Re: Radio button value null after post, but appears in post headers upon inspection
That looks like an unusual array dump to me. Multiple radio button fields with the same name should result in a single variable being passed back to the controller. Can you even build an array with multiple instances of the same key-value pairing as shown above? My guess would be that IF you have managed to get the above array passed to the controller that last instance of data[EdibleUndy] [underwear_type_id] would be the one read and that doesn't have a value. Can you show us the HTML code for the form including the radio buttons and the hidden field you mention, and also could you supply the PHP you are using to create the form? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: in the link have a parameter 'CAKEPHP'
That's cakePHP's session ID. It's appearing because the server believes you don't have cookies enabled. You can't turn it off as it is a php core behavior. On Sat, Jan 3, 2009 at 2:44 AM, Rimoe meiyo...@gmail.com wrote: hi, everyone, in my link have a parameter CAKEPHP. for example http://www.mysite.com/mails/add?CAKEPHP=3831a0f2e1fb52c9ae15d163f57c4695 how to remove it, recently I have read a topic about to remove the parameter, but I have forget to it. If you know, please tell me. Thank you. rimoe --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
show image from layout
How can one show image either logo or any other from layout file as this file is present in webroot folder so there is confution over image address. Thanks Piyush -- View this message in context: http://www.nabble.com/show-image-from-layout-tp21263881p21263881.html Sent from the CakePHP mailing list archive at Nabble.com. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Validation errors for the wrong form
Hi! Are the forms for both login and register differ. or you have included the login element within the registration form? thanks! On Jan 1, 9:33 am, mariacheu...@gmail.com mariacheu...@gmail.com wrote: Hi all, Apologies if this has been addressed before, but I've searched and can't find a thing. I have a register page, with a login element in the sidebar. The login element is on every page in the sidebar. When I try register with invalid data, the correct validation errors show for the register form. However, the errors are also showing on the login form, even though that hasn't been submitted. The login form is using jQuery Ajax to submit. Any ideas anyone? Thanks! --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Validation errors for the wrong form
They are both different forms, the register form is from a view and the login is part of a sidebar element. Both forms have different names too. Thanks for your reply! On Jan 3, 6:21 pm, Nature Lover nature_lover1...@yahoo.co.in wrote: Hi! Are the forms for both login and register differ. or you have included the login element within the registration form? thanks! On Jan 1, 9:33 am, mariacheu...@gmail.com mariacheu...@gmail.com wrote: Hi all, Apologies if this has been addressed before, but I've searched and can't find a thing. I have a register page, with a login element in the sidebar. The login element is on every page in the sidebar. When I try register with invalid data, the correct validation errors show for the register form. However, the errors are also showing on the login form, even though that hasn't been submitted. The login form is using jQuery Ajax to submit. Any ideas anyone? Thanks! --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Radio button value null after post, but appears in post headers upon inspection
Can you show us the HTML code for the form including the radio buttons
and the hidden field you mention, and also could you supply the PHP
you are using to create the form?
Can you show us the HTML code for the form including the radio buttons
and the hidden field you mention, and also could you supply the PHP
you are using to create the form?
Sure thing. In my add action:
$this-set('user', $this-user);
$underwear_types =
$this-EdibleUndy-UnderwearType-find('all');
$this-set(compact('underwear_types'));
Because I want to get all the fields in UnderwearType (id, name AND
image), I am doing passing 'all' to find instead of 'list'.
Then, in the view, since I want to customize the output of radio
buttons, I loop though the array and specify a single option each
time. Here's the view snippet:
?
echo $form-create('EdibleUndy', array('url'=$appCallbackUrl .
/ . $this-params['url']['url'])); //
echo 'table class=undietypes';
echo 'tr';
foreach($underwearTypes as $type){
echo 'td';
echo $form-input('underwear_type_id',
array('type'='radio',
'legend'=false,
'label'='',
'div'=false,
'options'=array($type['UnderwearType']['id']='')));
echo 'h2' .
$type['UnderwearType']['name'] . '/h2';
echo $html-image($appCallbackUrl .
'img/' . $type
['UnderwearType']['image']);
echo '/td';
}
echo '/tr';
echo 'tr';
echo 'td colspan=4';
echo $form-submit('Select', array
('type'='submit','div'='submit_container'));
echo '/td';
echo '/tr';
echo '/table';
echo $form-end();
?
The resulting HTML output is this:
form method=post action=http://www.mydomain.com/underwear/undies/
add
fieldset style=display:none;
input type=hidden name=_method value=POST /
/fieldset
table class=undietypes
tr
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId1 value=1 /
h2Boxers/h2
img src=http://www.syntax101.com/facebook/
edible_underwear/img/boxers.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId2 value=2 /
h2Briefs/h2
img src=http://www.syntax101.com/facebook/
edible_underwear/img/briefs.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId3 value=3 /
h2Panties/h2
img src=http://www.syntax101.com/facebook/
edible_underwear/img/panties.gif alt= /
/td
/tr
tr
td colspan=4
div class=submit_containerinput type=submit
type=submit value=Wear //div
/td
/tr
/table
/form
Thanks for your insight!
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Re: Radio button value null after post, but appears in post headers upon inspection
Can you show us the HTML code for the form including the radio buttons
and the hidden field you mention, and also could you supply the PHP
you are using to create the form?
Can you show us the HTML code for the form including the radio buttons
and the hidden field you mention, and also could you supply the PHP
you are using to create the form?
Sure thing. In my add action:
$this-set('user', $this-user);
$underwear_types =
$this-EdibleUndy-UnderwearType-find('all');
$this-set(compact('underwear_types'));
Because I want to get all the fields in UnderwearType (id, name AND
image), I am doing passing 'all' to find instead of 'list'.
Then, in the view, since I want to customize the output of radio
buttons, I loop though the array and specify a single option each
time. Here's the view snippet:
?
echo $form-create('EdibleUndy', array('url'=$appCallbackUrl .
/ . $this-params['url']['url'])); //
echo 'table class=undietypes';
echo 'tr';
foreach($underwearTypes as $type){
echo 'td';
echo $form-input('underwear_type_id',
array('type'='radio',
'legend'=false,
'label'='',
'div'=false,
'options'=array($type['UnderwearType']['id']='')));
echo 'h2' .
$type['UnderwearType']['name'] . '/h2';
echo $html-image($appCallbackUrl .
'img/' . $type
['UnderwearType']['image']);
echo '/td';
}
echo '/tr';
echo 'tr';
echo 'td colspan=4';
echo $form-submit('Select', array
('type'='submit','div'='submit_container'));
echo '/td';
echo '/tr';
echo '/table';
echo $form-end();
?
The resulting HTML output is this:
form method=post action=http://www.mydomain.com/underwear/undies/
add
fieldset style=display:none;
input type=hidden name=_method value=POST /
/fieldset
table class=undietypes
tr
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId1 value=1 /
h2Boxers/h2
img src=http://www.mydomain.com/edible_underwear/img/
boxers.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId2 value=2 /
h2Briefs/h2
img src=http://www.mydomain.com/edible_underwear/img/
briefs.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId3 value=3 /
h2Panties/h2
img src=http://www.mydomain.com/]/edible_underwear/
img/panties.gif alt= /
/td
/tr
tr
td colspan=4
div class=submit_containerinput type=submit
type=submit value=Wear //div
/td
/tr
/table
/form
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RE: Adding table row with AJAX
Thanks for the reply. This is mostly working but there is something I am
still missing.
I tried two different things.
