routes, controller name and model name

[Filip Camerman]

I'm making a site for an artist and when showing image galleries of
art collections I want url's like

www.site.com/collections/name-of-collection

Now I also have a database table for collections, and hence a
collection_controller and a collection model.

First I had my routes like this
- Router::connect('/collections/:action/*', array('controller' =
'collections'));
- Router::connect('/collections/*', array('controller' =
'collections', 'action' = 'index'));

I thought that the first route would only catch url's with existing
actions, but instead it also caught my /collections/name-of-
collection url's. So I changed it to:

- Router::connect('/coll/:action/*', array('controller' =
'collections'));

since the url's don't matter for my admin actions. But then I have to
change my forms because

?= $form-create('Collection');?

will post the form to /collections/...

Now I can solve this in several ways (change the model name; make
separate routes for all my admin actions, ...) but I was wondering if
there's an elegant way to deal with custom url's that start with a
controller/model name?
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Model Find With Conditions On Associated Models Not Working

[erturne]

I've seen similar questions raised but haven't found a solution that
works. Perhaps it's because I'm new to CakePHP and don't fully
understand unbindModel(), bindModel(), container behaviors, etc.
Anyway, I'll start simple and take suggestions.

I have the following models in which a Menu hasAndBelongsToMany Entree
and hasMany MenuPosting, and a MenuPosting belongsTo Menu:

class Menu extends AppModel {
var $name = 'Menu';
var $hasAndBelongsToMany = 'Entree';
var $hasMany = array(
'MenuPosting' = array( 'dependent' = true )
);
 }

class MenuPosting extends AppModel {
var $name = 'MenuPosting';
var $belongsTo = 'Menu';
}

class Entree extends AppModel {
var $name = 'Entree';
 }

In my MenusController I have a display_current() action to find all
menus that are currently posted:

class MenusController extends AppController {
var $name = 'Menus';
function display_current() {

// Limit the query results to the menu that is visible (i.e. is
// currently posted)
$conditions = array(
'MenuPosting.post_start =' = date('Y-m-d H:i:s', 
strtotime
(now)),
'MenuPosting.post_end ' = date('Y-m-d H:i:s', 
strtotime(now)),
);

// Find the menus that meets the conditions.
$this-set('menu', $this-Menu-find('all', array('conditions' 
=
$conditions)));
}
 }

The error I get from this is:
Warning (512): SQL Error: 1054: Unknown column
'MenuPosting.post_start' in 'where clause'

$sql=   SELECT `Menu`.`id`, `Menu`.`old_id`, `Menu`.`created`,
`Menu`.`modified`, `Menu`.`title` FROM `menus` AS `Menu`   WHERE
`MenuPosting`.`post_start` = '2008-12-06 22:02:57' AND
`MenuPosting`.`post_end`  '2008-12-06 22:02:57'   

If I remove the conditions from the find I am able to retrieve all
Menu and associated Entree, so I believe I have set up my database
tables correctly.

Any suggestions?

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Re: simple write to database

[Terrence]

Hi, try using

$this-User-save($this-data['User']).

The save function expects an array so that it can tell what db fields
it can map the values to. Passing it $this-data['User']['about_me']
will only pass the value of about me, and cake doesn't know what to
map it to.

Hope that helps.


mike wrote:
 I have a simple example below.  I just want a form that saves stuff to
 the database.  Its never getting to log test2 I guess the data is
 not in the right format but how am I supposed to do it?  I've already
 spent a considerable amount of time debugging this and other problems,
 please help!

 thanks.

 CREATE TABLE users
 (
 id int NOT NULL AUTO_INCREMENT,
 lastName varchar(255),
 firstName varchar(255),
 picture_id int,
 about_me varchar(255),
 PRIMARY KEY (id)
 );

 here's the model:
 class User extends AppModel
 {
 // Its always good practice to include this variable.
 var $name = 'User';

 // This is used for validation, see Chapter Data Validation.
 var $validate = array();
 }

 controller:
 class NewuserController extends AppController
 {
   var $name = 'Newuser';
   var $uses = Array('User');

   function index()
   {

   }

   function add() {
   $this-log('in newuser_controller add()');
   if(!empty($this-data)) {
   $this-log('test1');
   //If the form data can be validated and saved...
   if($this-User-save($this-data['User']['about_me'])) {
   $this-log('test2');
   //Set a session flash message and redirect.
   $this-Session-setFlash(User Saved!);
   $this-redirect('add');
   }
   }
   }
 }


 View:

 ...
 form action=?php echo $html-url('/newuser/add'); ?
 method=post
 ? //echo $form-create('User'); ?


 FONT CLASS=text12nbsp;BRDescribe yourself/FONT

 ?php echo $html-textarea('User/about_me')?

   ?php echo $html-submit('Add'); ?br/

 ? //echo $form-end('Add User'); ?
 /form

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View Caching

[NickABusey]

I've been through the manual and APIs more times than I can count, and
still have had no luck whatsoever getting view caching to work with
1.2, either on my local development box, or my server. I have an app/
tmp/cache directory that is 0777, I have
  Configure::write('Cache.check', true);
  Cache::config('default', array('engine' = 'File'));
in core.php, and I have
  var $helpers=array('Cache');
  var $cacheAction = 1 hour;
in my controller. However, no file is created, and no error is thrown.
What am I missing?

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Question about security level and session when browser is closed

[Milmar]

If the security level is set to medium and the browser is closed, the
user session is not deleted. So when opening the browser again after a
few minutes, the user still has access to the authorized pages without
logging in again.

My question is, if the browser is kept closed long enough, will the
session eventually expire if the security level is medium? How long
will it take?

Would it be possible to adjust the duration when the session expires
while keeping the same security level (medium)?
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Re: Ordering search results by best match

[Adam Royle]

Have a look at sphinx, and the SphinxBehaviour on the bakery.

Cheers,
Adam

On Dec 7, 7:22 am, Clay [EMAIL PROTECTED] wrote:
 My app has the following:

 Genus hasMany Species
 Species hasMany CommonName

 and the corresponding belongsTo relationships as well. Each of these
 models has a name field, and Genus and Species have several other
 fields.

 Say for example I have a Species Acer rubrum. This Species belongsTo
 Genus Acer and has a CommonName Red Maple.

 I want my users to be able to search the database for Red Maple and
 get as a result Species Acer rubrum (with the CommonName Red Maple
 highlighted to show that's what it matched. I can do this already.
 Yay.

 However if I have another Species with CommonName Bored Maple, I want
 to display this result too, but lower in the list than Red Maple.
 Since Bored comes before Red alphabetically I can't just order by
 CommonName.name to get the result I want.

 So what I know I could do is a series of finds:
 1) Find exact match: CommonName.name = $searchStr
 2) Find starting match: CommonName.name LIKE = $searchStr . '%'
 3) Find internal match: CommonName.name LIKE = '%' . $searchStr . '%'

 And then merge them into my results array.

 However, this seems very inefficient. Is there a way I can use cake's
 find() function to do this all in one go while ordering them in order
 of best match? I have thought about this for quite some time and done
 some research but I can't seem to find any clever solution.

 I might be able to do a custom query with a bunch of SELECTs as above
 combined with UNION or something, but this would be a pretty complex
 query and not really much more efficient than what I already know how
 to do.

 Thanks in advance any genius here who has a solution!

 -Clay
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Re: View Caching

[Adam Royle]

Maybe it was a typo, but view caches are stored inside tmp/cache/
views/ so make sure that directory is writable. Also, have you tried
with debug = 0?

Cheers,
Adam

On Dec 7, 5:44 pm, NickABusey [EMAIL PROTECTED] wrote:
 I've been through the manual and APIs more times than I can count, and
 still have had no luck whatsoever getting view caching to work with
 1.2, either on my local development box, or my server. I have an app/
 tmp/cache directory that is 0777, I have
   Configure::write('Cache.check', true);
   Cache::config('default', array('engine' = 'File'));
 in core.php, and I have
   var $helpers=array('Cache');
   var $cacheAction = 1 hour;
 in my controller. However, no file is created, and no error is thrown.
 What am I missing?
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Re: Dynamic Navigation in a Layout. How?

[Rob Wilkerson]

Rob Wilkerson wrote:
 I'm so new to Cake that I'm honestly not even sure how to best ask
 this question (much less search to see whether it's been asked), so
 I'm going to describe what I have and allow the question of how best
 to achieve the result to be implied.

 I have a layout (my default layout) that includes 3 different
 navigation menus (primary, secondary, tertiary).  Those menus are data
 driven - I have a nav_menus table that hasAndBelongsToMany
 nav_menu_items. I'd like to populate those menu instances dynamically
 in each place where they belong on the layout.  I'd like to share the
 logic that will retrieve the items based on the menu and use that to
 output the expected markup:

 ul class=alternate-nav
   li class=firstHome/li
   liDirections/li
   li class=lastContact/li
 /ul

 I'm not sure where to look to get started. I have dug around some, but
 I haven't seen anything that seems to address this particular need in
 a way that I can digest it as such.

