanel that extends Panel and calls super(form_panel); Because I've got different form panels, i.e. ContactFormPanel, GroupFormPanel, etc. I'm trying to replace the panels according to the user's actions. So... here's a small snippet of the replacement: MyFormPanelContact temp = new ContactFormPanel(new Contact()); temp.setOutputMarkupId(true); formPanel.replaceWith(temp); formPanel = temp; formPanel.setVisible(true); target.addComponent(formPanel); Note: formPanel is my initially created MyFormPanel, that has got setOutputPlaceHolderTag(true); and starts invisible. The first replacement is performed properly, but when a 'new group' button is hit for instance and I do the replacement in the very same manner, Ajax says unable to update markup - couldn't find component with id form_panel_some_number. I just can't figure out why this occurs. If someone gives a hint I would be grateful. Thank you in advance, Martin -Original Message- From: Eyal Golan [mailto:egola...@gmail.com] Sent: Monday, January 18, 2010 8:48 AM To: users@wicket.apache.org Subject: Re: WicketExtensions DataTable : Creating a Links using Model value? You can either override PropertyColumn's populateItem method, and add to the cell the link you want. Or you can use AbstractColumn. Eyal Golan egola...@gmail.com Visit: http://jvdrums.sourceforge.net/ LinkedIn: http://www.linkedin.com/in/egolan74 P Save a tree. Please don't print this e-mail unless it's really necessary On Mon, Jan 18, 2010 at 8:24 AM, Ashika Umanga Umagiliya auma...@biggjapan.com wrote: Greetings, I was going through WicketExtensions DataTable source and I was wondering how to create a hyperlink using PropertyColumn. In docs it says: columns[0] = new PropertyColumn(new Model(Family Id), familyId); What I want to do is, 1) create a Link using the field value familyId,and directing to another page which display SubFamilies related to selected 'familyId' 2) create a Link which direct to another website . eg: http://www.ncbi.org/family?id=$familyid My domain model is as: One 'Family' has many 'SubFamily' objects. || - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
submit a form from outside of it
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button. It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The code is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form) {
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button. It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form) {
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
Re: Escaping variable declaration/resolution in properties files
VariableInterpolator.java says:
$ is the escape char. Thus $${text} can be used to escape it (ignore
interpretation).
On Mon, Jan 18, 2010 at 10:46 AM, Joseph Pachod j...@thomas-daily.de wrote:
Hi
I would like to add some text containing ${variable} in a XML properties
file, for an error message. This text of a variable declaration should be
rendered as it is.
However, wicket always tries to resolve this variable, and thus I get this
exception :
WicketMessage: Exception 'java.lang.IllegalArgumentException: Value of
variable [[_variable]] could not be resolved while interpolating [[Input
cannot contains text like ${_variable}]]' occurred during validation
foo.steps.VariablesNotAcceptedValidator on component
1:generalForm:generalAttributes.jobName
Is there a clean way to avoid it ?
thanks in advance
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IT
THOMAS DAILY GmbH
Adlerstraße 19
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Deutschland
T + 49 761 3 85 59 310
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E joseph.pac...@thomas-daily.de
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WicketExtensions : DataTable : Sending 'id's between web pages?
My Domain model has 'Family' which has many 'SubFamily' objects. I managed to display the 'Family' information using a DataTable in 'FamilyPage'. Now I want to do is,when user clicks on a 'Family Id' the 'SubFamilyPage' should open and display relavant subfamilies in a datatable. I want to know how this events work together , I am wondering whether I have to use HTTP parameters like in JSP pages. Sorry for my ignorant , I am quite a newbie to Wicket. Regards. FamilyPage SubFamilesPage |FamiliesTable| |SubFamilesTable| |_| |___| | FamilidId 1 |-| sub families | | FamilidId 1 | | belong to id 1| - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: PageLink deprecated
I totally agree with Jeroen. The 3rd constructor is dangerous and should be removed. The other two, however, are lazy and create the page in the onClick (provided that the IPageLink interface is implemented correctly). Of course, it is possible to copy PageLink and IPageLink to wicket-security, but that would be a large API-incompatible change, for which I see no compelling reason. Emond On Monday 18 January 2010 08:44:28 Jeroen Steenbeeke wrote: Guys, no need to keep explaining what's wrong with passing a Page in the constructor, we understand that! Forget about that filthy 3rd constructor, I know it's wrong and I never used it anyway. That wasn't what my question was about. There are two more constructors: PageLink(String, Class) PageLink(String, IPageLink) Both of these do not replicate the dangerous behavior illustrated in this thread so far. I understand that we can easily create our own implementation that simulates the behavior we want. I just wanted to understand the reasoning for removing the whole class when only one of the constructors is dangerous. From what Martijn Dashorst just told me, it was a case of seeing as we already have Link and BookmarkablePageLink, we figured you could just use those instead. This is also the source of miscommunication so far. The Javadoc simply states what you should use instead, but does not explicitly state why. The assumption is that any behavior you can achieve with the PageLink/IPageLink combination can also be done with a simple Link. This does not take into account the use of the Page Identity for security checks however (mainly for determining link visibility, which, frankly, does not need an actual instance of the page in question), which brings us back to Emond's original point. On the other hand, one could argue that the only use for the page identity is for security purposes, and it would therefore be more at home in a specialized class in wicket-security. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Close window javascript
Hi,
I'm trying to close a popup window after the user has submitted a message.
I'm using a javascript to close the window but without any luck.
Can anyone see what goes wrong?
This is the AjaxSubmitLink that should submit the message and close the
window:
AjaxSubmitLink submit = new AjaxSubmitLink(submitLink) {
@Override
public void onSubmit(AjaxRequestTarget target, Form form) {
if (log.isDebugEnabled()) {
log.debug(Saveing comment + comment);
}
target.addComponent(response.setVisible(true));
target.addComponent(this.setVisible(false));
caseCommentService.create(...);
this.getPage().add(new
AbstractAjaxTimerBehavior(Duration.milliseconds(3000)) {
protected void onTimer(final AjaxRequestTarget target) {
stop();
target.prependJavascript(window.close(););
}
});
}
};
myForm.add(submit);
What i'm trying to achieve is to show a
message(response.setVisible(true))..) in three seconds before closing the
window.
But something it does not seem to work.
Best Regards
Muro
RE: submit a form from outside of it
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the constructor and
cannot be found in the hierarchy of the component this behavior is attached to
the form is located in the same page and displayed, but it's actually placed
within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button. It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form) {
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
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Re: Escaping variable declaration/resolution in properties files
kirillkh wrote:
VariableInterpolator.java says:
$ is the escape char. Thus $${text} can be used to escape it (ignore
interpretation).
Thanks a lot, it works fine.
--
Joseph Pachod
IT
THOMAS DAILY GmbH
Adlerstraße 19
79098 Freiburg
Deutschland
T + 49 761 3 85 59 310
F + 49 761 3 85 59 550
E joseph.pac...@thomas-daily.de
www.thomas-daily.de
Geschäftsführer/Managing Directors:
Wendy Thomas, Susanne Larbig
Handelsregister Freiburg i.Br., HRB 3947
Registrieren Sie sich unter www.signin.thomas-daily.de für die kostenfreien TD
Morning News, eine Auswahl aktueller Themen des Tages morgens um 9:00 in Ihrer
Mailbox.
Hinweis: Der Redaktionsschluss für unsere TD Morning News ist täglich um 8:30
Uhr. Es werden vorrangig Informationen berücksichtigt, die nach 16:00 Uhr des
Vortages eingegangen sind. Die Email-Adresse unserer Redaktion lautet
redakt...@thomas-daily.de.
To receive the free TD News International - a selection of the day's top issues
delivered to your mail box every day - please register at
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Please note: Information received for our TD News International after 4 p.m. will be given priority for publication the following day. The daily editorial deadline is 8:30 a.m. You can reach our editorial staff at redakt...@thomas-daily.de.
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Re: WicketExtensions : DataTable : Sending 'id's between web pages?
In the onClick() handler of your link, you can call setResponsePage()
with either the id in question, or pass in the full object:
add(new Link(showSubFamily){
private static final long serialVersionUID = 1L;
@Override
public void onClick() {
setResponsePage(new SubFamilyPage(id)); // or:
setResponsePage(new SubFamilyPage(family));
}
});
Of course, the SubFamilyPage needs a constructor that accepts the used
parameter.
Hope that helped.
Bert
On Mon, Jan 18, 2010 at 10:58, Ashika Umanga Umagiliya
auma...@biggjapan.com wrote:
My Domain model has 'Family' which has many 'SubFamily' objects.
I managed to display the 'Family' information using a DataTable in
'FamilyPage'.
Now I want to do is,when user clicks on a 'Family Id' the 'SubFamilyPage'
should open and display relavant subfamilies in a datatable.
I want to know how this events work together , I am wondering whether I have
to use HTTP parameters like in JSP pages.
Sorry for my ignorant , I am quite a newbie to Wicket.
Regards.