1) If I call the 'edit' action I get a jQuery error ( 'div is null' ' elem =
jQuery.makeArray( div.childNodes ); ')
I am not sure what div its looking for. The second problem with that is that
although I can see the table row returned at the start of the response
string the rest of the content for the view is also returned and I don't
know why.
2) If I call the 'add_ingredient_row' action I only get the content for the
view in the response but no new row.
So two questions:
1) Why is CakePHP returning all the content for the view?
2) Why is jQuery giving me that error?
Thanks again. If I can just get over this hurdle I will be off and running.
VIEW
h2Testing AJAX with CakePHP and Prototype and Scriptaculous/h2
table id=the_table
tr id=row1
tdRow1/td
/tr
/table
div id=mydiv
This is a div
/div
div id=the_buttonClick/div
CONTROLLER
class AjaxtestsController extends AppController
{
var $components = array( 'RequestHandler' );
function index()
{
$this-pageTitle = 'Index Page';
}
function view()
{
$this-pageTitle = 'View Page';
}
function edit()
{
$this-pageTitle = 'Edit Page';
if($this-RequestHandler-isAjax())
{
$this-layout = 'ajax';
echo 'trtdThis is a row/td/tr';
}
}
function add_ingredient_row()
{
echo 'trtdThis is a row/td/tr';
}
}
JAVASCRIPT
$(document).ready(function()
{
$('#the_button').click(function()
{
$.ajax({
url: 'add_ingredient_row/',
type: 'GET',
data: 'data string',
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#the_table').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of brian
Sent: Friday, January 02, 2009 8:26 PM
To: cake-php@googlegroups.com
Subject: Re: Adding table row with AJAX
On Fri, Jan 2, 2009 at 7:45 PM, Steven Wright rhythmicde...@gmail.com
wrote:
Hi Adam,
Thanks for writing back. How would you get the data back for the row
from CakePHP?
My row contains four columns with the following inputs:
amount [text]
measurement_type [select]
description [text]
ingredient [select]
This is normally rendered from an element as a table row. I would like
the output of that element appended to my table. So on the button
click I call the action add_ingredient_row and it should return the
content of the element. How do I get that content appended to my table.
I can do this without CakePHP. But for me this exercise is to use the
framework.
Being a jQuery user, I never bother with Cake's ajax() stuff. But jQuery is
dead simple (mostly) to use, so I don't mind having to write out a few lines
of JS. To do what you describe, I'd do something like
this:
$(document).ready()
{
$('#the_button').click(function()
{
$.ajax({
url: the_href,
type: 'GET',
data: the_data,
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
},
success: function(msg)
{
/* assuming your controller is returning
'tr.../tr'
*/
$('#the_table').append(msg);
}
});
});
});
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Re: $this-Auth-user not refreshing after edit
Here's how I do it:
private function reload () {
$this-Auth-login($this-User-read());
}
So just call $this-reload() from your controller.
--
View this message in context:
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Sent from the CakePHP mailing list archive at Nabble.com.
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Re: without default Database Configuration: Fatal error: ConnectionManager::getDataSource - Non-existent data source default
Sounds like you want to use multiple configurations based on the location of the app (live versus dev) There are a few ways to solve this. I wrote an article outlining my preferred method. You should read through the user comments as they add some helpful insight, including how to use this method while still allowing the bake commands to work. http://edwardawebb.com/programming/php-programming/cakephp/automatically-choose-database-connections-cakephp On Jan 2, 11:48 am, james revillini ja...@revillini.com wrote: I'm debating whether or not this is a bug. What do you think: If I use the bake script and create a database conf called local, then run it again and create one called live, then start trying to build models, it will fail with this error message: Fatal error: ConnectionManager::getDataSource - Non-existent data source default in C:\tmp\1.2.x.x\cake\libs\model \connection_manager.php on line 109 The exact step of model building that it will fail on is right after you answer the question about adding validation criteria for the model fields. I'm guessing cake is hard coded to use default somewhere, but I think it's a bug since it did ask me which db config to use and I told it to use the local one at the outset of running the bake script. Incidentally, the work around is to create a db config for default, and make sure it points to the database that has the correct structure for your bake. Another option is to turn it into a drinking game, where you need to chug every time you get a fatal error, and eventually the situation will smooth itself out. Thanks, Jim --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
RE: API
Not one really specific thing. In general the Cookbook does not seem to answer my questions so I find myself going to look at the API. Most methods seem to accept options but there is almost never a description of what those options are. I suppose given time I will figure this out, its just difficult for a new user to get past. Thanks for replying. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of Webweave Sent: Friday, January 02, 2009 11:43 PM To: CakePHP Subject: Re: API My experience is that you are much better off looking at the cookbook for that sort of information. In a few cases the API docs are very clear as to what the options are and how to use them, but often it requires a lot of digging to understand them. What part of the API are you having trouble with ? On Dec 29 2008, 6:18 pm, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: I am looking at the API trying to figure some stuff out. I keep getting stuck when they docs talk about options but dont actually give the options. How am I supposed to figure out what options are available for a particular class? Some of the time I can find the option in class or method definition but there is no real explanation of what it's for. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
How can we edit uploaded file
hi all On my edit view i have one option to update uploaded file but when i do this they are taking first character of file and in size also it will take character value so how can i edit uploaded file value in cake php --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Radio button value null after post, but appears in post headers upon inspection
The resulting HTML output is this:
form method=post action=http://www.mydomain.com/underwear/undies/
add
fieldset style=display:none;
input type=hidden name=_method value=POST /
/fieldset
table class=undietypes
tr
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId1 value=1 /
h2Boxers/h2
img src=http://www.mydomain.com/edible_underwear/img/
boxers.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId2 value=2 /
h2Briefs/h2
img src=http://www.mydomain.com/edible_underwear/img/
briefs.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId3 value=3 /
h2Panties/h2
img src=http://www.mydomain.com/]/edible_underwear/
img/panties.gif alt= /
/td
/tr
tr
td colspan=4
div class=submit_containerinput type=submit
type=submit value=Wear //div
/td
/tr
/table
/form
I have just created a list of radio buttons (was a select list in one
of my forms) using the form helper
echo $form-input('user_id', array('empty'='None', 'type'='radio'));
and it produced this ...
input name=data[Person][user_id] id=PersonUserId_ value=
type=hidden
input name=data[Person][user_id] id=PersonUserId1 value=1
type=radiolabel for=PersonUserId1PaulGardner/label
input name=data[Person][user_id] id=PersonUserId2 value=2
type=radiolabel for=PersonUserId2IanForster/label
input name=data[Person][user_id] id=PersonUserId3 value=3
type=radiolabel for=PersonUserId3SteveGinn/label
Your code is adding a hidden field for every radio option whereas mine
only has one at the very top. That could be interfering with things,
not that I'm sure why it adds a hidden field for every radio button?!?
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Re: Data validation doesn't work
Hi gearvOsh, Unfortunately nothing happens again :( I tryed with both options (required = true and notEmpty rule) and nothing happens. More tips? Thanks! On 29 dez 2008, 23:24, gearvOsh mileswjohn...@gmail.com wrote: You either need to have: required = true Or use the notEmpty rule. http://book.cakephp.org/view/127/One-Rule-Per-Fieldhttp://book.cakephp.org/view/740/notEmpty --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Data validation doesn't work
Hi gearvOsh, Unfortunately nothing happens again :( I tryed with both options (required = true and notEmpty rule) and nothing happens. More tips? Thanks! On 29 dez 2008, 23:24, gearvOsh mileswjohn...@gmail.com wrote: You either need to have: required = true Or use the notEmpty rule. http://book.cakephp.org/view/127/One-Rule-Per-Fieldhttp://book.cakephp.org/view/740/notEmpty --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Radio button value null after post, but appears in post headers upon inspection
WebbedIt:
When you submit this form does the selected radio button's valueshow
up or does it arrive at validation and the SQL as null like it does
for me?
