Okay, problem solved.
The _ext is not needed, removed it.
added the line:
$this->RequestHandler->renderAs($this, 'json');
Now renders as a JsonView and the _Serialize is automatically detected and
no template file required.
Regards,
--Kevin
On Wednesday, October 14, 2015 at 2:02:24 PM U
I discovered a bug in the Route.php file.
the reg expression used for finding extensions only works if no parameters
are passed in the URL.
I have temporarily modified my Routes.php file to find the extension
anywhere in the string and used str_replace to remove the .json from the
url and I no l
code that generated the URL:
$url = $this->Url->build(['controller' => 'Purchases', 'action' =>
'get_sales_people_select_list','_ext'=>'json']);
$url = $url . '/47';
The generated URL:
/purchases/get-sales-people-select-list.json/47
Is this not correct?
On Wed, Oct 14, 2015 at 4:40 PM, Anthony G
Did you correctly specify the data type json in your ajax call ?
Le 14 oct. 2015 09:02, "heavyKevy" a écrit :
> I have gone through the documentation, which is a bit too vague, and tried
> many things, but I still am getting an error that the template file is
> missing.
>
> I had this working sen
I have gone through the documentation, which is a bit too vague, and tried
many things, but I still am getting an error that the template file is
missing.
I had this working sending back a json response in 3.0 using ext='json',
yet after updating to 3.1 it is broken.
I checked the migration
Well, what you are doing there is pretty wrong in the first place.
Calling login($data) directly logs in whatever you send it (see the docs
for details), rendering a form login process invalid and insecure.
You should, just as with any other non-ajax request, properly use the Auth
adapters to
Hi All
I recently discovered that cake authentication class does not like multiple
ajax requests.
While recently testing my app, I found it would return a 403 error the
moment I do multiple ajax requests on a logged in user.
I managed to narrow it down to the following code:
$auth = $this->S
For better solution do not refresh the button but only the search result.
On 25-Sep-2013 6:47 PM, wrote:
> Hello,
> I have a problem with a search field on my site and I hope you can help
> me. I got a table called "Clients" with some information like firstname,
> lastname, adress etc. and all MV
You will need to rewrite the autocomplete code on page you are requesting
on the fly for search.
If I am correct your search is also Ajax based...
On 25-Sep-2013 6:47 PM, wrote:
> Hello,
> I have a problem with a search field on my site and I hope you can help
> me. I got a table called "Clients"
Hello,
I have a problem with a search field on my site and I hope you can help me.
I got a table called "Clients" with some information like firstname,
lastname, adress etc. and all MVC files. For testing I use the Client index
function. In my index view file I have a input field where I want to
Hello,
I have a problem with a search field on my site and I hope you can help me.
I got a table called "Clients" with some information like firstname,
lastname, adress etc. and all MVC files. For testing I use the Client index
function. In my view file I have a input field where I want to searc
So, I've got the following bit of code:
start('search'); ?>
Js->get('#ItemIndexForm')->serializeForm(array('isForm' =>
true, 'inline' => true));
$this->Js->get('#ItemIndexForm
#ItemName')->event('keyup',$this->Js->request(array('action' => 'index'),
array('update' => '#content', 'data' => $data
use debug and tail the debug.log file.
On Wednesday, February 9, 2011 12:43:42 PM UTC-6, cake-learner wrote:
>
> I set up one class to handle all ajax request called ajax_controll.
> The thing is everytime I develop a function and there is some kind of
> syntactic errors it r
Hi
I am currently using ajax in multiple controllers in my application. I
implemented the ajax functionality using the jQuery supported version.
$.ajax({
url:'getCountries',
type: "POST",
dataType: "html",
data:"data=" + result,
success: function(data){
//magic...
}
}
});
Is 1
more efficient over the other?
Thanks,
Dave
From: cake-php@googlegroups.com [mailto:cake-php@googlegroups.com] On Behalf
Of majna
Sent: Friday, October 19, 2012 8:46 AM
To: cake-php@googlegroups.com
Subject: Re: Render Element with AJAX Request / Response
Hi,
Controller::render() re
Hi,
Controller::render() returns CakeResponse object, so try
'html' => $this->render()->body()
On Friday, October 19, 2012 2:56:44 AM UTC+2, advantage+ wrote:
>
> My original function to return a JSON array no longer works with v2.1.3
>
>
>
> This worked in 1.3: loaded a form into a modal win
> $this->_ajaxReturn( $response ); //simply encodes and returns the
> $response echo json_encode($response);
>
Returns to what?