FamilyPage SubFamilesPage
|FamiliesTable| |SubFamilesTable|
|_| |___|
| FamilidId 1 |-| sub families |
| FamilidId 1 | | belong to id 1|
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Re: WicketExtensions DataTable : Creating a Links using Model value?
A somewhat similar question was asked just a couple of days ago [1] and, probably, has been asked and answered many times before on this list. Ernesto References [1] http://old.nabble.com/How-to-get-a-cell-work-as-a-link-in-AjaxFallbackDefaultDataTable-td27161378.html On Mon, Jan 18, 2010 at 7:24 AM, Ashika Umanga Umagiliya auma...@biggjapan.com wrote: Greetings, I was going through WicketExtensions DataTable source and I was wondering how to create a hyperlink using PropertyColumn. In docs it says: columns[0] = new PropertyColumn(new Model(Family Id), familyId); What I want to do is, 1) create a Link using the field value familyId,and directing to another page which display SubFamilies related to selected 'familyId' 2) create a Link which direct to another website . eg: http://www.ncbi.org/family?id=$familyid My domain model is as: One 'Family' has many 'SubFamily' objects. ||
Re: submit a form from outside of it
Not sure here, but the AjaxSubmitLink has an constructor that lets you pass in the form it should work on. Bert - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: Tree table with check box
Hey it worked,i wanted checkbox in other column
Above apparoach works perfectly.
Thanks
prati wrote:
Hi Peter,
Many thanks for your reply.My problem is exactly same as second one
discussed in the post.
I need checkbox in other column .
Code is
TreeTable.html
/tr
trtd wicket:id=treeTable class=my-tree/td
td /td
/tr
TreeTable.java
public TreeTablePage()
{
IColumn columns[] = new IColumn[] { col1(), col2() };
tree = new TreeTable(treeTable, createTreeModel(), columns);
tree.getTreeState().setAllowSelectMultiple(true);
add(tree);
tree.getTreeState().collapseAll();
}
private PropertyTreeColumn col1() {
return new PropertyTreeColumn(new
ColumnLocation(Alignment.MIDDLE, 5,
Unit.PROPORTIONAL), Check,
userObject.name);
}
private PropertyTreeColumn col2() {
return new PropertyTreeColumn(new
ColumnLocation(Alignment.LEFT,
7, Unit.EM), L2,
userObject.name) {
@Override
public Component newCell(MarkupContainer parent, String
id, TreeNode
node, int level) {
DefaultMutableTreeNode n =
(DefaultMutableTreeNode) node;
CheckBoxPanel boxPanel = new
CheckBoxPanel(mypanel);
return boxPanel;
}
};
}
I am getting no getter methods defined for ModelBean.
Am i doing it in a right way particularly HTML
Thanks
Prati
aldaris wrote:
Hi,
check out wicket-tree (http://code.google.com/p/wicket-tree/) and see
the example app, I think it will solve your problem.
Regards,
Peter
prati írta:
Hi,
I am also stuck in similar problem.If you can share your code snippets
of
how u did that will be of great help.
Thanks
Pratibha
vela wrote:
Hello again,
The links and nodes are added in the TreeFragment class. But the
TreeFragment is a private inner class in Treetable, could you tell how
to
use the TreeFragment to acheive this functionality
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RE: Close window javascript
See
ModalWindow.close(final AjaxRequestTarget target)
S
-Original Message-
From: Muro Copenhagen [mailto:copenha...@gmail.com]
Sent: Monday, January 18, 2010 11:12 AM
To: users@wicket.apache.org
Subject: Close window javascript
Hi,
I'm trying to close a popup window after the user has submitted a message.
I'm using a javascript to close the window but without any luck.
Can anyone see what goes wrong?
This is the AjaxSubmitLink that should submit the message and close the
window:
AjaxSubmitLink submit = new AjaxSubmitLink(submitLink) {
@Override
public void onSubmit(AjaxRequestTarget target, Form form) {
if (log.isDebugEnabled()) {
log.debug(Saveing comment + comment);
}
target.addComponent(response.setVisible(true));
target.addComponent(this.setVisible(false));
caseCommentService.create(...);
this.getPage().add(new
AbstractAjaxTimerBehavior(Duration.milliseconds(3000)) {
protected void onTimer(final AjaxRequestTarget target) {
stop();
target.prependJavascript(window.close(););
}
});
}
};
myForm.add(submit);
What i'm trying to achieve is to show a
message(response.setVisible(true))..) in three seconds before closing the
window.
But something it does not seem to work.
Best Regards
Muro
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CryptedUrlWebRequestCodingStrategy + WebRequestCodingStrategy = resource URLs are not encrypted (bug?).
Hi,
I started using Wicket rather recently. As part of our security
considerations, we do not want immediately expose the underlying
framework(s) we are using, so we went ahead with URL encryption. We used
standard approach as described in examples:
@Override
protected IRequestCycleProcessor newRequestCycleProcessor() {
return new WebRequestCycleProcessor(){
protected IRequestCodingStrategy newRequestCodingStrategy(){
return new CryptedUrlWebRequestCodingStrategy(new
WebRequestCodingStrategy());
}
};
}
Unfortunately I later discovered that this approach doesn't encrypt resource
URLs, e.g. from:
CSSPackageResource.getHeaderContribution(..);
or
link.add(new Image(logoImage));
What's worse such resource references include FQN of related classes.
After some investigation I found out that the problem is that
CryptedUrlWebRequestCodingStrategy only encrypts arguments string of the URL
and WebRequestCodingStrategy encodes resource references as path rather than
as argument.
I was able to get around this by subclassing WebRequestCodingStrategy and
overriding methods:
addResourceParameters(..);
encode(RequestCycle requestCycle, ISharedResourceRequestTarget
requestTarget);
to use arguments rather than path as resource reference.
However I'm unsure as to original reasoning behind original
CryptedUrlWebRequestCodingStrategy and WebRequestCodingStrategy. Is the
resource behaviour simply a bug? Or is it there for some reason that is
going to bite me down the road if I use my own 'fix'?
Best regards,
Sergey
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RE: submit a form from outside of it
Thanks Bert, but I use a button... I need to have the button in the page, because it does the same for any form. I want to have code re-usage. But however my abstract MyFormPanel does not have relevant markup, so I can't include the buttons there, it would be nice, but I can not. I just want to avoid having the very same buttons and markup in every subclass of MyFormPanel... Can anyone please give further assist? Thank you, Martin -Original Message- From: Bert [mailto:taser...@gmail.com] Sent: Monday, January 18, 2010 12:38 PM To: users@wicket.apache.org Subject: Re: submit a form from outside of it Not sure here, but the AjaxSubmitLink has an constructor that lets you pass in the form it should work on. Bert - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: submit a form from outside of it
I 'm not sure i completely understand your requirement, but the AjaxButton too has a constructor with a Form as parameter. Would it be possible to access the form when creating the button? Bert On Mon, Jan 18, 2010 at 13:11, Martin Asenov mase...@velti.com wrote: Thanks Bert, but I use a button... I need to have the button in the page, because it does the same for any form. I want to have code re-usage. But however my abstract MyFormPanel does not have relevant markup, so I can't include the buttons there, it would be nice, but I can not. I just want to avoid having the very same buttons and markup in every subclass of MyFormPanel... Can anyone please give further assist? - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: final in MarkupContainer#add(Component...) method
Hi,
since I work with Ilya on this project, I'd like to continue this topic as
it is something I'm keenly interested in.
First of all, thanks for your reply, Pedro.
However I must admit I don't understand your suggestion about lenient form
components. How is that supposed to work? And at any rate wouldn't it be too
much work (rewriting each component) for something that could be easily
achieved with minimal code if only the add(..) method wasn't final?
Your comment about RepeatingView seems incorrect (according to my
experience). As soon as you have something that is not numerical as a child
ID, the component complains loudly in logs (see
AbstractRepeater.onBeforeRender()). I also saw post by Igor somewhere where
he stated that numeric IDs were used for some purpose (hence the warning).
This kind of brings us back to original question -- why is
MarkupContainer.add(..) method final? Maybe it's something that needs to be
changed (or alternatively mechanism provided for hooking into this method)?
For our own purposes I already adjusted this method to be non-final (but
that means that upgrading Wicket version is going to be slightly more
problematic).
Declaring it non-final resolved the RepeaterView (actually our custom
component with slightly adjusted functionality) cleanly.
It also resolved the issue with page inheritance. As it stands right now (in
out-of-the-box Wicket), if wicket:child / is inside of any other tag with
wicket-id, then children pages can no longer use add(..) method directly (as
it would then add components in the wrong place in the hierarchy and page
rendering will crash). With 'final' gone, it is a simple matter to override
add(..) in the parent page so that it would add components properly -- thus
children need not be aware of the parent innards.
Best regards,
Sergey
Pedro H. O. dos Santos wrote:
This brings us to a suggested wrapping of the children in
WebMarkupContainers
This is not the only option you have, you can use lenient form components
like:
public class LenientTextField extends TextField
{
@Override
protected void onComponentTag(final ComponentTag tag)
{
tag.setName(input);
tag.put(type, text);
super.onComponentTag(tag);
}
}
so you have an text field that you can your for any tag you place on your
template.