Cheers!
On Jan 3, 11:01 am, WebbedIT p...@webbedit.co.uk wrote:
The resulting HTML output is this:
form method=post action=http://www.mydomain.com/underwear/undies/
add
fieldset style=display:none;
input type=hidden name=_method value=POST /
/fieldset
table class=undietypes
tr
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId1 value=1 /
h2Boxers/h2
img src=http://www.mydomain.com/edible_underwear/img/
boxers.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId2 value=2 /
h2Briefs/h2
img src=http://www.mydomain.com/edible_underwear/img/
briefs.gif alt= /
/td
td
input type=hidden name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId_ value= /
input type=radio name=data[EdibleUndy]
[underwear_type_id] id=EdibleUndyUnderwearTypeId3 value=3 /
h2Panties/h2
img src=http://www.mydomain.com/]/edible_underwear/
img/panties.gif alt= /
/td
/tr
tr
td colspan=4
div class=submit_containerinput type=submit
type=submit value=Wear //div
/td
/tr
/table
/form
I have just created a list of radio buttons (was a select list in one
of my forms) using the form helper
echo $form-input('user_id', array('empty'='None', 'type'='radio'));
and it produced this ...
input name=data[Person][user_id] id=PersonUserId_ value=
type=hidden
input name=data[Person][user_id] id=PersonUserId1 value=1
type=radiolabel for=PersonUserId1PaulGardner/label
input name=data[Person][user_id] id=PersonUserId2 value=2
type=radiolabel for=PersonUserId2IanForster/label
input name=data[Person][user_id] id=PersonUserId3 value=3
type=radiolabel for=PersonUserId3SteveGinn/label
Your code is adding a hidden field for every radio option whereas mine
only has one at the very top. That could be interfering with things,
not that I'm sure why it adds a hidden field for every radio button?!?
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Re: show image from layout
On Sat, Jan 3, 2009 at 6:07 AM, piyushsharmajec piyushsharma...@gmail.com wrote: How can one show image either logo or any other from layout file as this file is present in webroot folder so there is confution over image address. Thanks It does not matter that the layout file is stored in app/views/layouts/ and the images in app/webroot/img. The layouts directory is simply a folder to store template files; they are not served from there in the same manner as, say an HTML file would be. Any paths inside of a layout file should be relative to root (app/webroot/) not the layouts directory. So, if you have an image file at app/webroot/img/my_img.png you can put the following in your layout: img src=/img/my_img.png alt=... Likewise, if you have CSS rules which refer to images, you can either set the URLs to be relative from root - url(/img/my_img.png) or from the CSS directory - url(my_img_foloder/my_img.png) where the latter refers to an image stored at app/webroot/css/my_img_foloder/my_img.png and the former to the regular img directory. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: show image from layout
You can also use the image method from the html helper.
echo $html-image('my_logo.jpg');
On Sat, Jan 3, 2009 at 12:51 PM, brian bally.z...@gmail.com wrote:
On Sat, Jan 3, 2009 at 6:07 AM, piyushsharmajec
piyushsharma...@gmail.com wrote:
How can one show image either logo or any other from layout file as this
file
is present in webroot folder so there is confution over image address.
Thanks
It does not matter that the layout file is stored in
app/views/layouts/ and the images in app/webroot/img. The layouts
directory is simply a folder to store template files; they are not
served from there in the same manner as, say an HTML file would be.
Any paths inside of a layout file should be relative to root
(app/webroot/) not the layouts directory.
So, if you have an image file at app/webroot/img/my_img.png you can
put the following in your layout:
img src=/img/my_img.png alt=...
Likewise, if you have CSS rules which refer to images, you can either
set the URLs to be relative from root - url(/img/my_img.png) or
from the CSS directory - url(my_img_foloder/my_img.png)
where the latter refers to an image stored at
app/webroot/css/my_img_foloder/my_img.png and the former to the
regular img directory.
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Radio Button onClick
What Im trying to do is list some events as radio buttons. When one
of the buttons is clicked I wanted it to populate three textareas with
the data from that event record. ie: when clicking on wedding the
Event.name, Event.description, Event.information would be populated
into the three textareas.
The problem Im having is how to use the $form helper to get it to
work. The following are what I've tried
// Option #1
//
// controllers/events_controller.php
//
$events = $this-Event-find('list');
$this-set(compact('events'));
//
// views/events/create_event.ctp
//
echo $form-input('Event', array(
'div' = true,
'label' = true,
'type' = 'radio',
'options' = $events,
'onClick' = '$(EventName).value=aklsdjflkj'
));
problem with #1 is that I need to loop to access the correct
Event.name, Event.description, Event.information for the onClick
event. Right now with the find set to 'list', I don't have access to
that.
If I use the find query from step two here, I wouldn't know how to
access the correct Event.name, Event.description, Event.information
without being in a loop. Plus the label says array rather than the
appropriate Event.name.
// Option #2
//
// controllers/events_controller.php
//
$events = $this-Event-find('all',
array(
conditions = array(
Event.template = 1
)
)
);
$this-set(compact('events'));
//
// views/events/create_event.ctp
//
foreach($events as $event)
{
echo $form-input('Event.name', array(
'div' = true,
'label' = true,
'type' = 'radio',
'options' = $event[Event][name],
'value' = $event[Event][id]
));
}
// Problem with #2 is that when I put this in a loop, everything gets
messed up with the ids and the names of the inputs. Plus they are all
in there own div rather than being together.
Im pretty new at cake, but have been programming php for many years.
Any help is greatly appreciated.
Thanks so much,
Nick
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Re: show image from layout
On Sat, Jan 3, 2009 at 1:21 PM, Mike Bernat bern...@gmail.com wrote:
You can also use the image method from the html helper.
echo $html-image('my_logo.jpg');
Yes, I should have mentioned that. I was just trying to clarify the
paths issue. I suspect the OP is confused by the location of the
layouts dir, thinking of it as a normal directory under root from
whence HTML pages are served for a site.
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Re: API
Steven, if you could give some concrete examples perhaps we could improve the doc blocks :) -Mark On Jan 3, 11:36 am, Steven Wright rhythmicde...@gmail.com wrote: Not one really specific thing. In general the Cookbook does not seem to answer my questions so I find myself going to look at the API. Most methods seem to accept options but there is almost never a description of what those options are. I suppose given time I will figure this out, its just difficult for a new user to get past. Thanks for replying. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of Webweave Sent: Friday, January 02, 2009 11:43 PM To: CakePHP Subject: Re: API My experience is that you are much better off looking at the cookbook for that sort of information. In a few cases the API docs are very clear as to what the options are and how to use them, but often it requires a lot of digging to understand them. What part of the API are you having trouble with ? On Dec 29 2008, 6:18 pm, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: I am looking at the API trying to figure some stuff out. I keep getting stuck when they docs talk about options but dont actually give the options. How am I supposed to figure out what options are available for a particular class? Some of the time I can find the option in class or method definition but there is no real explanation of what it's for. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: $form-year not saving
In case someone stumbles upon this, an even more elegant solution
would be to ditch the controller code and use one line in your view:
? echo $form-input('movie_years', array('type' = 'date',
'dateFormat' = 'Y', 'minYear' = '1930', 'maxYear' = date('Y'))); ?