 Thanks.

I got pulled away from this for a while, but I'm trying to get back to
it so I wanted to revisit this question. validkeys suggested using a
component, but the more I look at it the less it seems right (at least
from a total n00b's perspective). I'm trying to insert data-driven
output into a _layout_. If I use a component, then I'd have to specify
the use of that component for every controller that uses that
template.

Is there not a more universal way to tell the template to include
dynamic content? It seems like an element (provided I can figure out
how to give it data access without breaking encapsulation) is the
right way to go.

Any thoughts?
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Re: simple write to database

[majna]

Change if($this-User-save($this-data['User']['about_me'])) {
to
if($this-User-save($this-data)) {

On Dec 6, 9:42 pm, mike [EMAIL PROTECTED] wrote:
 I have a simple example below.  I just want a form that saves stuff to
 the database.  Its never getting to log test2 I guess the data is
 not in the right format but how am I supposed to do it?  I've already
 spent a considerable amount of time debugging this and other problems,
 please help!

 thanks.

 CREATE TABLE users
 (
 id int NOT NULL AUTO_INCREMENT,
 lastName varchar(255),
 firstName varchar(255),
 picture_id int,
 about_me varchar(255),
 PRIMARY KEY (id)
 );

 here's the model:
 class User extends AppModel
 {
     // Its always good practice to include this variable.
     var $name = 'User';

     // This is used for validation, see Chapter Data Validation.
     var $validate = array();

 }

 controller:
 class NewuserController extends AppController
 {
         var $name = 'Newuser';
         var $uses = Array('User');

         function index()
         {

         }

         function add() {
                 $this-log('in newuser_controller add()');
                 if(!empty($this-data)) {
                         $this-log('test1');
                         //If the form data can be validated and saved...
                         
 if($this-User-save($this-data['User']['about_me'])) {
                         $this-log('test2');
                         //Set a session flash message and redirect.
                                 $this-Session-setFlash(User Saved!);
                                 $this-redirect('add');
                         }
                 }
         }

 }

 View:

 ...
 form action=?php echo $html-url('/newuser/add'); ?
 method=post
 ? //echo $form-create('User'); ?

 FONT CLASS=text12nbsp;BRDescribe yourself/FONT

 ?php echo $html-textarea('User/about_me')?

         ?php echo $html-submit('Add'); ?br/

 ? //echo $form-end('Add User'); ?
 /form
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Re: admin routes prefix routing paginator problems..

[majna]

You can play with $paginator in view:
  $paginator-options(array('prefix' =''));
  $paginator-options(array('url' =''));

On Dec 7, 1:34 am, Andras Kende [EMAIL PROTECTED] wrote:
 Hello,

 Im trying to do an app with 3 areas admin routes which requires login
 /admin/
 /nurse/
 /practice/

 routes.php:
  Router::connect(/admin/:controller/:action/*, array
 (prefix=admin, admin=true));
  Router::connect(/nurse/:controller/:action/*, array
 (prefix=nurse, nurse=true));
  Router::connect(/practice/:controller/:action/*, array
 (prefix=practice, practice=true));

 all seems working, except pagination, paginator links look like:

 http://www.domain.com/calls/nurse_index/page:2
 should be :http://www.domain.com/nurse/calls/index/page:2

 tried to add more more routes like
       Router::connect(/admin/:controller/:action/*, array
 (prefix=admin));
       Router::connect(/nurse/:controller/:action/*, array
 (prefix=nurse));
       Router::connect(/practice/:controller/:action/*, array
 (prefix=practice));

 but then it nurse index shows url 
 ashttp://www.domain.com/admin/calls/index/page:2
 , so somehow are overlapping..

 Is there a way to construct the url correctly ? maybe from a
 controller ?

 Thank you,

 Andras
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Re: store i18n data in database

[jitka (poLK)]

So, you are looking for web based editor of gettext files.
I am not aware of anything what can be easily used in cake app, but
fast googling told me there are some projects which one could learn
from, like GGTT, phpTranslator, PHPTrans, Pootle... Do not blame me,
if they will not suit your needs ;)
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Re: Multiple Instances of TinyMCE

[Arak Tai'Roth]

anyone able to help?

On Dec 6, 9:04 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:
 So I've been playing around with the TinyMCE Helper located in the
 bakery and so far it's been pretty awesome. However I do have one
 problem. In one of my areas there are two text boxes that need
 editing, and so I would like TinyMCE to control both text boxes. This
 is my code for that part of the form:

         echo $form-label('description', 'Description:');
         echo $tinymce-input('description', array('label' = false,
 'rows' = 20, 'cols' = 50));

         echo $form-label('tech_description', 'Technical
 Description:');
         echo $tinymce-input('tech_description', array('label' =
 false, 'rows' = 20, 'cols' = 50));

 However, only the first text box gets TinyMCE, the other one gets
 reduced to a normal textarea box. So I am just wondering if anyone
 knows how to make it produce two TinyMCE boxes so I can have multiple
 instances of it.
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Various named parameters

[navilon]

REF: http://book.cakephp.org/view/541/Named-parameters

I would like to pass optional named variables into my controller to
limit the data received from the database. (the variables shouldnt hve
to be passed in any particular order and each variable passed is
optional)

$this-Game-find(
'all',
array(
'conditions' =
array(
'Game.name LIKE ' = 
'%'.$this-passedArgs['name'].'%',
'Game.released' = '%' 
+ $this-passedArgs['released'] + '%',
'Game.date LIKE' = '%' 
+ $this-passedArgs['date'] + '%',
'Esrb.name LIKE' = '%' 
+ $this-passedArgs['esrb'] + '%'
)
)
)

If all variables are passed this works fine, however if only a few are
passed i get an error similar to this:
Notice (8): Undefined index:  released

Is there something I am missing? Or is there a better method?

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Sanity Check: Many-to-Many Relationship Not Saving

[Rob Wilkerson]

I have a simple many to many relationship scaffolded up that isn't
working the way it should. I'm sure I'm missing something, but I can't
see what that something is and could use a sanity check.

I have NavMenu and NavMenuItem models. Each of those
hasAndBelongsToMany of the other. In the database, I have a nav_menus
table, a nav_menu_items table and the linking table,
nav_menu_items_nav_menus.  As far as I can tell, all of my naming
conventions are correct.  Each of the primary tables has a field named
id and the linking table has the following fields: id, nav_menu_id,
nav_menu_item_id.

When I load the scaffolding page to edit a NavMenu (e.g. /nav_menus/
edit/4917839a-f064-447d-a492-0e733b196446), I see inputs for the
NavMenu properties and I also see a listbox to select the appropriate
menu items for the menu I'm editing. That would seem to indicate that
Cake is seeing everything correct, but when I select a few menu items
to be associated with the menu, the linking table isn't updated. All
other NavMenu fields can be updated as expected.

Obviously, this is my initial foray into CakePHP development, but I
can't figure out what I've done to screw this up.  Any help would be
appreciated.

Thanks.

Rob
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Re: Multiple Instances of TinyMCE

[Rob]

I'm not sure what the helper you're talking about does, but all you
need to do to get TinyMce to work is to reference the control you want
it applied to with JavaScript like:

script type=text/javascript src=your installation path/tiny_mce/
tiny_mce.js/script
script type=text/javascript
tinyMCE.init({
mode : textareas,
theme : simple,
editor_selector : mceSimple
});

tinyMCE.init({
mode : textareas,
theme : advanced,
editor_selector : mceAdvanced
});
/script

form method=post action=somepage
textarea name=content1 class=mceSimple style=width:100%
/textarea
textarea name=content2 class=mceAdvanced style=width:100%
/textarea
/form

(From http://tinymce.moxiecode.com/examples/example_04.php)


On Dec 7, 5:35 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:
 anyone able to help?

 On Dec 6, 9:04 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

  So I've been playing around with the TinyMCE Helper located in the
  bakery and so far it's been pretty awesome. However I do have one
  problem. In one of my areas there are two text boxes that need
  editing, and so I would like TinyMCE to control both text boxes. This
  is my code for that part of the form:

          echo $form-label('description', 'Description:');
          echo $tinymce-input('description', array('label' = false,
  'rows' = 20, 'cols' = 50));

          echo $form-label('tech_description', 'Technical
  Description:');
          echo $tinymce-input('tech_description', array('label' =
  false, 'rows' = 20, 'cols' = 50));

  However, only the first text box gets TinyMCE, the other one gets
  reduced to a normal textarea box. So I am just wondering if anyone
  knows how to make it produce two TinyMCE boxes so I can have multiple
  instances of it.
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Re: Dynamic Navigation in a Layout. How?

[Rob]

Elements are definitely the way to go. I make extensive use of
elements for common display items.

On the other hand, if you are just driving a common navigation, I'm
not sure you need an element since it can just go directly into the
layout.