About the meaningful wicket ids, you doesn't need to call rv.newchildid(),
only make sure to don't repeat the ids.
2010/1/14 Ilya German ilja.germ...@parex.lv
This brings us to a suggested wrapping of the children in
WebMarkupContainers, but we'd like to hide it to get rid of the
webmarkupcontainer item=new webmarkupcontainer(rv.newchildid());
part. To do this, it seems logical to extend the RepeatingView overriding
the add() method to be wrapping every component, however the add() method
is
final :(
Could anyone suggest some other way to resolve this situation? Or,
perhaps,
it could be acceptable to officially remove the final modifier from the
add() method?
Thanks in advance!
Ilya German.
--
Pedro Henrique Oliveira dos Santos
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list view
Hello,
I have a table that stores menutabs. In the list of tabs are main tabs
On the page there are main tabs and subtabs that would appear if the
tab that is click in the active one
for example home, applications,admin.
there could be subtabs under applications menu such as leave, loans etc.
there if applications menu is clicked and become the active tab
/page. i would like leave, loans tabs shown underneath it.
I have written something but the problem is when the subtabs do not show
i do not know if my approach is correct or maybe there is a better way
of achieveing the same thing.
code
menuListView=new ListView(menulist,menulist){
@Override
protected void populateItem(ListItem item) {
final AuthTabs menuitem = (AuthTabs)item.getModelObject();
// BookmarkablePageLink link =new
BookmarkablePageLink(link,PortalReflection.getClass(menuitem.getUrl()));
AjaxFallbackLink link =new AjaxFallbackLink (link) {
/* Link link =new Link (link) {
@Override
public void onClick() {
childreanList=getMenuList(menuitem.getTabid());
getChildListView( childreanList);
datatableContainer.replace(childListView);
//this navigates the page to the page class
setResponsePage(PortalReflection.getClass(menuitem.getUrl()));
}*/
@Override
public void onClick(AjaxRequestTarget target) {
// setClicked(true);
System.out.println(menuitem.getTabid() is:+menuitem.getTabid());
childreanList=getMenuList(menuitem.getTabid());
add(new AttributeModifier(class, true, new Model(hover)));
target.addComponent(this);
//this navigates the page to the page class
setResponsePage(PortalReflection.getClass(menuitem.getUrl()));
}
};
/*
link.add(new AttributeModifier(class, true, new
LoadableDetachableModel() {
private static final long serialVersionUID = 1L;
@Override
public final Object load() {
if(clicked ){
clicked=false;
return hover;
}
else {
return ;
}
}
}));
* */
link.setOutputMarkupId(true);
//link.add(new MenuLinkVisualBehavior(onclick, link));
link.add(new Label(caption, menuitem.getTabname()));
item.add(link);
}};
menuListView.setOutputMarkupId(true);
add(menuListView);
}
public ListView getChildListView(List childlist){
childListView=new ListView(childmenulist,childlist){
@Override
protected void populateItem(ListItem item) {
final AuthTabs menuitem = (AuthTabs)item.getModelObject();
BookmarkablePageLink zalink =new
BookmarkablePageLink(childlink,PortalReflection.getClass(menuitem.getUrl()));
zalink.add(new Label(childcaption, menuitem.getTabname()));
}
};
childListView.setOutputMarkupId(true);
return childListView;
}
/code
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Re: Close window javascript
Hi,
Thanks Sefan...that helped me on the way...
It's working now...
Best Regards
Muro
On Mon, Jan 18, 2010 at 1:03 PM, Stefan Droog sdr...@educator.eu wrote:
See
ModalWindow.close(final AjaxRequestTarget target)
S
-Original Message-
From: Muro Copenhagen [mailto:copenha...@gmail.com]
Sent: Monday, January 18, 2010 11:12 AM
To: users@wicket.apache.org
Subject: Close window javascript
Hi,
I'm trying to close a popup window after the user has submitted a message.
I'm using a javascript to close the window but without any luck.
Can anyone see what goes wrong?
This is the AjaxSubmitLink that should submit the message and close the
window:
AjaxSubmitLink submit = new AjaxSubmitLink(submitLink) {
@Override
public void onSubmit(AjaxRequestTarget target, Form form) {
if (log.isDebugEnabled()) {
log.debug(Saveing comment + comment);
}
target.addComponent(response.setVisible(true));
target.addComponent(this.setVisible(false));
caseCommentService.create(...);
this.getPage().add(new
AbstractAjaxTimerBehavior(Duration.milliseconds(3000)) {
protected void onTimer(final AjaxRequestTarget target) {
stop();
target.prependJavascript(window.close(););
}
});
}
};
myForm.add(submit);
What i'm trying to achieve is to show a
message(response.setVisible(true))..) in three seconds before closing the
window.
But something it does not seem to work.
Best Regards
Muro
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RE: submit a form from outside of it
Yes, Bert, it's accessible. The way you proposed comes up with: ERROR: Wicket.Ajax.Call.submitFormById: Trying to submit form with id 'form49a' that is not in document. in the ajax debug. But the form is in the document. Probably it has dynamic markup id and I should try making it persistent. Thanks, -Original Message- From: Bert [mailto:taser...@gmail.com] Sent: Monday, January 18, 2010 2:17 PM To: users@wicket.apache.org Subject: Re: submit a form from outside of it I 'm not sure i completely understand your requirement, but the AjaxButton too has a constructor with a Form as parameter. Would it be possible to access the form when creating the button? Bert On Mon, Jan 18, 2010 at 13:11, Martin Asenov mase...@velti.com wrote: Thanks Bert, but I use a button... I need to have the button in the page, because it does the same for any form. I want to have code re-usage. But however my abstract MyFormPanel does not have relevant markup, so I can't include the buttons there, it would be nice, but I can not. I just want to avoid having the very same buttons and markup in every subclass of MyFormPanel... Can anyone please give further assist? - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: final in MarkupContainer#add(Component...) method
About the lenient form component, I thought that you are having trouble
adding text fields to your form (for example), since they validate the
markup tag (nothing to do with add modifier). As you are adding this kind of
component dynamically, you can't write on development time on your markup,
the correct tag.
About the ids with digits for repeater, override the onBeforeRender method
and remove that validation if you need/want.
If you has some rules for your component, like wrap components, make more
sense you extend RepeatingView, create the addComponentAfterWrapHim method,
that delegate the call to add method after your customization is made. You
gain plainness by make clean that this component add method is special.
On Mon, Jan 18, 2010 at 10:52 AM, Sergejs Olefirs
sergejs.olef...@parex.lvwrote:
Hi,
since I work with Ilya on this project, I'd like to continue this topic as
it is something I'm keenly interested in.
First of all, thanks for your reply, Pedro.
However I must admit I don't understand your suggestion about lenient form
components. How is that supposed to work? And at any rate wouldn't it be
too
much work (rewriting each component) for something that could be easily
achieved with minimal code if only the add(..) method wasn't final?
Your comment about RepeatingView seems incorrect (according to my
experience). As soon as you have something that is not numerical as a child
ID, the component complains loudly in logs (see
AbstractRepeater.onBeforeRender()). I also saw post by Igor somewhere where
he stated that numeric IDs were used for some purpose (hence the warning).
This kind of brings us back to original question -- why is
MarkupContainer.add(..) method final? Maybe it's something that needs to be
changed (or alternatively mechanism provided for hooking into this method)?
For our own purposes I already adjusted this method to be non-final (but
that means that upgrading Wicket version is going to be slightly more
problematic).
Declaring it non-final resolved the RepeaterView (actually our custom
component with slightly adjusted functionality) cleanly.
It also resolved the issue with page inheritance. As it stands right now
(in
out-of-the-box Wicket), if wicket:child / is inside of any other tag with
wicket-id, then children pages can no longer use add(..) method directly
(as
it would then add components in the wrong place in the hierarchy and page
rendering will crash). With 'final' gone, it is a simple matter to override
add(..) in the parent page so that it would add components properly -- thus
children need not be aware of the parent innards.
Best regards,
Sergey
Pedro H. O. dos Santos wrote:
This brings us to a suggested wrapping of the children in
WebMarkupContainers
This is not the only option you have, you can use lenient form components
like:
public class LenientTextField extends TextField
{
@Override
protected void onComponentTag(final ComponentTag tag)
{
tag.setName(input);
tag.put(type, text);
super.onComponentTag(tag);
}
}
so you have an text field that you can your for any tag you place on your
template.
About the meaningful wicket ids, you doesn't need to call
rv.newchildid(),
only make sure to don't repeat the ids.
2010/1/14 Ilya German ilja.germ...@parex.lv
This brings us to a suggested wrapping of the children in
WebMarkupContainers, but we'd like to hide it to get rid of the
webmarkupcontainer item=new webmarkupcontainer(rv.newchildid());
part. To do this, it seems logical to extend the RepeatingView
overriding
the add() method to be wrapping every component, however the add()
method
is
final :(
Could anyone suggest some other way to resolve this situation? Or,
perhaps,
it could be acceptable to officially remove the final modifier from
the
add() method?