Much better!
--Mark
On Dec 22 2008, 8:19 pm, markfm m...@p-m-w.com wrote:
Never used $form-year before. Maybe somebody else can come up with a
fix for you.
However, here's how I would skin this cat...
In your controller, put in this code:
$yearsToDisplay = range('1930', date('Y'));
$movieYears = array_combine($yearsToDisplay, $yearsToDisplay);
$this-set('movieYears', $movieYears);
Then in your view:
?php echo $form-input('movie_year', array('type' = 'select',
'options' = $movieYears)); ?
Take care,
Mark
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Deleting from multiple models
I have a model Recipe that has a one to many relationship with a model
IngredientList
I want to delete a single recipe from the Recipe model which should
also delete all the records from the IngredientList model that have
the recipe's ID.
I would have thought that there would be a single line command like
there is for insert.
if ($this-Recipe-saveAll($this-data))
But this is only way I could find to do it:
function delete($id)
{
$this-IngredientList-deleteAll(IngredientList.recipe_id =
$id,
true);
$this-Recipe-del($id);
$this-flash('The recipe with id: '.$id.' has been deleted.', '/
recipes');
}
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Re: API
Ahh I dont want to be all whiny about it :) Essentially if there are options the options should be listed. On Jan 3, 1:50 pm, mark_story mark.st...@gmail.com wrote: Steven, if you could give some concrete examples perhaps we could improve the doc blocks :) -Mark On Jan 3, 11:36 am, Steven Wright rhythmicde...@gmail.com wrote: Not one really specific thing. In general the Cookbook does not seem to answer my questions so I find myself going to look at the API. Most methods seem to accept options but there is almost never a description of what those options are. I suppose given time I will figure this out, its just difficult for a new user to get past. Thanks for replying. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of Webweave Sent: Friday, January 02, 2009 11:43 PM To: CakePHP Subject: Re: API My experience is that you are much better off looking at the cookbook for that sort of information. In a few cases the API docs are very clear as to what the options are and how to use them, but often it requires a lot of digging to understand them. What part of the API are you having trouble with ? On Dec 29 2008, 6:18 pm, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: I am looking at the API trying to figure some stuff out. I keep getting stuck when they docs talk about options but dont actually give the options. How am I supposed to figure out what options are available for a particular class? Some of the time I can find the option in class or method definition but there is no real explanation of what it's for. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Cakephp on MAMP, Mac OSX
Thanks! This was the same issue i had. Once i copied over the .htaccess file to my cake app folder in the MAMP htdocs folder, all was good. Bob On Dec 26 2008, 2:45 pm, Chad Casselman ccassel...@gmail.com wrote: I am running the same setup. The only problem I ran into was, I extracted cake to my download folder and then copied and pasted it to where I wanted to code in it. Little did I know, the .htaccess files did not copy and paste even though I grabbed the whole directory. Extract cake to where you want to work on it. Chad On Fri, Dec 26, 2008 at 5:41 PM, Mickiii michael.la...@gmail.com wrote: Thanks for your reply, I have checked the httpd.conf located in the mamp folder /mamp/conf/apache/httpd.conf, and checked for the lines: LoadModule rewrite_module modules/mod_rewrite.so DocumentRoot /Users/user_name/Development/my_app and finally that this directory has AllowOverride set to All What else can be wrong? On Dec 26, 10:41 pm, Gwoo gwoo.cake...@gmail.com wrote: mod_rewrite needs to be enabled and AllowOverride All in the httpd.conf --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: API
On Sat, Jan 3, 2009 at 3:18 PM, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: Ahh I dont want to be all whiny about it :) Essentially if there are options the options should be listed. Agreed. The API docs are a misery to wade through at times. Especially for figuring out things like $options. It'd also be nice if there were a few comment blocks *within* some of the class methods. I do realise that that wouldn't be trivial to deal with, though. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
RE: API
Yeah. I am pretty sure the docs are auto created probably like Java Doc. But I cant really bitch too much about what is essentially a free framework. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of brian Sent: Saturday, January 03, 2009 3:43 PM To: cake-php@googlegroups.com Subject: Re: API On Sat, Jan 3, 2009 at 3:18 PM, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: Ahh I dont want to be all whiny about it :) Essentially if there are options the options should be listed. Agreed. The API docs are a misery to wade through at times. Especially for figuring out things like $options. It'd also be nice if there were a few comment blocks *within* some of the class methods. I do realise that that wouldn't be trivial to deal with, though. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
saveField with out modifying 'modified'
Is there any way to use saveField without having my 'modified' column
updated? Looking at the save() method, I thought, at first, that this
should work:
$this-User-saveField('last_login', date('Y-m-d H:i:s'),
array('fieldList' = array('last_login')));
However, it's disregarded. The block that sets the modified column begins with:
if ($this-hasField($updateCol) !in_array($updateCol, $fields)) {
... meaning, if I don't provide the modified column, it will be
updated. As I'm using saveField, there's no way to do so.
Shouldn't there be an option to override Cake's default behavior?
While I'd normally like 'modified' to be updated, it's superfluous in
this instance, as I don't consider this to be a modification to the
User model, per se.
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Re: API
Its not bitching if you are willing to point out where the problems are and perhaps offer some solutions to those problems. That's called contributing in my books. As for the API it is automatically generated via doxygen. But just because it is auto generated means it should be poor. So again, some places where you think the API is weak would help as you are probably not the only person who has experienced issues with it. -Mark On Jan 3, 3:47 pm, Steven Wright rhythmicde...@gmail.com wrote: Yeah. I am pretty sure the docs are auto created probably like Java Doc. But I cant really bitch too much about what is essentially a free framework. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of brian Sent: Saturday, January 03, 2009 3:43 PM To: cake-php@googlegroups.com Subject: Re: API On Sat, Jan 3, 2009 at 3:18 PM, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: Ahh I dont want to be all whiny about it :) Essentially if there are options the options should be listed. Agreed. The API docs are a misery to wade through at times. Especially for figuring out things like $options. It'd also be nice if there were a few comment blocks *within* some of the class methods. I do realise that that wouldn't be trivial to deal with, though. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Edit multiple models in the same page: possible?
Hi all,
I wonder if it's possible to have two different forms for two
different (but related) models in the same page.
Imagine a page where the logged in user can edit his profile (model
Profile) as well as to add a photo into his gallery (model Image).
?php echo $form-create('Profile'); ?
...
?php echo $form-create('Image'); ?
...
The problem here is that after submiting one of the forms, the user is
redirected to /model/edit/, and not back to the current page.
I cannot imagine how to fit both forms in the same page without
loosing the validationErrors info.
Any clue?
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Re: API
I usually open the files myself instead of using the API online. I usually have to go through the methods themselves to find out the options. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Deleting from multiple models
yes it is. check http://book.cakephp.org/view/80/hasOne for dependent
key
dependent: When the dependent key is set to true, and the model’s
delete() method is called with the cascade parameter set to true,
associated model records are also deleted. In this case we set it true
so that deleting a User will also delete her associated Profile.
On Jan 3, 9:56 pm, rhythmicde...@gmail.com rhythmicde...@gmail.com
wrote:
I have a model Recipe that has a one to many relationship with a model
IngredientList
I want to delete a single recipe from the Recipe model which should
also delete all the records from the IngredientList model that have
the recipe's ID.