On Dec 7, 5:24 am, Rob Wilkerson [EMAIL PROTECTED] wrote:
 Rob Wilkerson wrote:
  I'm so new to Cake that I'm honestly not even sure how to best ask
  this question (much less search to see whether it's been asked), so
  I'm going to describe what I have and allow the question of how best
  to achieve the result to be implied.

  I have a layout (my default layout) that includes 3 different
  navigation menus (primary, secondary, tertiary).  Those menus are data
  driven - I have a nav_menus table that hasAndBelongsToMany
  nav_menu_items. I'd like to populate those menu instances dynamically
  in each place where they belong on the layout.  I'd like to share the
  logic that will retrieve the items based on the menu and use that to
  output the expected markup:

  ul class=alternate-nav
     li class=firstHome/li
     liDirections/li
     li class=lastContact/li
  /ul

  I'm not sure where to look to get started. I have dug around some, but
  I haven't seen anything that seems to address this particular need in
  a way that I can digest it as such.

  Thanks.

 I got pulled away from this for a while, but I'm trying to get back to
 it so I wanted to revisit this question. validkeys suggested using a
 component, but the more I look at it the less it seems right (at least
 from a total n00b's perspective). I'm trying to insert data-driven
 output into a _layout_. If I use a component, then I'd have to specify
 the use of that component for every controller that uses that
 template.

 Is there not a more universal way to tell the template to include
 dynamic content? It seems like an element (provided I can figure out
 how to give it data access without breaking encapsulation) is the
 right way to go.

 Any thoughts?
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Re: Dynamic Navigation in a Layout. How?

[Rob Wilkerson]

On Dec 7, 10:59 am, Rob [EMAIL PROTECTED] wrote:
 Elements are definitely the way to go. I make extensive use of
 elements for common display items.

Okay, good. At least I'm not _way_ off base, then. Now, the follow-on
question is this: how do I make an element data-driven? In my case,
I'd want my element to have access to the NavMenu model in order to
retrieve the items in any given menu. I haven't found much (read:
anything) documented on this.

 On the other hand, if you are just driving a common navigation, I'm
 not sure you need an element since it can just go directly into the
 layout.

Except that it's data-driven and exists with slight variations in 3
different locations within the layout.

Thanks again.
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Re: Multiple Instances of TinyMCE

[Arak Tai'Roth]

yea, I was just hoping for something that would use the helper, as
that's the entire reason why I have it. But if I have to do this then
I guess I have to do it this way. I will try this out and see if it
works, don't see why it wouldn't.

On Dec 7, 9:55 am, Rob [EMAIL PROTECTED] wrote:
 I'm not sure what the helper you're talking about does, but all you
 need to do to get TinyMce to work is to reference the control you want
 it applied to with JavaScript like:

 script type=text/javascript src=your installation path/tiny_mce/
 tiny_mce.js/script
 script type=text/javascript
 tinyMCE.init({
         mode : textareas,
         theme : simple,
         editor_selector : mceSimple

 });

 tinyMCE.init({
         mode : textareas,
         theme : advanced,
         editor_selector : mceAdvanced});

 /script

 form method=post action=somepage
         textarea name=content1 class=mceSimple style=width:100%
         /textarea
         textarea name=content2 class=mceAdvanced style=width:100%
         /textarea
 /form

 (Fromhttp://tinymce.moxiecode.com/examples/example_04.php)

 On Dec 7, 5:35 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

  anyone able to help?

  On Dec 6, 9:04 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

   So I've been playing around with the TinyMCE Helper located in the
   bakery and so far it's been pretty awesome. However I do have one
   problem. In one of my areas there are two text boxes that need
   editing, and so I would like TinyMCE to control both text boxes. This
   is my code for that part of the form:

           echo $form-label('description', 'Description:');
           echo $tinymce-input('description', array('label' = false,
   'rows' = 20, 'cols' = 50));

           echo $form-label('tech_description', 'Technical
   Description:');
           echo $tinymce-input('tech_description', array('label' =
   false, 'rows' = 20, 'cols' = 50));

   However, only the first text box gets TinyMCE, the other one gets
   reduced to a normal textarea box. So I am just wondering if anyone
   knows how to make it produce two TinyMCE boxes so I can have multiple
   instances of it.
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Re: Multiple Instances of TinyMCE

[Arak Tai'Roth]

It appears to have worked, so I guess I will just end up doing it this
way. Thanks for your help.

On Dec 7, 10:24 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:
 yea, I was just hoping for something that would use the helper, as
 that's the entire reason why I have it. But if I have to do this then
 I guess I have to do it this way. I will try this out and see if it
 works, don't see why it wouldn't.

 On Dec 7, 9:55 am, Rob [EMAIL PROTECTED] wrote:

  I'm not sure what the helper you're talking about does, but all you
  need to do to get TinyMce to work is to reference the control you want
  it applied to with JavaScript like:

  script type=text/javascript src=your installation path/tiny_mce/
  tiny_mce.js/script
  script type=text/javascript
  tinyMCE.init({
          mode : textareas,
          theme : simple,
          editor_selector : mceSimple

  });

  tinyMCE.init({
          mode : textareas,
          theme : advanced,
          editor_selector : mceAdvanced});

  /script

  form method=post action=somepage
          textarea name=content1 class=mceSimple style=width:100%
          /textarea
          textarea name=content2 class=mceAdvanced style=width:100%
          /textarea
  /form

  (Fromhttp://tinymce.moxiecode.com/examples/example_04.php)

  On Dec 7, 5:35 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

   anyone able to help?

   On Dec 6, 9:04 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

So I've been playing around with the TinyMCE Helper located in the
bakery and so far it's been pretty awesome. However I do have one
problem. In one of my areas there are two text boxes that need
editing, and so I would like TinyMCE to control both text boxes. This
is my code for that part of the form:

        echo $form-label('description', 'Description:');
        echo $tinymce-input('description', array('label' = false,
'rows' = 20, 'cols' = 50));

        echo $form-label('tech_description', 'Technical
Description:');
        echo $tinymce-input('tech_description', array('label' =
false, 'rows' = 20, 'cols' = 50));

However, only the first text box gets TinyMCE, the other one gets
reduced to a normal textarea box. So I am just wondering if anyone
knows how to make it produce two TinyMCE boxes so I can have multiple
instances of it.
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Handling user privacy settings in conjunction with ACL and Auth

[Joe Critchley]

I'm currently developing an application that relies heavily on user
privacy.

Firstly, I'm going to set-up an ACL that includes guests being able to
view member's profiles and the website's homepage, but nothing else.
Registered members would have access to the other areas, along with
other ACL groups etc etc.

However, my problem occurs when, for example, a SPECIFIC USER (i.e.
NOT just an action in a controller) has decided to make their profile
private (or even certain parts of their profile).

What would be the best way to handle these privacy settings? My
initial thought is to place all the access logic inside the
beforeFilter() method, after all of the ACL and Auth logic. However,
would this significantly affect database performance? In fact, is
there a way for custom records to be handled through ACL?
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Set:extract retaining the original id

[mario]

Hello guys,

I'm having some problem with the Set::extract feature of cakephp:

Here is my array from $this-data that was already submitted:

[User] = Array
(
[User] = Array
(
[1] = Array
(
[0] = 3
)

[2] = Array
(
[0] = 2
)

[3] = Array
(
[0] = 4
)

)

)

and I want to have a result (array) like this:

[User] = Array
(
[User] = Array
(
[1] = 3
[2] = 2
[3] = 4
)

)


However, when i do this:

$this-data['User']['User'] = Set::extract($this-data['User']
['User'], '{n}.0');

The result is always like this:

[User] = Array
(
[User] = Array
(
[0] = 3
[1] = 2
[2] = 4
)

)


See the difference? Set:extract is always resetting (from zero and so
on) the ids of the root array I'm trying to extract.

I receive,

[0] = 3
[1] = 2
[2] = 4

instead of what I want which is:

[1] = 3
[2] = 2
[3] = 4


Please help. I really need the original ids for parsing purposes.


Thanks,

Mario


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Re: ORM, fetching related rows of corresponding model: undefined index..

[Tanay]

MODEL answers.php

?php
class Answer extends AppModel {
   var $name = 'Answer';
   var $belongsTo = array(
  'Question' = array(
 'className'= 'Question'
  )
   );
}
?

Model questions.php

?php
class Question extends AppModel {
   var $name = 'Question';
   var $hasMany = array(
  'Answer' = array(
 'className' = 'Answer'
  )
   );
}
?

===

On Dec 7, 11:00 am, thatsgreat2345 [EMAIL PROTECTED] wrote:
 Post your model code so we can see if you are doing your associations
 correctly, or if they are spelled correctly.