Your best bet is to use a JSON view:
http://book.cakephp.org/2.0/en/views/json-and-xml-views.html
--
Like Us on FaceBook h
My original function to return a JSON array no longer works with v2.1.3
This worked in 1.3: loaded a form into a modal window
Controller code snip:
if ( empty( $this->data ) ) {
$this->data = $record; //debug shows the $record so there is
data / debug is at 0 now
Hey
I'm using chosen for a dynamic multiselect, see
link<http://harvesthq.github.com/chosen/>
.
I works perfectly but when I put it in a view which a call with ajax the
javascript doesn't work anymore.
So how can I run a javascript after a ajax request.
the view with th
Hello everybody,
I've a problem with my JQuery Mobile Application. When I open my Web
Root I become redirected by Auth Component to my Login-Action. When I
try to request the same webroot via Ajax the response is empty. Jquery
mobile therefore writes Undifined as result in my browser window. Has
a
further reading
http://cakephp.lighthouseapp.com/projects/42648/tickets/1941-changing-debug-level-on-runtime
--
Our newest site for the community: CakePHP Video Tutorials
http://tv.cakephp.org
Check out the new CakePHP Questions site http://ask.cakephp.org and help others
with their CakePHP r
thanks this is a good point. i think it is a problem in generally, not
only on ajax requests...
if (isset($config['debug']) || isset($config['log'])) {
$reporting = 0;
if ($_this->debug) {
this lines from configure config['debug'] is
oops. it's not on boot strap only. my bad.
On Thu, Aug 25, 2011 at 12:47 PM, stas kim
wrote:
> it's probably because of this line
> https://github.com/cakephp/cakephp/blob/master/cake/libs/configure.php#L119
> which is executed on bootstrap. meaning even if you change debug level
> later in your
it's probably because of this line
https://github.com/cakephp/cakephp/blob/master/cake/libs/configure.php#L119
which is executed on bootstrap. meaning even if you change debug level
later in your controller it will only effect Debugger class behavior
cheers
On Wed, Aug 24, 2011 at 11:21 AM, Ben
Fixing warnings and notices is not the answer to his question--With debug
mode off, he should not be getting such messages displayed. At worse, he
should be seeing formatted Cake errors or simply a blank page if something
is wrong.
Perhaps the warnings/notices/errors are being thrown before you
yes you're right. but the debug mode is for developing purposes and my
question belongs to control the debug mode in app controller.
On 22 Aug., 10:28, "Dr. Loboto" wrote:
> The best way is to fix that warnings and notices. Good app never
> throws this kind of shit.
>
> On 18 авг, 20:18, p r
The best way is to fix that warnings and notices. Good app never
throws this kind of shit.
On 18 авг, 20:18, p r wrote:
> yes the Requesthandler is in use and the debug level is set correctly
> to 0 but errors and notices doesn't care.
> It seems the backtrace is added before the debug level is s
yes the Requesthandler is in use and the debug level is set correctly
to 0 but errors and notices doesn't care.
It seems the backtrace is added before the debug level is set in
AppController? When i set the debug level in core.php it works as i
desired.
On 18 Aug., 15:06, mi...@brightstorm.co.uk
> Hello,
>
> could anybody explain this behavior? i set in my app controller
>
> if($this->RequestHandler->isAjax()) {
> $this->layout = 'ajax';
> $this->autoRender = false;
> Configure::write('debug', 0);
> }
>
> but notices and err
> Hello,
>
> could anybody explain this behavior? i set in my app controller
>
> if($this->RequestHandler->isAjax()) {
> $this->layout = 'ajax';
> $this->autoRender = false;
> Configure::write('debug', 0);
> }
>
> but notices and err
Hello,
could anybody explain this behavior? i set in my app controller
if($this->RequestHandler->isAjax()) {
$this->layout = 'ajax';
$this->autoRender = false;
Configure::write('debug', 0);
}
but notices and errors are even
Hi there.
I've been looking for a way to detect on the client side that the ajax
request he performed was not successful because the user session has
expired.
My problem is not related with security level (seen too many people
with this one..).