I would have thought that there would be a single line command like
there is for insert.
if ($this-Recipe-saveAll($this-data))
But this is only way I could find to do it:
function delete($id)
{
$this-IngredientList-deleteAll(IngredientList.recipe_id =
$id,
true);
$this-Recipe-del($id);
$this-flash('The recipe with id: '.$id.' has been deleted.',
'/
recipes');
}
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Re: in the link have a parameter 'CAKEPHP'
Thank You. [?] 2009/1/3 Mike Bernat bern...@gmail.com That's cakePHP's session ID. It's appearing because the server believes you don't have cookies enabled. You can't turn it off as it is a php core behavior. On Sat, Jan 3, 2009 at 2:44 AM, Rimoe meiyo...@gmail.com wrote: hi, everyone, in my link have a parameter CAKEPHP. for example http://www.mysite.com/mails/add?CAKEPHP=3831a0f2e1fb52c9ae15d163f57c4695 how to remove it, recently I have read a topic about to remove the parameter, but I have forget to it. If you know, please tell me. Thank you. rimoe --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~--- inline: 000.gif
Re: API
On Sat, Jan 3, 2009 at 6:58 PM, mark_story mark.st...@gmail.com wrote: Its not bitching if you are willing to point out where the problems are and perhaps offer some solutions to those problems. That's called contributing in my books. As for the API it is automatically generated via doxygen. But just because it is auto generated means it should be poor. So again, some places where you think the API is weak would help as you are probably not the only person who has experienced issues with it. I think there's a source documentation problem throughout. I don't think it's a matter of any one file. Perhaps that's a decision the devs made to keep the files slim, although I'm unaware of any significant impact lots of comments has on performance. I doubt it's relevant. There have been times when I've been studying some method and, once the clue fairy appears, have thought that a particular inline comment or explanation at the top of the method would help considerably. But I doubt that Trac submissions just for source comment additions would be welcomed if too many people did so. Perhaps I'm wrong; I don't kow how the devs feel about that. As with any code source, though, a bit (well, more than a bit) of effort does, eventually, lead to a more comprehensive understanding of the framework. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: API
On Sat, Jan 3, 2009 at 7:36 PM, gearvOsh mileswjohn...@gmail.com wrote: I usually open the files myself instead of using the API online. I usually have to go through the methods themselves to find out the options. You can get to the source of any method by following the link, Definition at line [#] in the method description. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Edit multiple models in the same page: possible?
Use Ajax?
On Jan 3, 4:07 pm, Jaime ja...@iteisa.com wrote:
Hi all,
I wonder if it's possible to have two different forms for two
different (but related) models in the same page.
Imagine a page where the logged in user can edit his profile (model
Profile) as well as to add a photo into his gallery (model Image).
?php echo $form-create('Profile'); ?
...
?php echo $form-create('Image'); ?
...
The problem here is that after submiting one of the forms, the user is
redirected to /model/edit/, and not back to the current page.
I cannot imagine how to fit both forms in the same page without
loosing the validationErrors info.
Any clue?
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storing user_ids
I want to have the logged in user information stored with each record. Is there a way to set this on the controller or do I have to rely on hidden fields in the views? Thanks! --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
RE: API
Well here is one example. What are the options for this particular method? Return JavaScript text for an observer... Parameters: string $klass Name of JavaScript class string $name array $options Ajax options Returns: string Formatted JavaScript Definition at line 864 of file ajax.php. References remoteFunction(), and Helper::value(). Referenced by observeField(), and observeForm(). On php.net all the functions are very well documented for instance it's immediately clear what this function does and what it's parameters do and what they are. http://us.php.net/manual/en/function.mysql-affected-rows.php -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of mark_story Sent: Saturday, January 03, 2009 6:59 PM To: CakePHP Subject: Re: API Its not bitching if you are willing to point out where the problems are and perhaps offer some solutions to those problems. That's called contributing in my books. As for the API it is automatically generated via doxygen. But just because it is auto generated means it should be poor. So again, some places where you think the API is weak would help as you are probably not the only person who has experienced issues with it. -Mark On Jan 3, 3:47 pm, Steven Wright rhythmicde...@gmail.com wrote: Yeah. I am pretty sure the docs are auto created probably like Java Doc. But I cant really bitch too much about what is essentially a free framework. -Original Message- From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf Of brian Sent: Saturday, January 03, 2009 3:43 PM To: cake-php@googlegroups.com Subject: Re: API On Sat, Jan 3, 2009 at 3:18 PM, rhythmicde...@gmail.com rhythmicde...@gmail.com wrote: Ahh I dont want to be all whiny about it :) Essentially if there are options the options should be listed. Agreed. The API docs are a misery to wade through at times. Especially for figuring out things like $options. It'd also be nice if there were a few comment blocks *within* some of the class methods. I do realise that that wouldn't be trivial to deal with, though. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
RE: Deleting from multiple models
Thank you.
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of Alexandru
Sent: Saturday, January 03, 2009 7:42 PM
To: CakePHP
Subject: Re: Deleting from multiple models
yes it is. check http://book.cakephp.org/view/80/hasOne for dependent key
dependent: When the dependent key is set to true, and the model’s
delete() method is called with the cascade parameter set to true, associated
model records are also deleted. In this case we set it true so that deleting
a User will also delete her associated Profile.
On Jan 3, 9:56 pm, rhythmicde...@gmail.com rhythmicde...@gmail.com
wrote:
I have a model Recipe that has a one to many relationship with a model
IngredientList
I want to delete a single recipe from the Recipe model which should
also delete all the records from the IngredientList model that have
the recipe's ID.
I would have thought that there would be a single line command like
there is for insert.
if ($this-Recipe-saveAll($this-data))
But this is only way I could find to do it:
function delete($id)
{
$this-IngredientList-deleteAll(IngredientList.recipe_id = $id,
true);
$this-Recipe-del($id);
$this-flash('The recipe with id: '.$id.' has been
deleted.', '/ recipes');
}
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Re: storing user_ids
On Sat, Jan 3, 2009 at 8:18 PM, ymadh ad...@dogtrainersearch.com wrote: I want to have the logged in user information stored with each record. Is there a way to set this on the controller or do I have to rely on hidden fields in the views? Each record of what? Is there a user_id column in the table? Is the User model associated with it? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: storing user_ids
Each record of the associated tables. So if a client is created under a logged in user, I want it to autopopulate the db with user_id which is in the session. That way each user only sees their clients..make sense? My client table would have client info, and of course a user_id field. I didn't link the models ... Would that create extra overhead on every page that needs that user_id field?? Thanks! --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: storing user_ids
On Sat, Jan 3, 2009 at 9:02 PM, ymadh ad...@dogtrainersearch.com wrote:
Each record of the associated tables. So if a client is created under
a logged in user, I want it to autopopulate the db with user_id which
is in the session. That way each user only sees their clients..make
sense? My client table would have client info, and of course a user_id
field.
I didn't link the models ... Would that create extra overhead on every
page that needs that user_id field??
If you have a user_id column in a model's table, then you should
associated it with the User model. That way, you'll have to do very
little to store it.
In your view, you can add a hidden field:
echo $form-hidden('User.id', array('value' = $session-read('User.id')));
That will give your controller a $data array something along these lines:
array(
'YourModel' = array(
'some field' = 'some value',
...
),
'User' = array(
'id' = 42
)
)
That presupposes that only logged-in users will use that form. And,
again, that this model is properly associated with User.
As for overhead, you can use the ContainableBehavior to adjust what
Cake queries the DB for.