 On Dec 6, 9:50 pm, Tanay [EMAIL PROTECTED] wrote:



  it seems no relationships are working with my cakephp.
  has it got anything to do with my version of php5?

  i have got struck at this for a long time.
  help will be very greatly appreciated.

  here is the var_dump of $questions as seen in home.ctp
  array(2) { [0]= array(1) { [Question]= array(5) { [id]= string
  (1) 1 [question]= string(22) Why do you use
  CakePHP [questioner]= string(5) Ahsan [created]= string(19)
  2008-02-11 22:19:04 [modified]= string(19) 2008-02-11
  22:19:04 } } [1]= array(1) { [Question]= array(5) { [id]=
  string(1) 2 [question]= string(25) Why won't you use
  CakePHP [questioner]= string(5) Ahsan [created]= string(19)
  2008-02-11 22:22:06 [modified]= string(19) 2008-02-11
  22:22:06 } } }

  coresponding queries sql dump is
  DESCRIBE `questions`
  SELECT `Question`.`id`, `Question`.`question`,
  `Question`.`questioner`, `Question`.`created`, `Question`.`modified`
  FROM `questions` AS `Question` WHERE 1 = 1

  On Dec 7, 1:47 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

   I see you are using the Packt Publishing book, ensure that your model
   relations aren't misspelled or anything else it won't load the answers
   and thus won't exist, in the controller do a debug on the find, that
   or in the view do a pr($questions) before the foreach this will
   display the array and see what exactly what is in the array.

   On Dec 6, 9:40 am, Tanay [EMAIL PROTECTED] wrote:

Tanay wrote:
 i have two dbs questions , and answers..

 corresponding models

 questions has many answers
 answers belogns to questions

 in questions_controller, i have this home function
  function home() {
           $this-Question-recursive = 1;
           $this-set('questions', $this-Question-find('all'));
        }

 in my home.ctp, i tried to show the corresponding answers using the
 code below
 i took a question from an array of questions using foreach
 foreach question, $question['Answer'] is expected to give array of
 answers, but it says  Undefined index:  Answer
 THAT LIne is marked in code below

 home.ctp:

     ?php if(empty($questions)): ?
        p class=no_answerNo Questions yet. Be the first one to
                  post a Question!/p
     ?php else: ?
        dl
        ?php foreach ($questions as $question): ?
           dtspan?php e($question['Question']['questioner']);
            ?/span/dt
                      dd
              ?php e($html-link($question['Question']['question']
     .'?', array('action' = 'show', $question['Question']['id']))); ?
              ?php
                 $answer_count = count($question
 ['Answer']);   //  -ERROR HERE
                 if(!$answer_count)
                    e((no answers yet));
                 else if($answer_count == 1)
                    e((1 answer));
                 else
                    e((.$answer_count. answers));
               ?
           /dd
        ?php endforeach; ?
        /dl
     ?php endif; ?

 the above code is a part of project in book CakePHP application
 Development,
 the code is same as in book
 is it due to some different version of cakephp that author used?
 i use latest 1.2 version

. . . . . .
Still the same. But now it says
The new variable name $question['question']['answer'] is undefined
index.
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Re: View Caching

[NickABusey]

That was not a typo, that's what I was missing. Thanks very much!

On Dec 7, 5:09 am, Adam Royle [EMAIL PROTECTED] wrote:
 Maybe it was a typo, but view caches are stored inside tmp/cache/
 views/ so make sure that directory is writable. Also, have you tried
 with debug = 0?

 Cheers,
 Adam

 On Dec 7, 5:44 pm, NickABusey [EMAIL PROTECTED] wrote:

  I've been through the manual and APIs more times than I can count, and
  still have had no luck whatsoever getting view caching to work with
  1.2, either on my local development box, or my server. I have an app/
  tmp/cache directory that is 0777, I have
    Configure::write('Cache.check', true);
    Cache::config('default', array('engine' = 'File'));
  in core.php, and I have
    var $helpers=array('Cache');
    var $cacheAction = 1 hour;
  in my controller. However, no file is created, and no error is thrown.
  What am I missing?
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Re: Dynamic Navigation in a Layout. How?

[Filip Camerman]

What you need is indeed an element, and to query the database from
inside an element you can use requestAction in it (that's what it's
there for).

It's explained here:
http://book.cakephp.org/view/434/requestAction

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Re: Various named parameters

[brian]

$conditions = array();

if (isset($this-passedArgs['released']))
{
   $conditions['Game.released'] = '%' + $this-passedArgs['released'] + '%';
}

...

$this-Game-find('all', array('conditions' = $conditions));

You may want to sanitize.

On Sun, Dec 7, 2008 at 7:16 AM, navilon [EMAIL PROTECTED] wrote:

 REF: http://book.cakephp.org/view/541/Named-parameters

 I would like to pass optional named variables into my controller to
 limit the data received from the database. (the variables shouldnt hve
 to be passed in any particular order and each variable passed is
 optional)

$this-Game-find(
'all',
array(
'conditions' =
array(
'Game.name LIKE ' = 
 '%'.$this-passedArgs['name'].'%',
'Game.released' = '%' 
 + $this-passedArgs['released'] + '%',
'Game.date LIKE' = 
 '%' + $this-passedArgs['date'] + '%',
'Esrb.name LIKE' = 
 '%' + $this-passedArgs['esrb'] + '%'
)
)
)

 If all variables are passed this works fine, however if only a few are
 passed i get an error similar to this:
 Notice (8): Undefined index:  released

 Is there something I am missing? Or is there a better method?

 


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Re: Dynamic Navigation in a Layout. How?

[Rob]

Just do a find on your NavMenu in the model you are driving your view
from, and then you can access the data from your element.

I do this on my site in my jobs view, where I want to display a
couple differently filtered lists of job slots. I populate the lists
into $userSlots and $availableSlots in my jobs_controller, then in my
view, I use the same element to display them by passing in those two
data sets:

?php if (isset($userSlots)):?
h3?php __('You are signed up for');?/h3
?php echo $this-element('signup_list', array( 'slots' =
$userSlots )); ?
?php endif; ?
h3?php __('Available Time Slots for '. $job['Job']
['job_name']);?/h3
?php if (isset($availableSlots)):?
?php echo $this-element('signup_list', array( 'slots' =
$availableSlots )); ?
?php else: ?
h3Nobody signed up for this job yet/h3
?php endif; ?



On Dec 7, 8:08 am, Rob Wilkerson [EMAIL PROTECTED] wrote:
 On Dec 7, 10:59 am, Rob [EMAIL PROTECTED] wrote:

  Elements are definitely the way to go. I make extensive use of
  elements for common display items.

 Okay, good. At least I'm not _way_ off base, then. Now, the follow-on
 question is this: how do I make an element data-driven? In my case,
 I'd want my element to have access to the NavMenu model in order to
 retrieve the items in any given menu. I haven't found much (read:
 anything) documented on this.

  On the other hand, if you are just driving a common navigation, I'm
  not sure you need an element since it can just go directly into the
  layout.

 Except that it's data-driven and exists with slight variations in 3
 different locations within the layout.

 Thanks again.
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Re: Model Find With Conditions On Associated Models Not Working

[brian]

You need  a LEFT JOIN in your query. MenuPosting can't be in the WHERE
clause. Perhaps if you post the actual SQL query you're aiming for we
can figure out how to have Cake generate it.

What are post_start  post_end, anyway?

On Sat, Dec 6, 2008 at 10:06 PM, erturne [EMAIL PROTECTED] wrote:

 I've seen similar questions raised but haven't found a solution that
 works. Perhaps it's because I'm new to CakePHP and don't fully
 understand unbindModel(), bindModel(), container behaviors, etc.
 Anyway, I'll start simple and take suggestions.

 I have the following models in which a Menu hasAndBelongsToMany Entree
 and hasMany MenuPosting, and a MenuPosting belongsTo Menu:

 class Menu extends AppModel {
var $name = 'Menu';
var $hasAndBelongsToMany = 'Entree';
var $hasMany = array(
'MenuPosting' = array( 'dependent' = true )
);
  }

 class MenuPosting extends AppModel {
var $name = 'MenuPosting';
var $belongsTo = 'Menu';
 }

 class Entree extends AppModel {
var $name = 'Entree';
  }

 In my MenusController I have a display_current() action to find all
 menus that are currently posted:

 class MenusController extends AppController {
var $name = 'Menus';
function display_current() {

// Limit the query results to the menu that is visible (i.e. is
// currently posted)
$conditions = array(
'MenuPosting.post_start =' = date('Y-m-d H:i:s', 
 strtotime
 (now)),
'MenuPosting.post_end ' = date('Y-m-d H:i:s', 
 strtotime(now)),
);

// Find the menus that meets the conditions.
$this-set('menu', $this-Menu-find('all', array('conditions' 
 =
 $conditions)));
}
  }

 The error I get from this is:
 Warning (512): SQL Error: 1054: Unknown column
 'MenuPosting.post_start' in 'where clause'

 $sql=   SELECT `Menu`.`id`, `Menu`.`old_id`, `Menu`.`created`,
 `Menu`.`modified`, `Menu`.`title` FROM `menus` AS `Menu`   WHERE
 `MenuPosting`.`post_start` = '2008-12-06 22:02:57' AND
 `MenuPosting`.`post_end`  '2008-12-06 22:02:57'   

 If I remove the conditions from the find I am able to retrieve all
 Menu and associated Entree, so I believe I have set up my database
 tables correctly.