I've red this blog pos
Why are you mixing the (deprecated) Ajax-Helper with html code for a
form? Have you tried relying on the helper to produce the code for
your form?
On 14 Mrz., 14:21, heohni
wrote:
> Hi,
>
> I have this form:
> Ajax->Form('searchMember', 'post', array('update' =>
> 'target', 'url' => array('action
On Mon, Mar 14, 2011 at 9:21 AM, heohni
wrote:
> Hi,
>
> I have this form:
> Ajax->Form('searchMember', 'post', array('update' =>
> 'target', 'url' => array('action' => 'searchMember'))); ?>
>
> Suchwort:
>
>
>
>
> Ajax->Form->end(); ?>
>
> And everytime I press the submit button, the request
Hi,
I have this form:
Ajax->Form('searchMember', 'post', array('update' =>
'target', 'url' => array('action' => 'searchMember'))); ?>
Suchwort:
Ajax->Form->end(); ?>
And everytime I press the submit button, the request is send 2 times,
I can see this with firebug in my firefox.
Is there some
9, 2011 3:17 PM
> To: CakePHP
> Subject: Re: Is there any way to display error when working on ajax
> request?
>
> Since your method of request for the action is ajax, I'm assuming it's
> publicly accessible.
>
> So the best way to dev it is thru the browser then apply
hn818
Sent: Wednesday, February 09, 2011 3:17 PM
To: CakePHP
Subject: Re: Is there any way to display error when working on ajax request?
Since your method of request for the action is ajax, I'm assuming it's
publicly accessible.
So the best way to dev it is thru the browser then apply your aj
e:
> I set up one class to handle all ajax request called ajax_controll.
> The thing is everytime I develop a function and there is some kind of
> syntactic errors it returns empty. Usually take a couple of hours to
> develop
> one simple query because of this. Is it anyway to sp
I set up one class to handle all ajax request called ajax_controll.
The thing is everytime I develop a function and there is some kind of
syntactic errors it returns empty. Usually take a couple of hours to
develop
one simple query because of this. Is it anyway to spit out the php
error on
browser
Hi,
yesterday I noticed some odd behavior of my web app while testing it
in Safari:
Here is the line of code in my controller where it all the weirdness
happens:
if (!$this->RequestHandler->isAjax()) {
$this->Session->write('Auth.User.activeRecordId', NO_ID);
$this->Session
hi,
usually i do this to implement ajax feeling:
VIEW:
with jquery I call the controller functions and tell cake where (in a
DIV) to write the result
$('#attachments_count').load('countattachments');
CONTROLLER
then all the logic happens in my controller function,
countattachments, and I tell
I had weird problems with ajax calls happening twice (when deleting a
record in a modalbox). I'm pretty certain it came down to using
FireBug for me which sometimes, but not all the time, reloads the
request to get some extra information.
Paul.
Check out the new CakePHP Questions site http://cak
Check that you are returning false to the form itself, when you submit
the form with ajax, as else the form will think that it must also
submit!
Does this apply to your case?
Enjoy,
John
On May 22, 7:00 pm, Felipe Carballo
wrote:
> Hello guys! I'm having a problem using jQuery UI and Ajax Help
Hello guys! I'm having a problem using jQuery UI and Ajax Helper, with
Prototype. My form is opened in a dialog box using jQuery UI. When I
send the form, my action is executed twice. If I remove the library
jQuery UI, the action is performed correctly (once). Does anyone know
how I can fix this pr
ith" option you can add any JS code that should return well-
> > formed "name1=value1&name2=value2" string. Value of this option is not
> > escaped by Cake.
>
> > On Mar 22, 8:39 pm, Karsten Backhaus wrote:
>
> > > I wonder if it is possible to add pos
ption is not
> escaped by Cake.
>
> On Mar 22, 8:39 pm, Karsten Backhaus wrote:
>
> > I wonder if it is possible to add postdata to the ajax request before
> > it is send to the server. Currently I simply add hidden form-fields to
> > the formular which is supposed t
In "with" option you can add any JS code that should return well-
formed "name1=value1&name2=value2" string. Value of this option is not
escaped by Cake.