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saving to habtm models from checkbox groups.
I posted about this earlier but I'm still having trouble. I think I
have my models set up properly now:
class User extends AppModel
{
var $name = 'User';
var $hasAndBelongsToMany = array ( 'Ethnicity' = array(
'className'= 'Ethnicity',
'joinTable' =
'ethnicities_users',
'foreignKey'= 'user_id',
'associationForeignKey' =
'ethnicity_id'
)
);
}
class Ethnicity extends AppModel
{
var $name = 'Ethnicity';
var $hasAndBelongsToMany = array ( 'User' = array(
'className'= 'User',
'joinTable' =
'ethnicities_users',
'foreignKey'= 'ethnicity_id',
'associationForeignKey' = 'user_id'
)
);
}
I've been trying different stuff in the controller, this is what I
have now:
function add() {
$this-User-create();
if(!empty($this-data)) {
if($this-User-save($this-data)) {
$this-data['User']['Ethnicity']['user_id'] =
$this-User-
getLastInsertId();
$this-User-Ethnicity-save($this-data);
$this-Session-setFlash(User Saved!);
}
}
}
here is what I have in the view, which I don't think is correct:
? echo $form-create('User', array('action'='add')); ?
? echo $form-input('User.Ethnicity', array( 'type' = 'select',
'multiple' = 'checkbox' ),$ethnicities); ?
?php echo $form-end('Submit'); ?br/
form the research I've done, I feel like the input fieldname should be
'User.Ethnicity. .id' or something like that. However, anytime I add
the 'id' into the fieldname, the checkboxes don't get created in the
html anymore. here's how $ethnicities is populated in the controller,
if that matters:
$this-set('ethnicities',$this-Ethnicity-find('list', array('fields'
= array('id','ethnicityName';
the result right now is this:
[data] = Array
(
[User] = Array
(
[Ethnicity] = Array
(
[0] = 1
[1] = 3
[user_id] = 34
)
[age] = 50
[about_me] = sadfadfs
)
)
but my understanding is I need something like this:
[data] = Array
(
[User] = Array
(
[Ethnicity] = Array
(
[ethnicity_id] = 1
[user_id] = 34,
[ethnicity_id] = 2
[user_id] = 34
)
[age] = 50
[about_me] = sadfadfs
)
)
here is a link to my other thread for reference:
http://groups.google.com/group/cake-php/browse_thread/thread/2c74ee5184fe8d90/62788d3299b36a31?lnk=gstq=checkbox+groups#62788d3299b36a31
any help apprciated. thanks!
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-~--~~~~--~~--~--~---
Re: Adding table row with AJAX
Hi Brian,
I was wondering if you had an answer to this issue I am having. I dont
understand the error, or why the data is not appending.
Thanks.
On Jan 3, 9:43 am, Steven Wright rhythmicde...@gmail.com wrote:
Thanks for the reply. This is mostly working but there is something I am
still missing.
I tried two different things.
1) If I call the 'edit' action I get a jQuery error ( 'div is null' ' elem =
jQuery.makeArray( div.childNodes ); ')
I am not sure what div its looking for. The second problem with that is that
although I can see thetablerowreturned at the start of the response
string the rest of the content for the view is also returned and I don't
know why.
2) If I call the 'add_ingredient_row' action I only get the content for the
view in the response but no newrow.
So two questions:
1) Why is CakePHP returning all the content for the view?
2) Why is jQuery giving me that error?
Thanks again. If I can just get over this hurdle I will be off and running.
VIEW
h2Testing AJAX with CakePHP and Prototype and Scriptaculous/h2
tableid=the_table
tr id=row1
tdRow1/td
/tr
/table
div id=mydiv
This is a div
/div
div id=the_buttonClick/div
CONTROLLER
class AjaxtestsController extends AppController
{
var $components = array( 'RequestHandler' );
function index()
{
$this-pageTitle = 'Index Page';
}
function view()
{
$this-pageTitle = 'View Page';
}
function edit()
{
$this-pageTitle = 'Edit Page';
if($this-RequestHandler-isAjax())
{
$this-layout = 'ajax';
echo 'trtdThis is arow/td/tr';
}
}
function add_ingredient_row()
{
echo 'trtdThis is arow/td/tr';
}
}
JAVASCRIPT
$(document).ready(function()
{
$('#the_button').click(function()
{
$.ajax({
url: 'add_ingredient_row/',
type: 'GET',
data: 'data string',
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#the_table').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of brian
Sent: Friday, January 02, 2009 8:26 PM
To: cake-php@googlegroups.com
Subject: Re:Addingtablerowwith AJAX
On Fri, Jan 2, 2009 at 7:45 PM, Steven Wright rhythmicde...@gmail.com
wrote:
Hi Adam,
Thanks for writing back. How would you get the data back for therow
from CakePHP?
Myrowcontains four columns with the following inputs:
amount [text]
measurement_type [select]
description [text]
ingredient [select]
This is normally rendered from an element as atablerow. I would like
the output of that element appended to mytable. So on the button
click I call the action add_ingredient_row and it should return the
content of the element. How do I get that content appended to mytable.
I can do this without CakePHP. But for me this exercise is to use the
framework.
Being a jQuery user, I never bother with Cake's ajax() stuff. But jQuery is
dead simple (mostly) to use, so I don't mind having to write out a few lines
of JS. To do what you describe, I'd do something like
this:
$(document).ready()
{
$('#the_button').click(function()
{
$.ajax({
url: the_href,
type: 'GET',
data: the_data,
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
},
success: function(msg)
{
/* assuming your controller is returning
'tr.../tr'
*/
$('#the_table').append(msg);
}
});
});
});
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
CakePHP group.
To post to this group, send email to cake-php@googlegroups.com
To
RE: Adding table row with AJAX
In addition, just to prove I am not a complete dope. I get how this works,
just not with CakePHP.
AJAX_TEST.PHP
?php
$row_count = ($_POST['row_count'] + 1);
echo tr id=\row$row_count\tdContent for row $row_count/td/tr;
?
AJAX_TEST.HTML
html
head
title/title
script type=text/javascript src=javascript/jquery-1.2.6.js/script
script type=text/javascript
$(document).ready(function()
{
$('#button').click(function()
{
$.ajax({
url: 'ajax_test.php',
type: 'POST',
data: row_count= +
document.getElementById('mytable').rows.length,
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#mytable').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
/script
head
body
table id=mytable
tr id=row1
tdContent for row 1/td
/tr
/table
div id=buttonClick/div
/body
/html
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of rhythmicde...@gmail.com
Sent: Saturday, January 03, 2009 9:45 PM
To: CakePHP
Subject: Re: Adding table row with AJAX
Hi Brian,
I was wondering if you had an answer to this issue I am having. I dont
understand the error, or why the data is not appending.
Thanks.
On Jan 3, 9:43 am, Steven Wright rhythmicde...@gmail.com wrote:
Thanks for the reply. This is mostly working but there is something I
am still missing.
I tried two different things.
1) If I call the 'edit' action I get a jQuery error ( 'div is null' '
elem = jQuery.makeArray( div.childNodes ); ') I am not sure what div
its looking for. The second problem with that is that although I can
see thetablerowreturned at the start of the response string the rest
of the content for the view is also returned and I don't know why.
2) If I call the 'add_ingredient_row' action I only get the content
for the view in the response but no newrow.
So two questions:
1) Why is CakePHP returning all the content for the view?
2) Why is jQuery giving me that error?
Thanks again. If I can just get over this hurdle I will be off and
running.