 Any suggestions?

 


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Re: Multiple Instances of TinyMCE

[Rob]

I'd check with the author of the helper, or diagnose it yourself.

Basically you just need each element to have a different class/id in
order for TinyMCE to select it.

On Dec 7, 8:55 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:
 It appears to have worked, so I guess I will just end up doing it this
 way. Thanks for your help.

 On Dec 7, 10:24 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

  yea, I was just hoping for something that would use the helper, as
  that's the entire reason why I have it. But if I have to do this then
  I guess I have to do it this way. I will try this out and see if it
  works, don't see why it wouldn't.

  On Dec 7, 9:55 am, Rob [EMAIL PROTECTED] wrote:

   I'm not sure what the helper you're talking about does, but all you
   need to do to get TinyMce to work is to reference the control you want
   it applied to with JavaScript like:

   script type=text/javascript src=your installation path/tiny_mce/
   tiny_mce.js/script
   script type=text/javascript
   tinyMCE.init({
           mode : textareas,
           theme : simple,
           editor_selector : mceSimple

   });

   tinyMCE.init({
           mode : textareas,
           theme : advanced,
           editor_selector : mceAdvanced});

   /script

   form method=post action=somepage
           textarea name=content1 class=mceSimple style=width:100%
           /textarea
           textarea name=content2 class=mceAdvanced style=width:100%
           /textarea
   /form

   (Fromhttp://tinymce.moxiecode.com/examples/example_04.php)

   On Dec 7, 5:35 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

anyone able to help?

On Dec 6, 9:04 am, Arak Tai'Roth [EMAIL PROTECTED] wrote:

 So I've been playing around with the TinyMCE Helper located in the
 bakery and so far it's been pretty awesome. However I do have one
 problem. In one of my areas there are two text boxes that need
 editing, and so I would like TinyMCE to control both text boxes. This
 is my code for that part of the form:

         echo $form-label('description', 'Description:');
         echo $tinymce-input('description', array('label' = false,
 'rows' = 20, 'cols' = 50));

         echo $form-label('tech_description', 'Technical
 Description:');
         echo $tinymce-input('tech_description', array('label' =
 false, 'rows' = 20, 'cols' = 50));

 However, only the first text box gets TinyMCE, the other one gets
 reduced to a normal textarea box. So I am just wondering if anyone
 knows how to make it produce two TinyMCE boxes so I can have multiple
 instances of it.
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Re: routes, controller name and model name

[brian]

If you don't want to put an action before name-of-collection you'll
need to create a route for each action, AFAIK.

Router::connect('/showcases', array('controller' = 'showcases',
'action' = 'index'));
Router::connect('/showcases/edit/*', array('controller' =
'showcases', 'action' = 'edit'));

etc.

Router::connect('/collections/:slug',
array('controller' = 'collections', 'action' = 'view'),
array('slug' = '[-_A-Za-z0-9]+')
);


On Sun, Dec 7, 2008 at 5:56 AM, Filip Camerman [EMAIL PROTECTED] wrote:

 I'm making a site for an artist and when showing image galleries of
 art collections I want url's like

 www.site.com/collections/name-of-collection

 Now I also have a database table for collections, and hence a
 collection_controller and a collection model.

 First I had my routes like this
 - Router::connect('/collections/:action/*', array('controller' =
 'collections'));
 - Router::connect('/collections/*', array('controller' =
 'collections', 'action' = 'index'));

 I thought that the first route would only catch url's with existing
 actions, but instead it also caught my /collections/name-of-
 collection url's. So I changed it to:

 - Router::connect('/coll/:action/*', array('controller' =
 'collections'));

 since the url's don't matter for my admin actions. But then I have to
 change my forms because

 ?= $form-create('Collection');?

 will post the form to /collections/...

 Now I can solve this in several ways (change the model name; make
 separate routes for all my admin actions, ...) but I was wondering if
 there's an elegant way to deal with custom url's that start with a
 controller/model name?
 


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Re: ORM, fetching related rows of corresponding model: undefined index..

[Rob]

Well based on your post, you are not getting the related model, and it
appears you've only done a find on the Questions without setting
recursive in your find('all'), try this:

 $this-set('questions', $this-Question-find('all', array
('recursive' = 1)));

On Dec 7, 9:55 am, Tanay [EMAIL PROTECTED] wrote:
 MODEL answers.php

     ?php
     class Answer extends AppModel {
        var $name = 'Answer';
        var $belongsTo = array(
           'Question' = array(
              'className'    = 'Question'
           )
        );
     }
     ?

 Model questions.php

     ?php
     class Question extends AppModel {
        var $name = 'Question';
        var $hasMany = array(
           'Answer' = array(
              'className'     = 'Answer'
           )
        );
     }
     ?

 ===

 On Dec 7, 11:00 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

  Post your model code so we can see if you are doing your associations
  correctly, or if they are spelled correctly.

  On Dec 6, 9:50 pm, Tanay [EMAIL PROTECTED] wrote:

   it seems no relationships are working with my cakephp.
   has it got anything to do with my version of php5?

   i have got struck at this for a long time.
   help will be very greatly appreciated.

   here is the var_dump of $questions as seen in home.ctp
   array(2) { [0]= array(1) { [Question]= array(5) { [id]= string
   (1) 1 [question]= string(22) Why do you use
   CakePHP [questioner]= string(5) Ahsan [created]= string(19)
   2008-02-11 22:19:04 [modified]= string(19) 2008-02-11
   22:19:04 } } [1]= array(1) { [Question]= array(5) { [id]=
   string(1) 2 [question]= string(25) Why won't you use
   CakePHP [questioner]= string(5) Ahsan [created]= string(19)
   2008-02-11 22:22:06 [modified]= string(19) 2008-02-11
   22:22:06 } } }

   coresponding queries sql dump is
   DESCRIBE `questions`
   SELECT `Question`.`id`, `Question`.`question`,
   `Question`.`questioner`, `Question`.`created`, `Question`.`modified`
   FROM `questions` AS `Question` WHERE 1 = 1

   On Dec 7, 1:47 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

I see you are using the Packt Publishing book, ensure that your model
relations aren't misspelled or anything else it won't load the answers
and thus won't exist, in the controller do a debug on the find, that
or in the view do a pr($questions) before the foreach this will
display the array and see what exactly what is in the array.

On Dec 6, 9:40 am, Tanay [EMAIL PROTECTED] wrote:

 Tanay wrote:
  i have two dbs questions , and answers..

  corresponding models

  questions has many answers
  answers belogns to questions

  in questions_controller, i have this home function
   function home() {
            $this-Question-recursive = 1;
            $this-set('questions', $this-Question-find('all'));
         }

  in my home.ctp, i tried to show the corresponding answers using the
  code below
  i took a question from an array of questions using foreach
  foreach question, $question['Answer'] is expected to give array of
  answers, but it says  Undefined index:  Answer
  THAT LIne is marked in code below

  home.ctp:

      ?php if(empty($questions)): ?
         p class=no_answerNo Questions yet. Be the first one to
                   post a Question!/p
      ?php else: ?
         dl
         ?php foreach ($questions as $question): ?
            dtspan?php e($question['Question']['questioner']);
             ?/span/dt
                       dd
               ?php e($html-link($question['Question']['question']
      .'?', array('action' = 'show', $question['Question']['id']))); 
  ?
               ?php
                  $answer_count = count($question
  ['Answer']);   //  -ERROR HERE
                  if(!$answer_count)
                     e((no answers yet));
                  else if($answer_count == 1)
                     e((1 answer));
                  else
                     e((.$answer_count. answers));
                ?
            /dd
         ?php endforeach; ?
         /dl
      ?php endif; ?

  the above code is a part of project in book CakePHP application
  Development,
  the code is same as in book
  is it due to some different version of cakephp that author used?
  i use latest 1.2 version

 . . . . . .
 Still the same. But now it says
 The new variable name $question['question']['answer'] is undefined
 index.
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Re: Well formed XML for RESTful web services ...

[lightglitch]

Maybe this will help you:

http://www.firephp.org/Wiki/Libraries/CakePHP

On Dec 7, 4:32 am, Rob [EMAIL PROTECTED] wrote:
 No, I don't expect Cake to format the output at all since we are doing
 MVC.

 I want to use the debug output without it affecting my HTTP stream.

 The bug/problem is that if debug  1, the content type gets flipped to
 text/html, and the SQL debug gets appended to the XML (after the end
 of the HTML because it's being output directly from the dbo_source.php

 If we're truly doing MVC, there should never be debug information
 going directly to the view, and the debug information should be
 accessible in the model to the controller so we can decide how to deal
 with it.