On Mar 22, 8:39 pm, Karsten Backhaus wrote:
> I wonder if it is possible to add postdata to the ajax request before
I wonder if it is possible to add postdata to the ajax request before
it is send to the server. Currently I simply add hidden form-fields to
the formular which is supposed to be sent to the server though that's
pretty slow as the browser has to rebuild the dom everytime I add
something to the
Simple include of RequestHandler component will do all automagically.
On Feb 12, 9:17 am, NickPick wrote:
> thanks. that worked.
>
> On Feb 11, 7:38 pm, John Andersen wrote:
>
>
>
> > In your items controller, you have to set the layout, when you detect
> > that
thanks. that worked.
On Feb 11, 7:38 pm, John Andersen wrote:
> In your items controller, you have to set the layout, when you detect
> that it is an AJAX request that you are processing.
> See how to detect this
> athttp://book.cakephp.org/view/350/Obtaining-Request-Information
&g
In your items controller, you have to set the layout, when you detect
that it is an AJAX request that you are processing.
See how to detect this at
http://book.cakephp.org/view/350/Obtaining-Request-Information
And here for the layout
http://book.cakephp.org/view/51/Controller-Attributes#Page
The below ajax request works well, but it's using default.ctp as
layout file instad of ajax.ctp. How can I tell it to use ajax.ctp as
layout file?
thanks
array(
'controller'=>'items',
'action'=&g
Well, I don't know how I ever got along without Firebug! Thanks for
that Dave!!
The good news is I found a workaround (at the very bottom here). The
not so good news is that it looks like if you're updating multiple
elements as part of an ajax request and you're complete callback
Thanks, Dave! I'll check that out. Makes complete sense. Oh, I found
the $ajax->div() and $ajax->divEnd() methods in the cake 1.2 api
documentation. That was a little while ago, but if I remember
correctly, my ajax calls weren't updating the selection box data until
I enclosed them in ajax divs.
l
Scott, here is observeField() code below that triggers a call in the
controller to get product categories, classes, types, brands, styles
and sub-brands. And from David's response I need to change the
'loaded' option to 'complete' and figure out what kind of javascript
error I'm getting...
echo $f
f the select element
>>
>> if ( $("pstyle_sel").length == 0)
>> $("style").hide();
>>
>> this kind of works. The first time I select something that doesn't
>> have styles, it doesn't hide the row. Next I selected something t
h of the select element
>
> if ( $("pstyle_sel").length == 0)
> $("style").hide();
>
> this kind of works. The first time I select something that doesn't
> have styles, it doesn't hide the row. Next I selected something that
> did have styl
ave styles and it HID THE ROW. It's like it's seeing the data
> from the previous ajax request, like it's one step behind?
>
> I'm using 'loaded' instead of 'complete' because when I used
> 'complete' my ajax loading indicator just ke
").hide();
this kind of works. The first time I select something that doesn't
have styles, it doesn't hide the row. Next I selected something that
did have styles and it HID THE ROW. It's like it's seeing the data
from the previous ajax request, like it's one step behi
ajax->observeField. The update is working great,
> > but I would like to be able to hide the row if no styles are returned
> > by the Ajax request.
> >
> > My view code is:
> >
> >
> > Product Style:
> >
> >
he update is working great,
> but I would like to be able to hide the row if no styles are returned
> by the Ajax request.
>
> My view code is:
>
>
> Product Style:
>
>
I have a row that includes a form selection of product styles and is
being updated via an ajax->observeField. The update is working great,
but I would like to be able to hide the row if no styles are returned
by the Ajax request.
My view code is:
Product St
an
> > AJAX response?
>
> > I have a form that contains a dropdown and a group of radio buttons.
> > When a new item is selected from the dropdown, an XMLHTTPRequest is
> > sent to the server to get the new collection of radio buttons.
>
> > Right now, my contro
n a new item is selected from the dropdown, an XMLHTTPRequest is
> sent to the server to get the new collection of radio buttons.
>
> Right now, my controller determines if its an AJAX request, queries
> for the list of items, manually generates the HTML for the radio
> buttons, and sends
, my controller determines if its an AJAX request, queries
for the list of items, manually generates the HTML for the radio
buttons, and sends the response. I know that the View layer offers a
FormHelper to "automagically" create the form element ($form->radio
()), is there anything simi
To avoid such problems I set debug to 0 at the very end of AJAX view/
layout. So If there are errors - I see them and can fix, if there are
no errors - response is clear of debug data.