Thanks in advance!
Ilya German.
--
Pedro Henrique Oliveira dos Santos
--
View this message in context:
http://old.nabble.com/final-in-MarkupContainer-add%28Component...%29-method-tp27161187p27210046.html
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--
Pedro Henrique Oliveira dos Santos
Re: final in MarkupContainer#add(Component...) method
About the ids with digits for repeater, override the onBeforeRender method and remove that validation if you need/want. Impossible. I just filed https://issues.apache.org/jira/browse/WICKET-2684 - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: final in MarkupContainer#add(Component...) method
2010/1/14 Ilya German ilja.germ...@parex.lv: Hello! We're struggling with working around the final modifier for the MarkupContainer#add(Component ...) method. We have the following scenario: 1. We'd like to use a repeater to add some components to the form. 2. We'd like these components to work with CompoundPropertyModel, thus we need these to have meaningful wicket ids 3. We'd like not to find ourselves backstabbed by the non-numeric ids in the RepeatingView's children. This brings us to a suggested wrapping of the children in WebMarkupContainers, but we'd like to hide it to get rid of the webmarkupcontainer item=new webmarkupcontainer(rv.newchildid()); part. To do this, it seems logical to extend the RepeatingView overriding the add() method to be wrapping every component, however the add() method is final :( Could anyone suggest some other way to resolve this situation? Or, perhaps, it could be acceptable to officially remove the final modifier from the add() method? Thanks in advance! Ilya German. 1. You can use PropertyModel and numeric ids isn't a problem 2. You can wait for resolution on https://issues.apache.org/jira/browse/WICKET-2684 and (if it gets fixed) use non-numeric child ids in RepeatingView. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
RE: submit a form from outside of it
it's not working... The problem probably is that the form has it's own
generated markup id and the button looks for the form with markup id from the
one that comes in the AjaxButton's constructor as an argument. Because I have:
submitButton = new AjaxButton(submit_button, formPanel.getForm()) {
protected void onSubmit(ART target) {
do smth
}
}
and because I display different panels with different forms ( especially their
markupId ) it can't find the current one by it's markup id. I just can't think
of a workaround to solve this.
Please, if someone has a better idea, it would be great to share it...
Best Regards,
Martin
-Original Message-
From: Martin Asenov [mailto:mase...@velti.com]
Sent: Monday, January 18, 2010 3:33 PM
To: users@wicket.apache.org
Subject: RE: submit a form from outside of it
Yes, Bert, it's accessible. The way you proposed comes up with:
ERROR: Wicket.Ajax.Call.submitFormById: Trying to submit form with id 'form49a'
that is not in document.
in the ajax debug. But the form is in the document. Probably it has dynamic
markup id and I should try making it persistent.
Thanks,
-Original Message-
From: Bert [mailto:taser...@gmail.com]
Sent: Monday, January 18, 2010 2:17 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
I 'm not sure i completely understand your requirement, but the
AjaxButton too has a constructor
with a Form as parameter. Would it be possible to access the form when
creating the button?
Bert
On Mon, Jan 18, 2010 at 13:11, Martin Asenov mase...@velti.com wrote:
Thanks Bert, but I use a button... I need to have the button in the page,
because it does the same for any form. I want to have code re-usage. But
however my abstract MyFormPanel does not have relevant markup, so I can't
include the buttons there, it would be nice, but I can not. I just want to
avoid having the very same buttons and markup in every subclass of
MyFormPanel...
Can anyone please give further assist?
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Re: final in MarkupContainer#add(Component...) method
Thanks again for your reply. On the one hand, I agree with you, adding custom 'addComponent' method does make the mechanics of the code more clear. On the other hand, in practice (we used such methods for a couple of weeks), it results in the situations in application code where you sometimes have to use add(..) method and sometimes addComponent(..) method (and in case of addComponent(..) method, add(..) method is also available, so no compilation errors there). It invariably led to people forgetting to use addComponent(..) instead of add(..) and thus application crashing/behaving incorrectly at runtime. This was extremely annoying -- hence our decision to un-finalize add(..) method. On the yet another hand, OO programming kind of implies that objects are supposed to hide their implementation details, so even disregarding my previous usability comment, I'm not sure I agree that addComponent(..) is better than overridden add(..) method from the philosophical point of view. And if I may to try and adjust the course of this discussion a bit -- the issue with repeaters is not the only argument against final add(..) method. As I mentioned in passing, there's also issue of Page inheritance where subclasses with wicket:extend cannot use add(..) method if parent Page has wicket:child / nested into some other tag with wicket:id -- leading to the very similar issues. I'm sure there can be other examples as well. So perhaps add(..) method doesn't need to be final after all? Or would it break something else that I don't know about? If someone could explain reasons behind add(..) being final it would be most appreciated -- that might change my opinion on the whole subject. Best regards, Sergey Pedro H. O. dos Santos wrote: If you has some rules for your component, like wrap components, make more sense you extend RepeatingView, create the addComponentAfterWrapHim method, that delegate the call to add method after your customization is made. You gain plainness by make clean that this component add method is special. On Mon, Jan 18, 2010 at 10:52 AM, Sergejs Olefirs sergejs.olef...@parex.lvwrote: This kind of brings us back to original question -- why is MarkupContainer.add(..) method final? Maybe it's something that needs to be changed (or alternatively mechanism provided for hooking into this method)? For our own purposes I already adjusted this method to be non-final (but that means that upgrading Wicket version is going to be slightly more problematic). Declaring it non-final resolved the RepeaterView (actually our custom component with slightly adjusted functionality) cleanly. It also resolved the issue with page inheritance. As it stands right now (in out-of-the-box Wicket), if wicket:child / is inside of any other tag with wicket-id, then children pages can no longer use add(..) method directly (as it would then add components in the wrong place in the hierarchy and page rendering will crash). With 'final' gone, it is a simple matter to override add(..) in the parent page so that it would add components properly -- thus children need not be aware of the parent innards. -- View this message in context: http://old.nabble.com/final-in-MarkupContainer-add%28Component...%29-method-tp27161187p27211055.html Sent from the Wicket - User mailing list archive at Nabble.com. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: final in MarkupContainer#add(Component...) method
And if I may to try and adjust the course of this discussion a bit -- the issue with repeaters is not the only argument against final add(..) method. As I mentioned in passing, there's also issue of Page inheritance where subclasses with wicket:extend cannot use add(..) method if parent Page has wicket:child / nested into some other tag with wicket:id -- leading to the very similar issues. I'm sure there can be other examples as well. This thread already get discussed in somewhere, and and custom addComponentOnChild(...) is an good option. So perhaps add(..) method doesn't need to be final after all? Or would it break something else that I don't know about? If someone could explain reasons behind add(..) being final it would be most appreciated -- that might change my opinion on the whole subject. There is no api break, IMO it is an good framework frozen spot, since situation on link[1] are avoided. [1] http://markmail.org/search/?q=list:org.apache.wicket.users+from:Pedro+Santos+override+date:200910+ajax#query:list%3Aorg.apache.wicket.users%20from%3A%22Pedro%20Santos%22%20override%20date%3A200910%20ajax+page:1+mid:u5uemowe7tqujn3o+state:resultshttp://markmail.org/search/?q=list:org.apache.wicket.users+from:%22Pedro+Santos%22+override+date:200910+ajax#query:list%3Aorg.apache.wicket.users%20from%3A%22Pedro%20Santos%22%20override%20date%3A200910%20ajax+page:1+mid:u5uemowe7tqujn3o+state:results On Mon, Jan 18, 2010 at 12:15 PM, Sergejs Olefirs sergejs.olef...@parex.lvwrote: Thanks again for your reply. On the one hand, I agree with you, adding custom 'addComponent' method does make the mechanics of the code more clear. On the other hand, in practice (we used such methods for a couple of weeks), it results in the situations in application code where you sometimes have to use add(..) method and sometimes addComponent(..) method (and in case of addComponent(..) method, add(..) method is also available, so no compilation errors there). It invariably led to people forgetting to use addComponent(..) instead of add(..) and thus application crashing/behaving incorrectly at runtime. This was extremely annoying -- hence our decision to un-finalize add(..) method. On the yet another hand, OO programming kind of implies that objects are supposed to hide their implementation details, so even disregarding my previous usability comment, I'm not sure I agree that addComponent(..) is better than overridden add(..) method from the philosophical point of view. And if I may to try and adjust the course of this discussion a bit -- the issue with repeaters is not the only argument against final add(..) method. As I mentioned in passing, there's also issue of Page inheritance where subclasses with wicket:extend cannot use add(..) method if parent Page has wicket:child / nested into some other tag with wicket:id -- leading to the very similar issues. I'm sure there can be other examples as well. So perhaps add(..) method doesn't need to be final after all? Or would it break something else that I don't know about? If someone could explain reasons behind add(..) being final it would be most appreciated -- that might change my opinion on the whole subject. Best regards, Sergey Pedro H. O. dos Santos wrote: If you has some rules for your component, like wrap components, make more sense you extend RepeatingView, create the addComponentAfterWrapHim method, that delegate the call to add method after your customization is made. You gain plainness by make clean that this component add method is special. On Mon, Jan 18, 2010 at 10:52 AM, Sergejs Olefirs sergejs.olef...@parex.lvwrote: This kind of brings us back to original question -- why is MarkupContainer.add(..) method final? Maybe it's something that needs to be changed (or alternatively mechanism provided for hooking into this method)? For our own purposes I already adjusted this method to be non-final (but that means that upgrading Wicket version is going to be slightly more problematic). Declaring it non-final resolved the RepeaterView (actually our custom component with slightly adjusted functionality) cleanly. It also resolved the issue with page inheritance. As it stands right now (in out-of-the-box Wicket), if wicket:child / is inside of any other tag with wicket-id, then children pages can no longer use add(..) method directly (as it would then add components in the wrong place in the hierarchy and page rendering will crash). With 'final' gone, it is a simple matter to override add(..) in the parent page so that it would add components properly -- thus children need not be aware of the parent innards. -- View this message in context: http://old.nabble.com/final-in-MarkupContainer-add%28Component...%29-method-tp27161187p27211055.html Sent from the Wicket - User mailing list archive at Nabble.com.