 Say for instance I have debug set to 1 to catch errors. Now in my
 controller code, I have to turn off debug for XML, so I don't get any
 chance to catch those errors, which may or may not need to be dealt
 with.

 I know the workaround is to detect that I'm doing XML and set debug to
 0, but that doesn't help in development in any way.

 Formatting the debug data and spitting it into the view like it does
 is violating the MVC design (while it is nice when you're debugging
 HTML).

 On Dec 6, 6:31 pm, James K [EMAIL PROTECTED] wrote:

  Not sure I understand the problem. Do you expect Cake to format it's
  debug information based on the content type? It's debug information -
  it's meant for debugging.

  You can type Configure::write('debug', 0) at the top of any action in
  order to lower the debug level selectively per action without having
  to shut if off for the entire app.

  I have lots of views in a particular application that return XML or
  JSON, and of course having debug info at the bottom of that response
  will break those requests. Once I confirm the response is returning
  correctly formatted XML or JSON, I turn the debugging off for that
  action and only turn it back on in that action in the event I need
  to debug :P

  Good luck,
  James

  On Dec 6, 6:24 pm, Rob [EMAIL PROTECTED] wrote:

   I'll probably get no response or flames telling me to just set debug
   to 1, but 

   I've set up my VolunteerCake code to be a RESTful web service by using
   the Router::parseExtensions() and Router::mapResources() magic. This
   works fine as long as debug is not greater than 1.

   When debug is greater than 1 however, the XML that gets returned is
   broken due to problems with the way the SQL debug is spit out.

   I traced this down to a call In the close() function of
   dbo_source.php, to showLog() which spits out the SQL in an HTML table

   I call the web service and get the XML as expected. The closing XML is
   followed by the dump of the SQL like:

         /VolunteerCake!-- 0.9699s --table class=cake-sql-log
   id=cakeSqlLog_1228604311493b0397a19568_61229664 summary=Cake SQL
   Log cellspacing=0 border = 0
         caption(default) 36 queries took 127 ms/caption

   Other web service tools I've worked with operate by giving you access
   to the debug data so you can spit it out as part of the XML if you
   want to, so you end up with something like:

   VolunteerCakemyDatanon-debug data/myDatadebugdebug data
   (usually wrapped in a CDATA)/debug/myData/VolunteerCake

   Ideally Cake shouldn't be changing the content-type and spitting out
   HTML after the well-formed document has been created anyway.


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Can/Should I use 1.2RC3 in a production environment

[Ed Howland]

HI, cake noob here. I know 1.2 isn't stable yet, but does anyone have
any experience or advice wrt usin 1.2RC3 in a production environment?
We are migrating a Perl CGI project over to Cake and first decided to
use 1.1 because of the stable tag on it.

But if 1.2 is going stable in January/early '09, should we develop on
it? I only want to do TDD work and have read that using Simpletest in
1.1 is problematic.

Thanks
Ed

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Re: Custom Helper not Working

[Sapslabs]

gervOsh,

Thanks for replying. I do have the PHP tags I just didn't post them on
my code.

The complete code looks like this:

?php

class ElapsedHelper extends AppHelper {
function test(){
return $this-output('test');
}
}

?

On Dec 6, 9:04 pm, gearvOsh [EMAIL PROTECTED] wrote:
 Missing php open/close tags?

 On Dec 6, 1:00 pm, Sapslabs [EMAIL PROTECTED] wrote:

  I am creating a custom helper like this

  class ElapsedHelper extends AppHelper {
      function test(){
               return $this-output('test');
      }

  }

  Which is save under views/helpers/elapsed.php

  I am including this on my controller using

  var $helpers = array('Elapsed');

  but when I try it I get this error:

  class ElapsedHelper extends AppHelper {   function test(){ return
  $this-output('test');} }

  Missing Helper Class

  Error: The helper class ElapsedHelper can not be found or does not
  exist.

  Error: Create the class below in file: www/views/helpers/elapsed.php

  ?php
  class ElapsedHelper extends AppHelper {

  }

  ?

  Any idea on what I am doing wrong?

  Thanks

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Re: Can/Should I use 1.2RC3 in a production environment

[thatsgreat2345]

1.2 unstable, where did you get that from?

On Dec 7, 9:59 am, Ed Howland [EMAIL PROTECTED] wrote:
 HI, cake noob here. I know 1.2 isn't stable yet, but does anyone have
 any experience or advice wrt usin 1.2RC3 in a production environment?
 We are migrating a Perl CGI project over to Cake and first decided to
 use 1.1 because of the stable tag on it.

 But if 1.2 is going stable in January/early '09, should we develop on
 it? I only want to do TDD work and have read that using Simpletest in
 1.1 is problematic.

 Thanks
 Ed
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Re: Can/Should I use 1.2RC3 in a production environment

[lightglitch]

Despite being in RC3 I already use it in some productions projects
without
any problems. It may have some bugs in some features but the main 1.2
core
is stable enough to start development.

And 1.2 is a big evolution over 1.1 version.


On Dec 7, 5:59 pm, Ed Howland [EMAIL PROTECTED] wrote:
 HI, cake noob here. I know 1.2 isn't stable yet, but does anyone have
 any experience or advice wrt usin 1.2RC3 in a production environment?
 We are migrating a Perl CGI project over to Cake and first decided to
 use 1.1 because of the stable tag on it.

 But if 1.2 is going stable in January/early '09, should we develop on
 it? I only want to do TDD work and have read that using Simpletest in
 1.1 is problematic.

 Thanks
 Ed
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Re: Can/Should I use 1.2RC3 in a production environment

[Ed Howland]

Thanks to all.

That gives me comfort esp. wrt to unit testing.

Ed

On Sun, Dec 7, 2008 at 2:02 PM, lightglitch [EMAIL PROTECTED] wrote:

 Despite being in RC3 I already use it in some productions projects
 without
 any problems. It may have some bugs in some features but the main 1.2
 core
 is stable enough to start development.

 And 1.2 is a big evolution over 1.1 version.


 On Dec 7, 5:59 pm, Ed Howland [EMAIL PROTECTED] wrote:
 HI, cake noob here. I know 1.2 isn't stable yet, but does anyone have
 any experience or advice wrt usin 1.2RC3 in a production environment?
 We are migrating a Perl CGI project over to Cake and first decided to
 use 1.1 because of the stable tag on it.

 But if 1.2 is going stable in January/early '09, should we develop on
 it? I only want to do TDD work and have read that using Simpletest in
 1.1 is problematic.

 Thanks
 Ed
 




-- 
Ed Howland
http://greenprogrammer.blogspot.com
http://twitter.com/ed_howland

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Re: Can/Should I use 1.2RC3 in a production environment

[aranworld]

For the last 6 months or so, most of the changes made to the 1.2 core
code have been with regards to test coverage.  Very little (if any) of
the most commonly used API has changed, and I think the developers
have been pretty clear that not much will change either.  For the most
part, websites I put together in March 2008 still function perfectly
fine with the latest version of the core code.

Upgrading from 1.1 to 1.2, however, does require a bit of work,
because of the changes made to the way the HTML helper forms work.
Also, so many more people are focused on 1.2, that if there are any
major vulnerabilities or performance problems discovered, they will be
found quickly making its use relatively safe from a project management
standpoint.

-Aran

On Dec 7, 12:26 pm, Ed Howland [EMAIL PROTECTED] wrote:
 Thanks to all.

 That gives me comfort esp. wrt to unit testing.

 Ed



 On Sun, Dec 7, 2008 at 2:02 PM, lightglitch [EMAIL PROTECTED] wrote:

  Despite being in RC3 I already use it in some productions projects
  without
  any problems. It may have some bugs in some features but the main 1.2
  core
  is stable enough to start development.

  And 1.2 is a big evolution over 1.1 version.

  On Dec 7, 5:59 pm, Ed Howland [EMAIL PROTECTED] wrote:
  HI, cake noob here. I know 1.2 isn't stable yet, but does anyone have
  any experience or advice wrt usin 1.2RC3 in a production environment?
  We are migrating a Perl CGI project over to Cake and first decided to
  use 1.1 because of the stable tag on it.

  But if 1.2 is going stable in January/early '09, should we develop on
  it? I only want to do TDD work and have read that using Simpletest in
  1.1 is problematic.

  Thanks
  Ed

 --
 Ed Howlandhttp://greenprogrammer.blogspot.comhttp://twitter.com/ed_howland
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Re: Ordering search results by best match

[Johnathan Henderson]

I searched for sphinx at the bakery and found nothing  Adam, do
you have a direct link?

THANKS!

-John

On Dec 7, 7:05 am, Adam Royle [EMAIL PROTECTED] wrote:
 Have a look at sphinx, and the SphinxBehaviour on the bakery.