On Oct 29, 5:55 pm, Thanos Atskakanis
wrote:
> it was set to 2 but AjaxHandler component was setting it back to
it was set to 2 but AjaxHandler component was setting it back to 0 -
disabled it and found out that the problem was coming from a missing file in
the live server - rookie error :o)
thanks a lot for that
On Thu, Oct 29, 2009 at 4:04 AM, Dr. Loboto wrote:
>
> Set debug to at least 1 (and no mat
Set debug to at least 1 (and no matter that it is ajax) to see error.
On Oct 28, 10:34 pm, Thanos Atskakanis
wrote:
> Hi all,
>
> I am currently building a CMS and my system does some ajax requests which
> return some html: simple and nice.
> When I run it locally it all works fine but when I pu
Hi all,
I am currently building a CMS and my system does some ajax requests which
return some html: simple and nice.
When I run it locally it all works fine but when I put it live instead of
the expected html I get "null". Not "Not found" or any other error, just the
word "null".
All I use is the
er-22-09 7:42 AM
To: CakePHP
Subject: Re: Is Ajax request
How is the HTML looking? Does it look the same as the other links?
Enjoy,
John
On Oct 21, 6:59 pm, "Dave Maharaj :: WidePixels.com"
wrote:
> I have this function where its first called normally by a link (http
>
How is the HTML looking? Does it look the same as the other links?
Enjoy,
John
On Oct 21, 6:59 pm, "Dave Maharaj :: WidePixels.com"
wrote:
> I have this function where its first called normally by a link (http
> request). Once loaded the pagination links are done via Ajax. I have 'all',
> 'ne
I have this function where its first called normally by a link (http
request). Once loaded the pagination links are done via Ajax. I have 'all',
'new', 'archive' links which all get called to paginate the result using
Ajax and jquery. Problem is when i go to posts/index it works fine, click
'all'
A timeout where? In Cake, or javascript? I don't kow about Prototype
(what Cake's AjaxHelper uses), but jQuery allows to set a timeout.
But you should ensure that you have debug set to 0, first of all. That
will always slow things down somewhat. Also, look into
caching--perhaps the problem is due
Does nobody know how to solve this problem?
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hey DatacenterHellas,
Your best bet is to not login via an AJAX call. Just set
"standardSubmit: true" in your ExtJS login FormPanel config and call
"MyFormPanel.getForm().submit({method: 'POST'});". Now you can let
CakePHP handler your redirects server-side. On failure, you login page
will just r
Thanks all of you for your assistance :)
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c
You can't redirect from the server (PHP) when making an AJAX request.
You can only send a response back to the client. If your server code
determines that the user needs to be redirected, it should inform your
client-side code--the javascript function making the AJAX
request--that *it* s
Ok . . .
this is one aproach of the problem but I think is not the most
correct.
I mean what about if the user will not enter correct data ? ? ?
Have I to redirect him to an action and then will return again to
login ? ? ?
Is not so nice.
My project is based on CakePHP and ExtJS.
For the Log
window.location = url
On Thu, Jul 23, 2009 at 1:16 PM,
DatacenterHellas wrote:
>
> Hello all.
>
> How can I redirect after AJAX login to another action
> >
>
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You received this message because you are subscribed to the Google Groups
"CakePHP"
Hello all.
How can I redirect after AJAX login to another action
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You received this message because you are subscribed to the Google Groups
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To unsubscribe from thi
Hey,
I'm new to the CakePHP-Group. I joined, because I wasn't able to find
any help on this.
I've built a small project management interface on Cake. All actions
are done with Ajax (jQuery), so that the page doesn't have to be
reloaded. Normally the website is started at morning and remains
open
d, Jul 15, 2009 at 5:01 AM, Luke wrote:
>
> Hi,
>
> I am still working on my first Cake Project and I am now stuck in some
> AJAX Request where I would need some feedback.
>
> I am trying to achieve the following.
>
> I have recipes everybody can see. There is a link above the
Hello,
> I have tried this out to understand the structure behind it but get
>
> Query: removeFavorite
>
> So it must be the below line, but what is the issue with this code?
> Why is it using the method as a query?