Heap space issue
I am monitoring a Wicket 1.4.5 application running on Tomcat 6.0 which accesses a SQL Server 2000 database. Periodically the application becomes unresponsive due to a lack of heap space, and I have to bounce Tomcat. I'm trying to figure out what sort of errors could cause this to happen. Any suggestions? I've looked in the Tomcat error logs, but the only errors I see prior to the out-of-heapspace error are a small number of SQL errors that result from bad input data. The Wicket application handles these errors by switching the user to a standard error page. I suppose it's a bad architecture to rely on SQL errors reported by the database rather than checking the data, but does this result in a memory leak? Frank Silbermann - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: submit a form from outside of it
I think you have to pass the form to the behavior in some way or you can do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n +
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the constructor
and cannot be found in the hierarchy of the component this behavior is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button. It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form) {
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
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Re: Heap space issue
Use jmap -histo pid or a memory profiler (yourkit) or visualvm to look at the heap. It also helps to use jstat -gc pid 1000 in cases when you have low heap availability before killing your darlings. It might not be a memory leak but possibly a connection pool running out of connections. Any number of programming errors can lead to memory leaks, including exceptions. Without actual profiling you won't be able to find the leaks. Martijn On Mon, Jan 18, 2010 at 4:14 PM, Frank Silbermann frank.silberm...@fedex.com wrote: I am monitoring a Wicket 1.4.5 application running on Tomcat 6.0 which accesses a SQL Server 2000 database. Periodically the application becomes unresponsive due to a lack of heap space, and I have to bounce Tomcat. I'm trying to figure out what sort of errors could cause this to happen. Any suggestions? I've looked in the Tomcat error logs, but the only errors I see prior to the out-of-heapspace error are a small number of SQL errors that result from bad input data. The Wicket application handles these errors by switching the user to a standard error page. I suppose it's a bad architecture to rely on SQL errors reported by the database rather than checking the data, but does this result in a memory leak? Frank Silbermann - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org -- Become a Wicket expert, learn from the best: http://wicketinaction.com Apache Wicket 1.4 increases type safety for web applications Get it now: http://www.apache.org/dyn/closer.cgi/wicket/1.4.4 - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: wicketstuff push, publishing event in a page2 and component installed with channel listener in page1
On Fri, Jan 15, 2010 at 5:24 AM, vineet semwal vineetsemwal1...@gmail.comwrote: Sorry ,a little late .. push is a great project,thanks for your efforts. i am a little confused, 1)does the time out only happens after a remove event is published or apart from this, there is another timeout which happens when server is finished pushing into the client? Here are the configuration options for the Jetty implementation of cometd. You can change the connection timeout value to notice disconects sooner (at the cost of ineffiency) http://cometd.org/documentation/cometd-java/server/configuration You can check the bayeux specificition for the details. ( http://svn.cometd.com/trunk/bayeux/bayeux.html) 2)i see some problems when using more than one listener on one component, i tried reproducing the problem by a little tinkering in your example , currently the example in the quickstart i am attaching has two listeners on different components ,you can reproduce the problem by adding listeners to the same component. a event in one channel is caught by channel listener meant for another channel. Great, I'll look into this. thanks again .. On Sat, Dec 26, 2009 at 11:15 PM, Rodolfo Hansen kry...@gmail.com wrote: Regarding remove listeners: Most browsers fail to report the remove event. Only firefox reports removal immediately, all other browsers depend on the timeout for a comet reconnect to notice and fire the remove event; you may need to lower the timeout for the cometd connections. Also,can i install more than one channel listener on a component? Never tried it, but there should be no problem, can you write a quickstart with your use cases, so I can flesh any bugs out? On Thu, Dec 24, 2009 at 10:03 AM, vineet semwal vineetsemwal1...@gmail.comwrote: Hellos, recently i started using wicketstuff push ,i have few doubts as following .. i have a situation where i need to publish a event in page 2 and add the channel listener in page 1 . for eg. a sign out event published in page 2 which i do using a remove listener. Also,can i install more than one channel listener on a component? -- regards, Vineet Semwal -- Rodolfo Hansen CTO, KindleIT Software Development Email: rhan...@kindleit.net Mobile: +1 (809) 860-6669 -- regards, Vineet Semwal - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org -- Rodolfo Hansen CTO, KindleIT Software Development Email: rhan...@kindleit.net Mobile: +1 (809) 860-6669
Re: PageLink deprecated
well, if the functionality can be accomplished using either BookmarkablePageLink or Link, why do we need yet another way to do it? -igor On Sun, Jan 17, 2010 at 11:44 PM, Jeroen Steenbeeke j.steenbeeke...@gmail.com wrote: Guys, no need to keep explaining what's wrong with passing a Page in the constructor, we understand that! Forget about that filthy 3rd constructor, I know it's wrong and I never used it anyway. That wasn't what my question was about. There are two more constructors: PageLink(String, Class) PageLink(String, IPageLink) Both of these do not replicate the dangerous behavior illustrated in this thread so far. I understand that we can easily create our own implementation that simulates the behavior we want. I just wanted to understand the reasoning for removing the whole class when only one of the constructors is dangerous. From what Martijn Dashorst just told me, it was a case of seeing as we already have Link and BookmarkablePageLink, we figured you could just use those instead. This is also the source of miscommunication so far. The Javadoc simply states what you should use instead, but does not explicitly state why. The assumption is that any behavior you can achieve with the PageLink/IPageLink combination can also be done with a simple Link. This does not take into account the use of the Page Identity for security checks however (mainly for determining link visibility, which, frankly, does not need an actual instance of the page in question), which brings us back to Emond's original point. On the other hand, one could argue that the only use for the page identity is for security purposes, and it would therefore be more at home in a specialized class in wicket-security. -- Jeroen Steenbeeke www.fortuityframework.com - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: CryptedUrlWebRequestCodingStrategy + WebRequestCodingStrategy = resource URLs are not encrypted (bug?).
the original design goal of the crypted strategy was to only encrypt
what the user sees in the url bar. since they never see resource urls
there was no reason to encrypt those.
re fqns, you can add class aliases into SharedResources to hide those.
-igor
2010/1/18 Sergejs Olefirs sergejs.olef...@parex.lv:
Hi,
I started using Wicket rather recently. As part of our security
considerations, we do not want immediately expose the underlying
framework(s) we are using, so we went ahead with URL encryption. We used
standard approach as described in examples:
@Override
protected IRequestCycleProcessor newRequestCycleProcessor() {
return new WebRequestCycleProcessor(){
protected IRequestCodingStrategy newRequestCodingStrategy(){
return new CryptedUrlWebRequestCodingStrategy(new
WebRequestCodingStrategy());
}
};
}
Unfortunately I later discovered that this approach doesn't encrypt resource
URLs, e.g. from:
CSSPackageResource.getHeaderContribution(..);
or
link.add(new Image(logoImage));
What's worse such resource references include FQN of related classes.
After some investigation I found out that the problem is that
CryptedUrlWebRequestCodingStrategy only encrypts arguments string of the URL
and WebRequestCodingStrategy encodes resource references as path rather than
as argument.
I was able to get around this by subclassing WebRequestCodingStrategy and
overriding methods:
addResourceParameters(..);
encode(RequestCycle requestCycle, ISharedResourceRequestTarget
requestTarget);
to use arguments rather than path as resource reference.
However I'm unsure as to original reasoning behind original
CryptedUrlWebRequestCodingStrategy and WebRequestCodingStrategy. Is the
resource behaviour simply a bug? Or is it there for some reason that is
going to bite me down the road if I use my own 'fix'?