 Cheers,
 Adam

 On Dec 7, 7:22 am, Clay [EMAIL PROTECTED] wrote:

  My app has the following:

  Genus hasMany Species
  Species hasMany CommonName

  and the corresponding belongsTo relationships as well. Each of these
  models has a name field, and Genus and Species have several other
  fields.

  Say for example I have a Species Acer rubrum. This Species belongsTo
  Genus Acer and has a CommonName Red Maple.

  I want my users to be able to search the database for Red Maple and
  get as a result Species Acer rubrum (with the CommonName Red Maple
  highlighted to show that's what it matched. I can do this already.
  Yay.

  However if I have another Species with CommonName Bored Maple, I want
  to display this result too, but lower in the list than Red Maple.
  Since Bored comes before Red alphabetically I can't just order by
  CommonName.name to get the result I want.

  So what I know I could do is a series of finds:
  1) Find exact match: CommonName.name = $searchStr
  2) Find starting match: CommonName.name LIKE = $searchStr . '%'
  3) Find internal match: CommonName.name LIKE = '%' . $searchStr . '%'

  And then merge them into my results array.

  However, this seems very inefficient. Is there a way I can use cake's
  find() function to do this all in one go while ordering them in order
  of best match? I have thought about this for quite some time and done
  some research but I can't seem to find any clever solution.

  I might be able to do a custom query with a bunch of SELECTs as above
  combined with UNION or something, but this would be a pretty complex
  query and not really much more efficient than what I already know how
  to do.

  Thanks in advance any genius here who has a solution!

  -Clay
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Re: Model Find With Conditions On Associated Models Not Working

[erturne]

On Dec 7, 1:54 pm, brian [EMAIL PROTECTED] wrote:
 You need  a LEFT JOIN in your query. MenuPosting can't be in the WHERE
 clause. Perhaps if you post the actual SQL query you're aiming for we
 can figure out how to have Cake generate it.


Essentially I want information about menus and the entrees on them.
Something like:

SELECT Menus.title, Entrees.name FROM menus Menus
INNER JOIN entrees_menus EntreesMenus ON EntreesMenus.menu_id=Menus.id
INNER JOIN entrees Entrees ON EntreesMenus.entree_id=Entrees.id
INNER JOIN menu_postings MenuPostings ON MenuPostings.menu_id=Menus.id
WHERE MenuPostings.post_start = NOW() AND MenuPostings.post_end  NOW
();

Of course I'll eventually want more than just the menu title and
entree names, but that's good enough for now.

I don't think LEFT JOIN is correct for what I need because I'm not
interested in a menu without entrees! INNER JOIN is the way to go
here. Not sure if we can make CakePHP do that using find() or
Containable or unbindModel()/bindModel(). I'm definitely open to
suggestions.

 What are post_start  post_end, anyway?

Records in the menu_postings table define when each menu is visible on
the website. We might have a particular menu visible on the website at
various times throughout the year (defined by post_start and
post_end).


 On Sat, Dec 6, 2008 at 10:06 PM, erturne [EMAIL PROTECTED] wrote:

  I've seen similar questions raised but haven't found a solution that
  works. Perhaps it's because I'm new to CakePHP and don't fully
  understand unbindModel(), bindModel(), container behaviors, etc.
  Anyway, I'll start simple and take suggestions.

  I have the following models in which a Menu hasAndBelongsToMany Entree
  and hasMany MenuPosting, and a MenuPosting belongsTo Menu:

  class Menu extends AppModel {
         var $name = 'Menu';
         var $hasAndBelongsToMany = 'Entree';
         var $hasMany = array(
                 'MenuPosting' = array( 'dependent' = true )
         );
   }

  class MenuPosting extends AppModel {
         var $name = 'MenuPosting';
         var $belongsTo = 'Menu';
  }

  class Entree extends AppModel {
         var $name = 'Entree';
   }

  In my MenusController I have a display_current() action to find all
  menus that are currently posted:

  class MenusController extends AppController {
         var $name = 'Menus';
         function display_current() {

                 // Limit the query results to the menu that is visible (i.e. 
  is
                 // currently posted)
                 $conditions = array(
                         'MenuPosting.post_start =' = date('Y-m-d H:i:s', 
  strtotime
  (now)),
                         'MenuPosting.post_end ' = date('Y-m-d H:i:s', 
  strtotime(now)),
                 );

                 // Find the menus that meets the conditions.
                 $this-set('menu', $this-Menu-find('all', 
  array('conditions' =
  $conditions)));
         }
   }

  The error I get from this is:
  Warning (512): SQL Error: 1054: Unknown column
  'MenuPosting.post_start' in 'where clause'

  $sql    =       SELECT `Menu`.`id`, `Menu`.`old_id`, `Menu`.`created`,
  `Menu`.`modified`, `Menu`.`title` FROM `menus` AS `Menu`   WHERE
  `MenuPosting`.`post_start` = '2008-12-06 22:02:57' AND
  `MenuPosting`.`post_end`  '2008-12-06 22:02:57'   

  If I remove the conditions from the find I am able to retrieve all
  Menu and associated Entree, so I believe I have set up my database
  tables correctly.

  Any suggestions?
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Re: API documentation tool

[Everton Yoshitani]

It's doxygen :)



2008/12/5 Olivier Percebois-Garve [EMAIL PROTECTED]:
 yeah after some research it seems to be doxygen.

 thx. Dont forgot to have a nice weekend.

 Olivier

 On Fri, Dec 5, 2008 at 2:29 PM, dr. Hannibal Lecter [EMAIL PROTECTED]
 wrote:

 Good point... probably Doxygen then..seems like it (look at the
 examples on Doxygen site, especially the tabs)..

 On Dec 5, 2:14 pm, Olivier Percebois-Garve [EMAIL PROTECTED]
 wrote:
  But it looks visually so different to what I know from phpDocumentor. No
  left column, etc...
 
  On Fri, Dec 5, 2008 at 1:47 PM, dr. Hannibal Lecter
  [EMAIL PROTECTED]wrote:
 
 
 
   I'm not sure on this, but I believe it could be phpDocumentor (http://
   phpdoc.org/)?
 
   On Dec 5, 1:03 pm, Olivier Percebois-Garve [EMAIL PROTECTED]
   wrote:
Hi
 
What is used to generate the API doc of cakephp ?
 
thanks
 
Olivier



 




-- 
Att,
Everton Yoshitani

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Re: Model Find With Conditions On Associated Models Not Working

[brian]

I misinterpreted MenuPostings. I thought it was a table of comments or
something like that. So, it seems that, though there may be many
MenuPostings for a particularMenu, only a single one is really
active at a given time. Is that the case? If so, why not just add
columns post_start  post_end to the menus table and modify them
whenever the posting dates for a menu need to be changed? That would
make your query much simpler.

On Sun, Dec 7, 2008 at 5:27 PM, erturne [EMAIL PROTECTED] wrote:



 On Dec 7, 1:54 pm, brian [EMAIL PROTECTED] wrote:
 You need  a LEFT JOIN in your query. MenuPosting can't be in the WHERE
 clause. Perhaps if you post the actual SQL query you're aiming for we
 can figure out how to have Cake generate it.


 Essentially I want information about menus and the entrees on them.
 Something like:

 SELECT Menus.title, Entrees.name FROM menus Menus
 INNER JOIN entrees_menus EntreesMenus ON EntreesMenus.menu_id=Menus.id
 INNER JOIN entrees Entrees ON EntreesMenus.entree_id=Entrees.id
 INNER JOIN menu_postings MenuPostings ON MenuPostings.menu_id=Menus.id
 WHERE MenuPostings.post_start = NOW() AND MenuPostings.post_end  NOW
 ();

 Of course I'll eventually want more than just the menu title and
 entree names, but that's good enough for now.

 I don't think LEFT JOIN is correct for what I need because I'm not
 interested in a menu without entrees! INNER JOIN is the way to go
 here. Not sure if we can make CakePHP do that using find() or
 Containable or unbindModel()/bindModel(). I'm definitely open to
 suggestions.

 What are post_start  post_end, anyway?

 Records in the menu_postings table define when each menu is visible on
 the website. We might have a particular menu visible on the website at
 various times throughout the year (defined by post_start and
 post_end).


 On Sat, Dec 6, 2008 at 10:06 PM, erturne [EMAIL PROTECTED] wrote:

  I've seen similar questions raised but haven't found a solution that
  works. Perhaps it's because I'm new to CakePHP and don't fully
  understand unbindModel(), bindModel(), container behaviors, etc.
  Anyway, I'll start simple and take suggestions.