VIEW
h2Testing AJAX with CakePHP and Prototype and Scriptaculous/h2
tableid=the_table
tr id=row1
tdRow1/td
/tr
/table
div id=mydiv
This is a div
/div
div id=the_buttonClick/div
CONTROLLER
class AjaxtestsController extends AppController {
var $components = array( 'RequestHandler' );
function index()
{
$this-pageTitle = 'Index Page';
}
function view()
{
$this-pageTitle = 'View Page';
}
function edit()
{
$this-pageTitle = 'Edit Page';
if($this-RequestHandler-isAjax())
{
$this-layout = 'ajax';
echo 'trtdThis is arow/td/tr';
}
}
function add_ingredient_row()
{
echo 'trtdThis is arow/td/tr';
}
}
JAVASCRIPT
$(document).ready(function()
{
$('#the_button').click(function()
{
$.ajax({
url: 'add_ingredient_row/',
type: 'GET',
data: 'data string',
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#the_table').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On
Behalf
Of brian
Sent: Friday, January 02, 2009 8:26 PM
To: cake-php@googlegroups.com
Subject: Re:Addingtablerowwith AJAX
On Fri, Jan 2, 2009 at 7:45 PM, Steven Wright
rhythmicde...@gmail.com
wrote:
Hi Adam,
Thanks for writing back. How would you get the data back for therow
from CakePHP?
Myrowcontains four columns with the following inputs:
amount [text]
measurement_type [select]
description
HABTM retrieving help
Heres my tables and relations:
Team
Team-belongsTo: Country, State
Team-hasAndBelongsToMany: User (the players on a team, table =
teams_players)
Team:
CREATE TABLE IF NOT EXISTS `teams` (
`id` int(10) NOT NULL auto_increment,
`status` enum('approved','pending') NOT NULL default 'pending',
`name` varchar(50) NOT NULL,
`tag` varchar(10) NOT NULL,
`urlName` varchar(30) NOT NULL,
`website` varchar(50) NOT NULL,
`irc` varchar(20) NOT NULL,
`about` varchar(255) NOT NULL,
`servers` varchar(100) NOT NULL,
`logo` varchar(100) NOT NULL,
`createDate` int(10) unsigned NOT NULL default '0',
`state_id` int(10) unsigned NOT NULL default '0',
`country_id` int(10) unsigned NOT NULL default '0',
`created` datetime default NULL,
`modified` datetime default NULL,
PRIMARY KEY (`id`),
KEY `state_id` (`state_id`),
KEY `country_id` (`country_id`)
)
TeamsPlayer
CREATE TABLE IF NOT EXISTS `teams_players` (
`id` int(11) NOT NULL auto_increment,
`team_id` int(10) unsigned NOT NULL,
`user_id` int(10) unsigned NOT NULL,
`role` enum('leader','captain','member') NOT NULL default 'member',
`status` enum('approved','pending') NOT NULL default 'pending',
`joinDate` int(10) NOT NULL,
PRIMARY KEY (`id`),
KEY `team_id` (`team_id`),
KEY `user_id` (`user_id`)
)
Ok if I do a basic Team-find() all the data is returned correctly.
Now the problem im running into is how do I grab all teams that I am
part of, and paginate it at the same time? Since the user_id values is
in TeamsPlayer, im getting confused on this. Any help would be
appreciated.
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
CakePHP group.
To post to this group, send email to cake-php@googlegroups.com
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RE: Adding table row with AJAX
One way that I found to make this work is to use a dummy.ctp file which is
empty. Then I set the debug level to 0 to suppress all output from CakePHP.
I then get just output I need. The use of a dummy file seems like a hack
workaround to me though but at this point I don’t have a better option.
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of rhythmicde...@gmail.com
Sent: Saturday, January 03, 2009 9:45 PM
To: CakePHP
Subject: Re: Adding table row with AJAX
Hi Brian,
I was wondering if you had an answer to this issue I am having. I dont
understand the error, or why the data is not appending.
Thanks.
On Jan 3, 9:43 am, Steven Wright rhythmicde...@gmail.com wrote:
Thanks for the reply. This is mostly working but there is something I
am still missing.
I tried two different things.
1) If I call the 'edit' action I get a jQuery error ( 'div is null' '
elem = jQuery.makeArray( div.childNodes ); ') I am not sure what div
its looking for. The second problem with that is that although I can
see thetablerowreturned at the start of the response string the rest
of the content for the view is also returned and I don't know why.
2) If I call the 'add_ingredient_row' action I only get the content
for the view in the response but no newrow.
So two questions:
1) Why is CakePHP returning all the content for the view?
2) Why is jQuery giving me that error?
Thanks again. If I can just get over this hurdle I will be off and
running.
VIEW
h2Testing AJAX with CakePHP and Prototype and Scriptaculous/h2
tableid=the_table
tr id=row1
tdRow1/td
/tr
/table
div id=mydiv
This is a div
/div
div id=the_buttonClick/div
CONTROLLER
class AjaxtestsController extends AppController {
var $components = array( 'RequestHandler' );
function index()
{
$this-pageTitle = 'Index Page';
}
function view()
{
$this-pageTitle = 'View Page';
}
function edit()
{
$this-pageTitle = 'Edit Page';
if($this-RequestHandler-isAjax())
{
$this-layout = 'ajax';
echo 'trtdThis is arow/td/tr';
}
}
function add_ingredient_row()
{
echo 'trtdThis is arow/td/tr';
}
}
JAVASCRIPT
$(document).ready(function()
{
$('#the_button').click(function()
{
$.ajax({
url: 'add_ingredient_row/',
type: 'GET',
data: 'data string',
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#the_table').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On
Behalf
Of brian
Sent: Friday, January 02, 2009 8:26 PM
To: cake-php@googlegroups.com
Subject: Re:Addingtablerowwith AJAX
On Fri, Jan 2, 2009 at 7:45 PM, Steven Wright
rhythmicde...@gmail.com
wrote:
Hi Adam,
Thanks for writing back. How would you get the data back for therow
from CakePHP?
Myrowcontains four columns with the following inputs:
amount [text]
measurement_type [select]
description [text]
ingredient [select]
This is normally rendered from an element as atablerow. I would like
the output of that element appended to mytable. So on the button
click I call the action add_ingredient_row and it should return the
content of the element. How do I get that content appended to mytable.
I can do this without CakePHP. But for me this exercise is to use
the framework.
Being a jQuery user, I never bother with Cake's ajax() stuff. But
jQuery is dead simple (mostly) to use, so I don't mind having to write
out a few lines of JS. To do what you describe, I'd do something like
this:
$(document).ready()
{
$('#the_button').click(function()
{
$.ajax({
url: the_href,
type: 'GET',
data: the_data,
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
},
RE: Adding table row with AJAX
Actually this works in place of the dummy file
$this-autoRender = false;
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On Behalf
Of rhythmicde...@gmail.com
Sent: Saturday, January 03, 2009 9:45 PM
To: CakePHP
Subject: Re: Adding table row with AJAX
Hi Brian,
I was wondering if you had an answer to this issue I am having. I dont
understand the error, or why the data is not appending.
Thanks.
On Jan 3, 9:43 am, Steven Wright rhythmicde...@gmail.com wrote:
Thanks for the reply. This is mostly working but there is something I
am still missing.
I tried two different things.
1) If I call the 'edit' action I get a jQuery error ( 'div is null' '
elem = jQuery.makeArray( div.childNodes ); ') I am not sure what div
its looking for. The second problem with that is that although I can
see thetablerowreturned at the start of the response string the rest
of the content for the view is also returned and I don't know why.