Best regards,
Sergey
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Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you can do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n +
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
RE: submit a form from outside of it
Thank you both for the replies! After all I put the buttons in myFormPanel. But
this way I can't know when the submit and cancel buttons are pressed so that I
can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you can do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n +
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form? form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
Re: PageLink deprecated
Because neither has a getPageClass() method? 2010/1/18 Igor Vaynberg igor.vaynb...@gmail.com: well, if the functionality can be accomplished using either BookmarkablePageLink or Link, why do we need yet another way to do it? -igor On Sun, Jan 17, 2010 at 11:44 PM, Jeroen Steenbeeke j.steenbeeke...@gmail.com wrote: Guys, no need to keep explaining what's wrong with passing a Page in the constructor, we understand that! Forget about that filthy 3rd constructor, I know it's wrong and I never used it anyway. That wasn't what my question was about. There are two more constructors: PageLink(String, Class) PageLink(String, IPageLink) Both of these do not replicate the dangerous behavior illustrated in this thread so far. I understand that we can easily create our own implementation that simulates the behavior we want. I just wanted to understand the reasoning for removing the whole class when only one of the constructors is dangerous. From what Martijn Dashorst just told me, it was a case of seeing as we already have Link and BookmarkablePageLink, we figured you could just use those instead. This is also the source of miscommunication so far. The Javadoc simply states what you should use instead, but does not explicitly state why. The assumption is that any behavior you can achieve with the PageLink/IPageLink combination can also be done with a simple Link. This does not take into account the use of the Page Identity for security checks however (mainly for determining link visibility, which, frankly, does not need an actual instance of the page in question), which brings us back to Emond's original point. On the other hand, one could argue that the only use for the page identity is for security purposes, and it would therefore be more at home in a specialized class in wicket-security. -- Jeroen Steenbeeke www.fortuityframework.com - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org -- Jeroen Steenbeeke www.fortuityframework.com - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n +
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference
to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form?
form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
RE: submit a form from outside of it
Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the parent
page when the submit button is pressed , so that I can say
target.addComponent(feed); when feed is in parent page...
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:37 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n +
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main difference
to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page. The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form?
form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
-
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For additional commands, e-mail: users-h...@wicket.apache.org
Re: submit a form from outside of it
Ah - I see. Yeah - you just have to roll your own option for this now. I'd
just recommend some sort of listener pattern - something like adding a void
formSubmitted(form, requesttarget) method on the page that child forms can
call. Then the pages can do whatever they need with it.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:40 AM, Martin Asenov mase...@velti.com wrote:
Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the
parent page when the submit button is pressed , so that I can say
target.addComponent(feed); when feed is in parent page...
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:37 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in
myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed
so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you
can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n
+
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior
is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main
difference
to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page.
The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form?
form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org
-
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RE: submit a form from outside of it
Yes, but in my case the panels don't even know about the parent - they don't
have a reference to it. I've got to think of some workaround on this one.
Thanks for your time,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:45 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Ah - I see. Yeah - you just have to roll your own option for this now. I'd
just recommend some sort of listener pattern - something like adding a void
formSubmitted(form, requesttarget) method on the page that child forms can
call. Then the pages can do whatever they need with it.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:40 AM, Martin Asenov mase...@velti.com wrote:
Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the
parent page when the submit button is pressed , so that I can say
target.addComponent(feed); when feed is in parent page...
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:37 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in
myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed
so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you
can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n
+
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior
is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form that has validation and so on, but the main
difference
to
ordinary forms is that my form does not contain it's submit button.
It's
located in a parent, in my case a web page.
I'm wondering how can I force the form submitting from the page.
The
code
is
submitButton = new AjaxButton(submit_button) {
protected void onSubmit(AjaxRequestTarget target, Form?
form)
{
myForm.processForm();
}
};
The method processForm() in myForm just calls process();
But nothing happens, looks like I'm missing something...
Thanks,
Martin
-
Re: submit a form from outside of it
IFeedbackProvider {
FeedbackPanel getFeedbackPanel();
}
MyPage extends Webpage implements IFeedbackProvider {
public MyPage() {
add(new FeedbackPanel(feedback));
add(new MySubPanel(panel, this));
}
@Override
FeedbackPanel getFeedbackPanel() {
return get(feedback);
}
}
MySubPanel extends Panel {
private IFeedbackProvider feedback;
public MySubPanel(String id, IFeedbackProvider provider) {
super(id);
this.feedback = provider;
... some ajax handler ...
onSubmit(AjaxRequestTarget t) {
t.addComponent(feedback.getFeedbackPanel());
}
}
}
On Mon, Jan 18, 2010 at 5:52 PM, Martin Asenov mase...@velti.com wrote:
Yes, but in my case the panels don't even know about the parent - they don't
have a reference to it. I've got to think of some workaround on this one.
Thanks for your time,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:45 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Ah - I see. Yeah - you just have to roll your own option for this now. I'd
just recommend some sort of listener pattern - something like adding a void
formSubmitted(form, requesttarget) method on the page that child forms can
call. Then the pages can do whatever they need with it.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:40 AM, Martin Asenov mase...@velti.com wrote:
Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the
parent page when the submit button is pressed , so that I can say
target.addComponent(feed); when feed is in parent page...
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:37 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in
myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed
so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you
can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n
+
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior
is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Hi,
Try this:
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
onclick) {
protected void onSubmit(AjaxRequestTarget target) {
//do what you have to do
}
};
Button submitButton = new Button(submitButton);
submitButton.add(behave);
...
Alexandru
2010/1/18 Martin Asenov mase...@velti.com
Hello, everyone!
I have a form
Re: filtering a datatable
Hi, You'll find everything you need in the phonebook application : https://wicket-stuff.svn.sf.net/svnroot/wicket-stuff/trunk/wicketstuff-core/phonebook/ https://wicket-stuff.svn.sf.net/svnroot/wicket-stuff/trunk/wicketstuff-core/phonebook/ Regards, Gabriel. julien Graglia wrote: I you have a piece of code of how to use filter and IFilterStateLocator... -- View this message in context: http://old.nabble.com/filtering-a-datatable-tp23062814p27214593.html Sent from the Wicket - User mailing list archive at Nabble.com. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Getting control of Wicket Session.
Hi, I am a Wicket fresher. I am still getting the hang of this. I want to develop a web application where I would like to show the user a message informing him that his session is about to expire in x minutes. If he wants to save his session then can click on a button which appears along with this message. Also, using javascript I would like to 'sense' that the user is typing-in or moving his mouse over the webpage, which is evident enough that he is using the session. I will somehow figure out the js code needed but *how will I tell Wicket to keep the session alive? Also how can I be notified x minutes before Wicket expires the session?* Regards, Apple Grew my blog @ http://blog.applegrew.com/
Popup New Window After Submit
I need to process my Form and then popup a report in a new browser
window with the result of the Form processing. I would like to do this
with one click from the user. Any ideas if this solution is already
built? I tried to use Link with PopupSettings, but the form submit does
not trigger and results in the form not being validated and the form
component models not being updated with the user input.
Here is a simplified example, in Form.onSubmit():
{
int sum = userInput1 + userInput2;
PageParameters params = new PageParameters();
params.add(Total, String.valueOf(sum));
// I want the response in a new window
setResponse(new ReportPage(params));
}
--
Jered Myers
Programmer/Project Analyst
jer...@maplewoodsoftware.com mailto:jer...@maplewoodsoftware.com
Maplewood Software
508 W. 6th Ave, Suite 900
Spokane, WA 99204
(509) 252-3550 ext. 109
www.maplewoodsoftware.com
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root context, IE, home page is not found
all,
my wicket 1.4.5 application is configured to run in root context. for some
reason, when it sets response page to the home page (which is not mounted), the
webserver produces error: The requested resource () is not available.
here's the code:
formFooter.add(new LinkVoid(ID_LINK_HOME)
{
private static final long serialVersionUID = 1L;
@Override
public void onClick()
{
setResponsePage(getApplication().getHomePage());
}
});
it happens only with IE (6, 7), only with root context, with Tomcat 6 and Sun's
Glassfish servers. looks like URL generated in that case has '.' appended to
it: http://localhost/.;.
if i simply hit F5 in the browser after the error shows up, it opens home page
just fine.
if i use bookmarkable link, it works fine: formFooter.add(new
BookmarkablePageLinkHomePage(ID_LINK_HOME,
getApplication().getHomePage()).setAutoEnable(true));
the issue looks very similar to http://issues.apache.org/jira/browse/WICKET-1449
is there a workaround?
Thanks,
Vadim
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Re: yui context menu after ajax request
rerender the menu javascripcode in ajaxevent. alex - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: OnChangeAjaxBehavior only first time
Do you mean the issue is that the onchange behavior is not called without
taking the focus off of the textfield? If you want something that is going
to work without taking focus off the textfield, hook into the onkeyup (or
down) event instead.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 3:03 PM, Antoine van Wel
antoine.van@gmail.comwrote:
Hi,
Does anybody have ideas how to solve this by *not* updating the form?
