Hi,
it really was problem with AjaxSubmitButton, can you please test if it
works now for you?
-Matej
ChuckDeal wrote:
> I don't mean to bump this, but was not quickstart not good enough to show the
> problem? If necessary, I can take another stab at it to make it more
> useful.
>
> Chuck
>
>
bumping is ok if something slips through the cracks like it has now, matej
is looking into it right now...
-igor
On 1/18/07, ChuckDeal <[EMAIL PROTECTED]> wrote:
I don't mean to bump this, but was not quickstart not good enough to show
the
problem? If necessary, I can take another stab at i
I don't mean to bump this, but was not quickstart not good enough to show the
problem? If necessary, I can take another stab at it to make it more
useful.
Chuck
ChuckDeal wrote:
>
> Here is a test case the shows the problem. I have two identical pages.
> The only difference is that one page
Here is a test case the shows the problem. I have two identical pages. The
only difference is that one page has the submit button contained by the form
in the second page it is outside of the Form.
http://www.nabble.com/file/5295/quickstart-ajaxbutton.zip
quickstart-ajaxbutton.zip (this contai
The AjaxSubmitButton specifically says that it does NOT have to be attached
to a form; from the javadoc
* A button that submits the form via ajax. Since this button takes the form
as
* a constructor argument it does not need to be added to it unlike the
* [EMAIL PROTECTED] Button} component.
a button should be in a form (one of its parents)
or a button should get the form through its constructor.
So what goes wrong exactly?
Can you have a small test case?
johan
On 12/18/06, ChuckDeal <[EMAIL PROTECTED]> wrote:
It gets there fine, its this line:
if (submit.getForm() == Form.thi
It gets there fine, its this line:
if (submit.getForm() == Form.this
that is the problem. submit (which is the button) does not belong to a
Form, it is supposed to contain a reference to a Form instead. SInce it
can't find a Form object for the button it throws an exception
throw new WicketR
It should work AjaxSubmitButton is a IFormSubmittingComponent component so
this code:
IFormSubmittingComponent submit =
(IFormSubmittingComponent)getPage().visitChildren(
IFormSubmittingComponent.class, new IVisitor()
{
public Object compone
try changing the id of your button to something other then "submit"
-igor
On 12/13/06, ChuckDeal <[EMAIL PROTECTED]> wrote:
I have a Page that has a Form with some select boxes in it and, outside
the
Form, I have three buttons (submit[AjaxSubmitButton], reset[AjaxLink],
cancel[AjaxLink]).
I
I don't mean to bump this, but I do want to make sure that I am using the
AjaxSubmitButton properly. In each instance that I had previously used the
button, it was OUTSIDE the Form, I had to move it inside the Form in order
to avoid the exception. Was this a known side-effect of the change? If
I have a Page that has a Form with some select boxes in it and, outside the
Form, I have three buttons (submit[AjaxSubmitButton], reset[AjaxLink],
cancel[AjaxLink]).
In Wicket 1.2.3 (Actually, I have been using the latest 1.2-SNAPSHOT
releases) The submit button works as advertised and submits t
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