>
> $success = $this->Product->User->removeFavorite($id);
>
>
> I can't see what
Hi,
Oh, so I can not re-send the complete Page when I realise the User is
not logged in!? So I will have to re-think my idea. I really thought
that there is a possibility, but good that you have given a clear "No,
not possible"
I have now tried something out from the IBM Tutorial, its a bit of
i,
>
> I am still working on my first Cake Project and I am now stuck in some
> AJAX Request where I would need some feedback.
>
> I am trying to achieve the following.
>
> I have recipes everybody can see. There is a link above the recipe to
> add it to your favorites. If you
Hi,
I am still working on my first Cake Project and I am now stuck in some
AJAX Request where I would need some feedback.
I am trying to achieve the following.
I have recipes everybody can see. There is a link above the recipe to
add it to your favorites. If you are logged in the page should
It can't be done from the server if it's an asynchronous request.
You'll have to do it with javascript. Perhaps you could send back a
particular response if you wish to redirect and have your JS callback
look for it. If it's found, use document.location = ...
On Sun, Jun 7, 2009 at 9:34 AM, Mar
Hi everyone.
I have a site that is farely heavy on AJAX.
Sometimes it happens that the Session has expired and the user needs
to log in again.
Is there a way that I can detect that and send the surfer to the log
in page instead of the login page loading inside the dom container?
--~--~-~-
well, i haven't done this specifically so I'm not sure I can
demonstrate exactly how, but you probably either have to allow
('addToWatchlist') in your before filter then instead test if the
request comes from a logged in user in that action before you add the
link to the watch list. If it does NOT
ler.
>
> How would I implement the idea you mention?
>
> Thanks,
>
> Dave
>
> -Original Message-
> From: Dr. Loboto [mailto:drlob...@gmail.com]
> Sent: May-20-09 12:58 PM
> To: CakePHP
> Subject: Re: If not Ajax request
>
> I almost can swear tha
I have an ajax request which works fine. I am testing out secnarios that the
end user may try in advance to see what would happen.
My link in the view;
link(' ',
array('controller'=>'journal','action'=>'quicksave',$journal['Journal&
Hi,
I am not sure what else I could copy & paste from the sourcecode. I
only have a normal login action, in my user controller, which only
saves the last login and thats pretty much it. An Login AJAX action?
How would this look?
Have you got an example?
Thanks. Luke
On 20 Mai, 22:18, christo w
all I can think of is that you don't have a login AJAX action set up
so the request dies without redirecting to a log in if you try to
access the link from not logged in? I dunno it's hard to tell from
your description. pastebin maybe?
On May 20, 10:58 am, Luke wrote:
> Hi,
>
> I have again re
hered from the Auth id and the variable passed in the URL
link in the controller.
How would I implement the idea you mention?
Thanks,
Dave
-Original Message-
From: Dr. Loboto [mailto:drlob...@gmail.com]
Sent: May-20-09 12:58 PM
To: CakePHP
Subject: Re: If not Ajax request
I almost can
Hi,
I have again read backwards and forwards through the www to find a
solution for my problem. I am sure that there is other people running
into the same issue. How do you guys solve it?
All I want is a link saying "Add article to watchlist" which than, if
the User is logged in writes it into t
Ajax request
I almost can swear that you tested it in IE and after save no actual request
to server was done at all (can be checked in Apache access
logs) - you saw cached result of previous save call. If yes, only addition
of any random param to form action (or whatever you request on "
I almost can swear that you tested it in IE and after save no actual
request to server was done at all (can be checked in Apache access
logs) - you saw cached result of previous save call. If yes, only
addition of any random param to form action (or whatever you request
on "save" click) will save
I am trying to make sure the requested function is Ajax, if not
redirect...so if someone tries to manually enter it into the url it
redirects them back to the index page but not working the way i want.
This is what happens:
Scenario 1:
original URL: localhost/testsite/journal/256
If someone goes to
Hi brian,
thanks for your response. This is not really a solution I am looking
for. I think the link should still be visible all the time and I can't
imagine that cake and AJAX would not be able to handle this somehow. I
am sure that someone has faced a similar issue before and I am hoping
that t
An AJAX request returns a chunk of HTML (or other data) and isn't
going to handle the redirect properly. The simplest way to deal with
this is to only display the link if the user is logged in.
On Tue, May 19, 2009 at 8:36 AM, Luke wrote:
>
> Hi,
>
> I am still pretty new to
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