I'm running into the same issue when putting an OnChangeAjaxBehavior
on a TextField inside a form (triggered only once). The component to
be updated is outside the form.
Updating the form like suggested solves the issue; however, when the
user types in text in this textfield and the form is constantly
updated, this means text will get lost. Adjusting the throttle delay
helps, but not enough. I just cannot update the form part via AJAX. So
are there any other ways to fix this problem?
Antoine.
On Wed, Mar 18, 2009 at 3:22 PM, Filippo De Luca
dl.fili...@filosganga.it wrote:
Hi wicketers,
I have fixed this issue by including form in target component:
private void registerBehaviors() {
userAgentField.add(new OnChangeAjaxBehavior() {
@Override
protected void onUpdate(AjaxRequestTarget target) {
logger.debug(AJAX matching user agent);
matchUserAgent();
target.addComponent(resultPanel);
target.addComponent(matchingForm);
}
});
}
I've found this mail:
http://www.nabble.com/OnChangeAjaxBehavior-receives-only-first-input-character-td20207317.html
,
it is the same issue.
It is this behavior default? or it is an issue?
2009/3/18 Filippo De Luca dl.fili...@filosganga.it
Hi,
I have a panel MatchingPanel with a form and a result div:
wicket:panel
form wicket:id=form action=#
div class=widget
labelwicket:message
key=form.userAgent[form.userAgent]/wicket:message/label
input wicket:id=userAgent class=userAgent
type=text maxlength=255 /
/div
div class=buttons
input type=submit value=Match
wicket:message=value:form.match /
/div
/form
div wicket:id=result
div
labelwicket:message
key=result.matchedUserAgent[result.matchedUserAgent]/wicket:message/label
span
wicket:id=matchedUserAgent[sourceUserAgent]/span
/div
/div
/wicket:panel
while my java code is:
class MatchingForm extends FormObject {
...
private void registerBehaviors() {
userAgentField.add(new OnChangeAjaxBehavior() {
@Override
protected void onUpdate(AjaxRequestTarget target) {
logger.debug(AJAX matching user agent);
matchUserAgent();
target.addComponent(resultPanel);
}
});
}
...
private void matchUserAgent() {
String userAgent = userAgentModel.getObject();
WURFLManager wurflManager = getWurflManager();
MatchingResult result = null;
if (StringUtils.isNotBlank(userAgent)) {
Instant start = new Instant();
Device device =
wurflManager.getDeviceForRequest(userAgent);
Instant end = new Instant();
Period duration = new Period(start, end);
result = new MatchingResult(userAgent, device, (long)
duration
.getMillis());
}
resultPanel.setResult(result);
}
}
The OnChangeBehaviour is called only the first time the userAgent field
take the focus. To call other time, i have to lost focus and give focus
to
userAgent input text. Why? The target component is outside the frm, it
is an
issue?
Thank you
--
Filippo De Luca
--
Email: dl.fili...@filosganga.it
Web: http://www.filosganga.it
LinkedIn: http://www.linkedin.com/in/filippodeluca
mobile: +393395822588
--
Filippo De Luca
--
Email: dl.fili...@filosganga.it
Web: http://www.filosganga.it
LinkedIn: http://www.linkedin.com/in/filippodeluca
mobile: +393395822588
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Re: yui context menu after ajax request
for my problem with dojo this works:
public void refreshMenu(AjaxRequestTarget pTarget) {
String js = dojo.addOnLoad(function(){\n + menu.generateJS()
+ \n}); ;
pTarget.appendJavascript(sss);
}
and i know the same stuff works for jquery and yui. the problem is that the
header javascript brokes after rerendering the component (i think its a problem
with object references because most browsers replace the object by creating a
new one and the javascript stuff bind on the dom-bject). What seems to work is
to put the javascript stuff into the body tag using the onComponentRendered()
to reactivate the behavior.
the ugly is you have to reload the menu in every ajax-event reredering the
component with menu.
hth
alex
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Re: yui context menu after ajax request
Interesting... I'll give that a try.. Thanks...
On Mon, Jan 18, 2010 at 6:06 PM, Alexander Elsholz
alexander.elsh...@widas.de wrote:
for my problem with dojo this works:
public void refreshMenu(AjaxRequestTarget pTarget) {
String js = dojo.addOnLoad(function(){\n + menu.generateJS()
+ \n}); ;
pTarget.appendJavascript(sss);
}
and i know the same stuff works for jquery and yui. the problem is that the
header javascript brokes after rerendering the component (i think its a
problem
with object references because most browsers replace the object by creating
a
new one and the javascript stuff bind on the dom-bject). What seems to work
is
to put the javascript stuff into the body tag using the
onComponentRendered()
to reactivate the behavior.
the ugly is you have to reload the menu in every ajax-event reredering the
component with menu.
hth
alex
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--
Dave Kallstrom
RE: root context, IE, home page is not found
looks like BookmarkablePageRequestTarget.respond(RequestCycle requestCycle) has
code to strip ./ if url starts with it.
can it be changed to strip . as well? will it be the right fix?
From: vad...@hotmail.com
To: users@wicket.apache.org
Subject: root context, IE, home page is not found
Date: Mon, 18 Jan 2010 20:36:55 +
all,
my wicket 1.4.5 application is configured to run in root context. for some
reason, when it sets response page to the home page (which is not mounted),
the webserver produces error: The requested resource () is not available.
here's the code:
formFooter.add(new LinkVoid(ID_LINK_HOME)
{
private static final long serialVersionUID = 1L;
@Override
public void onClick()
{
setResponsePage(getApplication().getHomePage());
}
});
it happens only with IE (6, 7), only with root context, with Tomcat 6 and
Sun's Glassfish servers. looks like URL generated in that case has '.'
appended to it: http://localhost/.;.
if i simply hit F5 in the browser after the error shows up, it opens home
page just fine.
if i use bookmarkable link, it works fine: formFooter.add(new
BookmarkablePageLinkHomePage(ID_LINK_HOME,
getApplication().getHomePage()).setAutoEnable(true));
the issue looks very similar to
http://issues.apache.org/jira/browse/WICKET-1449
is there a workaround?
Thanks,
Vadim
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Re: Getting control of Wicket Session.
@Pedro Thanks for the response. Now I have two questions. 1. How do I get reference of HttpSession? In Wicket I seem to get reference of WebSession. 2. How do I notify the servlet container (I guess Wicket is not in-charge of maintaining the HttpSession), that the user is active? If I make any Http request from client's end (AJAX), even if that request is simply ignored by my application, will that help to keep the session alive? @Steve Yup this will be AJAX. When my solution is ready, will share it, definitely. :) Regards, Apple Grew my blog @ http://blog.applegrew.com/ On Tue, Jan 19, 2010 at 8:57 AM, Steve Swinsburg steve.swinsb...@gmail.comwrote: Presumably, if the user is typing a long document and hasn't clicked on anything for a while, you'll need to indicate to Wicket/HttpSession (probably via AJAX), that the session is to be kept alive. I believe this is how the autosave functions of things like rich text editors work, keeping the session from timing out. Interested to see your solution. cheers, Steve On 19/01/2010, at 10:19 AM, Pedro Santos wrote: Use attributes from HttpSession like getLastAccessedTime() and getMaxInactiveInterval(). You can get access through WebRequest that holds an HttpServletRequest object. When HttpSession ends, the wicket session ends too. On Mon, Jan 18, 2010 at 5:33 PM, Apple Grew appleg...@gmail.com wrote: Hi, I am a Wicket fresher. I am still getting the hang of this. I want to develop a web application where I would like to show the user a message informing him that his session is about to expire in x minutes. If he wants to save his session then can click on a button which appears along with this message. Also, using javascript I would like to 'sense' that the user is typing-in or moving his mouse over the webpage, which is evident enough that he is using the session. I will somehow figure out the js code needed but *how will I tell Wicket to keep the session alive? Also how can I be notified x minutes before Wicket expires the session?* Regards, Apple Grew my blog @ http://blog.applegrew.com/ -- Pedro Henrique Oliveira dos Santos - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: Getting control of Wicket Session.