2) If I call the 'add_ingredient_row' action I only get the content
for the view in the response but no newrow.
So two questions:
1) Why is CakePHP returning all the content for the view?
2) Why is jQuery giving me that error?
Thanks again. If I can just get over this hurdle I will be off and
running.
VIEW
h2Testing AJAX with CakePHP and Prototype and Scriptaculous/h2
tableid=the_table
tr id=row1
tdRow1/td
/tr
/table
div id=mydiv
This is a div
/div
div id=the_buttonClick/div
CONTROLLER
class AjaxtestsController extends AppController {
var $components = array( 'RequestHandler' );
function index()
{
$this-pageTitle = 'Index Page';
}
function view()
{
$this-pageTitle = 'View Page';
}
function edit()
{
$this-pageTitle = 'Edit Page';
if($this-RequestHandler-isAjax())
{
$this-layout = 'ajax';
echo 'trtdThis is arow/td/tr';
}
}
function add_ingredient_row()
{
echo 'trtdThis is arow/td/tr';
}
}
JAVASCRIPT
$(document).ready(function()
{
$('#the_button').click(function()
{
$.ajax({
url: 'add_ingredient_row/',
type: 'GET',
data: 'data string',
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
alert('There has been an error');
},
success: function(msg)
{
//alert(msg);
$('#the_table').append(msg);
}
}); // ends ajax
}); // ends click
}); // ends document.ready
-Original Message-
From: cake-php@googlegroups.com [mailto:cake-...@googlegroups.com] On
Behalf
Of brian
Sent: Friday, January 02, 2009 8:26 PM
To: cake-php@googlegroups.com
Subject: Re:Addingtablerowwith AJAX
On Fri, Jan 2, 2009 at 7:45 PM, Steven Wright
rhythmicde...@gmail.com
wrote:
Hi Adam,
Thanks for writing back. How would you get the data back for therow
from CakePHP?
Myrowcontains four columns with the following inputs:
amount [text]
measurement_type [select]
description [text]
ingredient [select]
This is normally rendered from an element as atablerow. I would like
the output of that element appended to mytable. So on the button
click I call the action add_ingredient_row and it should return the
content of the element. How do I get that content appended to mytable.
I can do this without CakePHP. But for me this exercise is to use
the framework.
Being a jQuery user, I never bother with Cake's ajax() stuff. But
jQuery is dead simple (mostly) to use, so I don't mind having to write
out a few lines of JS. To do what you describe, I'd do something like
this:
$(document).ready()
{
$('#the_button').click(function()
{
$.ajax({
url: the_href,
type: 'GET',
data: the_data,
dataType: 'html',
timeout: 2000,
error: function(XMLHttpRequest, textStatus,
errorThrown)
{
// ...
},
success: function(msg)
{
/* assuming your controller is
returning 'tr.../tr'
*/
Re: in the class of [MailsController] use the [$this-User-create()]
var $uses = array('Mail','User');
2009/1/4 Rimoe meiyo...@gmail.com
hi,everyone.
in the class of [MailsController]
we can use the [$this-Mail-create();]
but I want to use the [$this-User-create();]
what should I do?
It's possible?
rimoe
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Auth Broken.
Hey Guys,
I am trying to use the Auth component, but I'm pretty sure there's
some bug preventing me from doing so. Here is what my
app_controller.php looks like:
?php
class AppController extends Controller {
var $components = array('Auth');
function beforeFilter(){
$this-Auth-allow('*');
$this-Auth-logoutRedirect = '/';
$this-Auth-loginRedirect = '/';
$this-Auth-authorize = 'controller';
}
function isAuthorized(){
die('isAuthorized Called!');
}
}
?
Whenever I browse to a view, I get redirected to the login page. I'd
expect the behavior to allow access to the view or at least the string
'isAuthorized Called!' to be printed, but it seems the isAuthorized
method is never called. The following works, though:
$this-Auth-allowedActions = array('*');
I am using PHP4 on XAMPP/Windows.
Can someone try to reproduce and/or tell me what I'm doing wrong?
Thanks,
Matt
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Return compressed XML
Hi,
I want to return unique compressed XML for each user request.
Something like this in the controller:
public function download()
{
$filename = 'some_temporary_file_name.gz';
$this-view = 'Media';
$params = array(
'id' = $filename,
'name' = 'some_temporary_file_name',
'download' = true,
'extension' = 'gz',
'path' = ''
);
// open file for writing with maximum compression
$zp = gzopen($filename, w9);
// write string to file
gzwrite($zp, $unique_xml_here);
// close file
gzclose($zp);
}
How can I create a unique file for each request and have the file be
deleted after the user downloaded it? I'm mostly concerned with the
deleting part unless there is a convenient CakePHP for the unique file
part.
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'php' is not recognized as an internal or external command, operable program or batch file'
Hi All, If I run *php -v* command to know php version on command line, it shows error like 'php' is not recognized as an internal or external command, operable program or batch file' what does this error means? I'm using xampp server on windows-xp(service pack-3); please help me -- --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Auth Broken.
P.S. I'm using release 1.2.0.7962.
On Sun, Jan 4, 2009 at 1:14 AM, Matt Williamson dawsdes...@gmail.comwrote:
Hey Guys,
I am trying to use the Auth component, but I'm pretty sure there's
some bug preventing me from doing so. Here is what my
app_controller.php looks like:
?php
class AppController extends Controller {
var $components = array('Auth');
function beforeFilter(){
$this-Auth-allow('*');
$this-Auth-logoutRedirect = '/';
$this-Auth-loginRedirect = '/';
$this-Auth-authorize = 'controller';
}
function isAuthorized(){
die('isAuthorized Called!');
}
}
?
Whenever I browse to a view, I get redirected to the login page. I'd
expect the behavior to allow access to the view or at least the string
'isAuthorized Called!' to be printed, but it seems the isAuthorized
method is never called. The following works, though:
$this-Auth-allowedActions = array('*');
I am using PHP4 on XAMPP/Windows.
Can someone try to reproduce and/or tell me what I'm doing wrong?
Thanks,
Matt
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CakePHP group.
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Re: saveField with out modifying 'modified'
On Sun, Jan 4, 2009 at 1:07 AM, Sergei yatse...@gmail.com wrote: Yes, this sucks. I had to read OLD modified date, and save it using save(data[] array with modified). I did this in cake1.1. I also just noticed that my last_login field is set on add(). Argghhh! --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: Adding table row with AJAX
On Sat, Jan 3, 2009 at 10:05 PM, Steven Wright rhythmicde...@gmail.com wrote: In addition, just to prove I am not a complete dope. I get how this works, just not with CakePHP. AJAX_TEST.PHP ?php $row_count = ($_POST['row_count'] + 1); echo tr id=\row$row_count\tdContent for row $row_count/td/tr; ? With Cake, you'd find the row_count value in $this-params['url']['row_count'] Debugger::dump($this-params); or debug($this-params); or pr() or anything else to get a look at that array. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
Re: 'php' is not recognized as an internal or external command, operable program or batch file'
You should set \xampp\php in the path environment variable. On Jan 4, 2:49 pm, amar goud amardeen...@gmail.com wrote: Hi All, If I run *php -v* command to know php version on command line, it shows error like 'php' is not recognized as an internal or external command, operable program or batch file' what does this error means? I'm using xampp server on windows-xp(service pack-3); please help me -- --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups CakePHP group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~--~~~~--~~--~--~---