((WebRequest)getRequest()).getHttpServletRequest().getSession() On Jan 18, 2010, at 10:12 PM, Apple Grew wrote: @Pedro Thanks for the response. Now I have two questions. 1. How do I get reference of HttpSession? In Wicket I seem to get reference of WebSession. 2. How do I notify the servlet container (I guess Wicket is not in-charge of maintaining the HttpSession), that the user is active? If I make any Http request from client's end (AJAX), even if that request is simply ignored by my application, will that help to keep the session alive? @Steve Yup this will be AJAX. When my solution is ready, will share it, definitely. :) Regards, Apple Grew my blog @ http://blog.applegrew.com/ On Tue, Jan 19, 2010 at 8:57 AM, Steve Swinsburg steve.swinsb...@gmail.comwrote: Presumably, if the user is typing a long document and hasn't clicked on anything for a while, you'll need to indicate to Wicket/HttpSession (probably via AJAX), that the session is to be kept alive. I believe this is how the autosave functions of things like rich text editors work, keeping the session from timing out. Interested to see your solution. cheers, Steve On 19/01/2010, at 10:19 AM, Pedro Santos wrote: Use attributes from HttpSession like getLastAccessedTime() and getMaxInactiveInterval(). You can get access through WebRequest that holds an HttpServletRequest object. When HttpSession ends, the wicket session ends too. On Mon, Jan 18, 2010 at 5:33 PM, Apple Grew appleg...@gmail.com wrote: Hi, I am a Wicket fresher. I am still getting the hang of this. I want to develop a web application where I would like to show the user a message informing him that his session is about to expire in x minutes. If he wants to save his session then can click on a button which appears along with this message. Also, using javascript I would like to 'sense' that the user is typing-in or moving his mouse over the webpage, which is evident enough that he is using the session. I will somehow figure out the js code needed but *how will I tell Wicket to keep the session alive? Also how can I be notified x minutes before Wicket expires the session?* Regards, Apple Grew my blog @ http://blog.applegrew.com/ -- Pedro Henrique Oliveira dos Santos - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: wicketstuff push, publishing event in a page2 and component installed with channel listener in page1
thanks, i will take a look at them. On Mon, Jan 18, 2010 at 9:32 PM, Rodolfo Hansen kry...@gmail.com wrote: On Fri, Jan 15, 2010 at 5:24 AM, vineet semwal vineetsemwal1...@gmail.comwrote: Sorry ,a little late .. push is a great project,thanks for your efforts. i am a little confused, 1)does the time out only happens after a remove event is published or apart from this, there is another timeout which happens when server is finished pushing into the client? Here are the configuration options for the Jetty implementation of cometd. You can change the connection timeout value to notice disconects sooner (at the cost of ineffiency) http://cometd.org/documentation/cometd-java/server/configuration You can check the bayeux specificition for the details. ( http://svn.cometd.com/trunk/bayeux/bayeux.html) 2)i see some problems when using more than one listener on one component, i tried reproducing the problem by a little tinkering in your example , currently the example in the quickstart i am attaching has two listeners on different components ,you can reproduce the problem by adding listeners to the same component. a event in one channel is caught by channel listener meant for another channel. Great, I'll look into this. thanks again .. On Sat, Dec 26, 2009 at 11:15 PM, Rodolfo Hansen kry...@gmail.com wrote: Regarding remove listeners: Most browsers fail to report the remove event. Only firefox reports removal immediately, all other browsers depend on the timeout for a comet reconnect to notice and fire the remove event; you may need to lower the timeout for the cometd connections. Also,can i install more than one channel listener on a component? Never tried it, but there should be no problem, can you write a quickstart with your use cases, so I can flesh any bugs out? On Thu, Dec 24, 2009 at 10:03 AM, vineet semwal vineetsemwal1...@gmail.comwrote: Hellos, recently i started using wicketstuff push ,i have few doubts as following .. i have a situation where i need to publish a event in page 2 and add the channel listener in page 1 . for eg. a sign out event published in page 2 which i do using a remove listener. Also,can i install more than one channel listener on a component? -- regards, Vineet Semwal -- Rodolfo Hansen CTO, KindleIT Software Development Email: rhan...@kindleit.net Mobile: +1 (809) 860-6669 -- regards, Vineet Semwal - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org -- Rodolfo Hansen CTO, KindleIT Software Development Email: rhan...@kindleit.net Mobile: +1 (809) 860-6669 -- regards, Vineet Semwal
Property Expression Language : Combing bean value and string ?
Greetings all,
In my Datatable , I use PropertyColumn to create my columns.
I want to know whether using Wicket EL , I can do something like following:
Input object has 'name' field.
EL String : Your name is ${name}
Output String : Your name is Tom
Thanks in advance.
-
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Re: PageLink deprecated
That's what I'm trying to say: it can't be accomplished by either BookmarkablePageLink or Link. Link does not have a getPageIdentity method and BookmarkablePageLink only works for bookmarkable links (duh). So Link is never an option because of the missing getPageIdentity method and BookmarkablePageLink only works for bookmarkable pages. What about links to pages that are not bookmarkable? Emond On Monday 18 January 2010 17:12:49 Igor Vaynberg wrote: well, if the functionality can be accomplished using either BookmarkablePageLink or Link, why do we need yet another way to do it? -igor On Sun, Jan 17, 2010 at 11:44 PM, Jeroen Steenbeeke j.steenbeeke...@gmail.com wrote: Guys, no need to keep explaining what's wrong with passing a Page in the constructor, we understand that! Forget about that filthy 3rd constructor, I know it's wrong and I never used it anyway. That wasn't what my question was about. There are two more constructors: PageLink(String, Class) PageLink(String, IPageLink) Both of these do not replicate the dangerous behavior illustrated in this thread so far. I understand that we can easily create our own implementation that simulates the behavior we want. I just wanted to understand the reasoning for removing the whole class when only one of the constructors is dangerous. From what Martijn Dashorst just told me, it was a case of seeing as we already have Link and BookmarkablePageLink, we figured you could just use those instead. This is also the source of miscommunication so far. The Javadoc simply states what you should use instead, but does not explicitly state why. The assumption is that any behavior you can achieve with the PageLink/IPageLink combination can also be done with a simple Link. This does not take into account the use of the Page Identity for security checks however (mainly for determining link visibility, which, frankly, does not need an actual instance of the page in question), which brings us back to Emond's original point. On the other hand, one could argue that the only use for the page identity is for security purposes, and it would therefore be more at home in a specialized class in wicket-security. -- Jeroen Steenbeeke www.fortuityframework.com - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
RE: submit a form from outside of it
Hi namesake! :-)
Unfortunately this is not what I need since as I said panels do not have
reference to their parent page. Anyway thanks for your time!
After all I put the feedbackpanel in every single subpanel (well , only in the
markup, because my abstract parent panel defines it in code).
Thank you all!
Best Regards,
Martin
-Original Message-
From: Martijn Dashorst [mailto:martijn.dasho...@gmail.com]
Sent: Monday, January 18, 2010 7:12 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
IFeedbackProvider {
FeedbackPanel getFeedbackPanel();
}
MyPage extends Webpage implements IFeedbackProvider {
public MyPage() {
add(new FeedbackPanel(feedback));
add(new MySubPanel(panel, this));
}
@Override
FeedbackPanel getFeedbackPanel() {
return get(feedback);
}
}
MySubPanel extends Panel {
private IFeedbackProvider feedback;
public MySubPanel(String id, IFeedbackProvider provider) {
super(id);
this.feedback = provider;
... some ajax handler ...
onSubmit(AjaxRequestTarget t) {
t.addComponent(feedback.getFeedbackPanel());
}
}
}
On Mon, Jan 18, 2010 at 5:52 PM, Martin Asenov mase...@velti.com wrote:
Yes, but in my case the panels don't even know about the parent - they don't
have a reference to it. I've got to think of some workaround on this one.
Thanks for your time,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:45 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Ah - I see. Yeah - you just have to roll your own option for this now. I'd
just recommend some sort of listener pattern - something like adding a void
formSubmitted(form, requesttarget) method on the page that child forms can
call. Then the pages can do whatever they need with it.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:40 AM, Martin Asenov mase...@velti.com wrote:
Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the
parent page when the submit button is pressed , so that I can say
target.addComponent(feed); when feed is in parent page...
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:37 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
What do you mean - you can't tell which button was pressed?
Just add an onSubmit to the button and inside of it, add your feedback
message. or call getPage().info(...), etc. Then allow the onSubmit of
the form to do its thing.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov mase...@velti.com wrote:
Thank you both for the replies! After all I put the buttons in
myFormPanel.
But this way I can't know when the submit and cancel buttons are pressed
so
that I can render the feedback that is located in the parent page.
Should I think of some listener?
BR,
Martin
-Original Message-
From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
Sent: Monday, January 18, 2010 6:31 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
Or wrap the outer page in a form so that any nested forms work with your
out-of-place submit button.
--
Jeremy Thomerson
http://www.wickettraining.com
On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat
alexandrubar...@gmail.com
wrote:
I think you have to pass the form to the behavior in some way or you
can
do
something like this..but it is ugly in some way :)
1. in the form panel
AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
onchange) {
...
public void renderHead(IHeaderResponse response) {
super.renderHead(response);
response.renderJavascript(function submit_my_form(){\n
+
getEventHandler().toString() + \n}, submit_my_form);
}
};
form.add(behave);
...
2. anywhere in the page
and your button will look like this:
input type=button value=my_button onclick=submit_my_form()/
On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov mase...@velti.com
wrote:
Hi, Alexandru, thanks for the quick reply.
I get
java.lang.IllegalStateException: form was not specified in the
constructor
and cannot be found in the hierarchy of the component this behavior
is
attached to
the form is located in the same page and displayed, but it's actually
placed within a panel that is a child of the page. :-(
BR,
-Original Message-
From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
Sent: Monday, January 18, 2010 11:26 AM
To: users@wicket.apache.org