  I have the following models in which a Menu hasAndBelongsToMany Entree
  and hasMany MenuPosting, and a MenuPosting belongsTo Menu:

  class Menu extends AppModel {
 var $name = 'Menu';
 var $hasAndBelongsToMany = 'Entree';
 var $hasMany = array(
 'MenuPosting' = array( 'dependent' = true )
 );
   }

  class MenuPosting extends AppModel {
 var $name = 'MenuPosting';
 var $belongsTo = 'Menu';
  }

  class Entree extends AppModel {
 var $name = 'Entree';
   }

  In my MenusController I have a display_current() action to find all
  menus that are currently posted:

  class MenusController extends AppController {
 var $name = 'Menus';
 function display_current() {

 // Limit the query results to the menu that is visible 
  (i.e. is
 // currently posted)
 $conditions = array(
 'MenuPosting.post_start =' = date('Y-m-d H:i:s', 
  strtotime
  (now)),
 'MenuPosting.post_end ' = date('Y-m-d H:i:s', 
  strtotime(now)),
 );

 // Find the menus that meets the conditions.
 $this-set('menu', $this-Menu-find('all', 
  array('conditions' =
  $conditions)));
 }
   }

  The error I get from this is:
  Warning (512): SQL Error: 1054: Unknown column
  'MenuPosting.post_start' in 'where clause'

  $sql=   SELECT `Menu`.`id`, `Menu`.`old_id`, `Menu`.`created`,
  `Menu`.`modified`, `Menu`.`title` FROM `menus` AS `Menu`   WHERE
  `MenuPosting`.`post_start` = '2008-12-06 22:02:57' AND
  `MenuPosting`.`post_end`  '2008-12-06 22:02:57'   

  If I remove the conditions from the find I am able to retrieve all
  Menu and associated Entree, so I believe I have set up my database
  tables correctly.

  Any suggestions?
 


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Re: Ordering search results by best match

[thatsgreat2345]

I guess it is pending approval but here is a cached version from
google

http://209.85.173.132/search?q=cache:NzoXH-E2eZwJ:bakery.cakephp.org/articles/view/sphinx-behavior+SphinxBehavior+cakephphl=enct=clnkcd=1gl=usclient=firefox-a

On Dec 7, 2:23 pm, Johnathan Henderson [EMAIL PROTECTED]
wrote:
 I searched for sphinx at the bakery and found nothing  Adam, do
 you have a direct link?

 THANKS!

 -John

 On Dec 7, 7:05 am, Adam Royle [EMAIL PROTECTED] wrote:

  Have a look at sphinx, and the SphinxBehaviour on the bakery.

  Cheers,
  Adam

  On Dec 7, 7:22 am, Clay [EMAIL PROTECTED] wrote:

   My app has the following:

   Genus hasMany Species
   Species hasMany CommonName

   and the corresponding belongsTo relationships as well. Each of these
   models has a name field, and Genus and Species have several other
   fields.

   Say for example I have a Species Acer rubrum. This Species belongsTo
   Genus Acer and has a CommonName Red Maple.

   I want my users to be able to search the database for Red Maple and
   get as a result Species Acer rubrum (with the CommonName Red Maple
   highlighted to show that's what it matched. I can do this already.
   Yay.

   However if I have another Species with CommonName Bored Maple, I want
   to display this result too, but lower in the list than Red Maple.
   Since Bored comes before Red alphabetically I can't just order by
   CommonName.name to get the result I want.

   So what I know I could do is a series of finds:
   1) Find exact match: CommonName.name = $searchStr
   2) Find starting match: CommonName.name LIKE = $searchStr . '%'
   3) Find internal match: CommonName.name LIKE = '%' . $searchStr . '%'

   And then merge them into my results array.

   However, this seems very inefficient. Is there a way I can use cake's
   find() function to do this all in one go while ordering them in order
   of best match? I have thought about this for quite some time and done
   some research but I can't seem to find any clever solution.

   I might be able to do a custom query with a bunch of SELECTs as above
   combined with UNION or something, but this would be a pretty complex
   query and not really much more efficient than what I already know how
   to do.

   Thanks in advance any genius here who has a solution!

   -Clay
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Re: Can/Should I use 1.2RC3 in a production environment

[Ed Howland]

On Sun, Dec 7, 2008 at 4:09 PM, aranworld [EMAIL PROTECTED] wrote:

 For the last 6 months or so, most of the changes made to the 1.2 core
 code have been with regards to test coverage.  Very little (if any) of
 the most commonly used API has changed, and I think the developers
 have been pretty clear that not much will change either.  For the most
 part, websites I put together in March 2008 still function perfectly
 fine with the latest version of the core code.

 Upgrading from 1.1 to 1.2, however, does require a bit of work,

Since this project is Greenfield, it shouldn't be a problem. All I
have now is the login page, home page and various user add,edit, list
stuff. Did note the file extension changed from .thtml to .ctp.

 because of the changes made to the way the HTML helper forms work.
 Also, so many more people are focused on 1.2, that if there are any
 major vulnerabilities or performance problems discovered, they will be
 found quickly making its use relatively safe from a project management
 standpoint.


Also most of the on-line discussion seems to assume 1.2. With the
release being nearly 2 years old, I;d say it is mature enough for our
use. After all I'm writing this on GMail(beta)!

Thanks
Ed

-- 
Ed Howland
http://greenprogrammer.blogspot.com
http://twitter.com/ed_howland

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How can I view the screencasts on Linux

[Ed Howland]

I am having trouble playing the screencasts on the cakephp.org site. I
am using Ubuntu 8.10 w/FF.

Does anyone have any suggestions?

Thanks
Ed

-- 
Ed Howland
http://greenprogrammer.blogspot.com
http://twitter.com/ed_howland

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saving into multiple table with multiple data

[ridwan arifandi]

i have 5 tables like these

1). purchases

2). purchase_details
 - purchase_id
 - item_id
3). items

4). purchase_item_details
 - purchase_detail_id
 - item_detail_id
5). item_details
 - item_id

in an action, i have to do more than one save action. rule

1). save data into purchases
2). get last insert id from table purchases

3). save data into items
4). get last insert id from table items

5). save data into purchase_details with those two variables
( purchase_id  item_id )
6). get last insert id from table purchase_details

7). save data into item_details with item_id
8). get last insert id from table item_details

9). save data into purchase_item_detail with those two variables
( purchase_detail_id  item_detail_id )

so.. i get a problem at rule #3... it cannot save, dont know why, i
have no error message at my web

items table


CREATE TABLE IF NOT EXISTS `items` (
  `id` int(11) NOT NULL auto_increment,
  `item_type_id` int(11) NOT NULL,
  `item_brand_type_id` int(11) NOT NULL,
  `desc` text NOT NULL,
  `info` text NOT NULL,
  `default_price` int(11) NOT NULL,
  `entry_date` date NOT NULL,
  `entry_by` int(11) NOT NULL,
  `update_date` date NOT NULL,
  `update_by` int(11) NOT NULL,
  `f_expired` enum('t','f') NOT NULL,
  `f_return` enum('t','f') NOT NULL,
  `f_active` enum('t','f') NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

please help me...
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Re: Model Find With Conditions On Associated Models Not Working

[erturne]

On Dec 7, 5:48 pm, brian [EMAIL PROTECTED] wrote:
 I misinterpreted MenuPostings. I thought it was a table of comments or
 something like that. So, it seems that, though there may be many
 MenuPostings for a particularMenu, only a single one is really
 active at a given time. Is that the case? If so, why not just add
 columns post_start  post_end to the menus table and modify them
 whenever the posting dates for a menu need to be changed? That would
 make your query much simpler.


They want to set up their menu schedule ahead of time, and not have to
remember to set the next posting each time one expires. I can also
envision other situations like this in other applications, and as this
is my first Cake app I'd like to learn what the limitations of the
technology are. In this case I COULD attack the problem from a
different angle, starting with MenuPosting rather than Menu so that I
can specify conditions on the post_start and post_end, but I don't
like the idea changing simply because I don't understand how to make
it work the way I want to. I'd really like to find out if it's
possible to do what I want (obtain menus filtered based on the
criteria of an associated model). Or if it's not possible, why.

 On Sun, Dec 7, 2008 at 5:27 PM, erturne [EMAIL PROTECTED] wrote:

  On Dec 7, 1:54 pm, brian [EMAIL PROTECTED] wrote:
  You need  a LEFT JOIN in your query. MenuPosting can't be in the WHERE
  clause. Perhaps if you post the actual SQL query you're aiming for we
  can figure out how to have Cake generate it.

  Essentially I want information about menus and the entrees on them.
  Something like:

  SELECT Menus.title, Entrees.name FROM menus Menus
  INNER JOIN entrees_menus EntreesMenus ON EntreesMenus.menu_id=Menus.id
  INNER JOIN entrees Entrees ON EntreesMenus.entree_id=Entrees.id
  INNER JOIN menu_postings MenuPostings ON MenuPostings.menu_id=Menus.id
  WHERE MenuPostings.post_start = NOW() AND MenuPostings.post_end  NOW
  ();

  Of course I'll eventually want more than just the menu title and
  entree names, but that's good enough for now.

  I don't think LEFT JOIN is correct for what I need because I'm not
  interested in a menu without entrees! INNER JOIN is the way to go
  here. Not sure if we can make CakePHP do that using find() or
  Containable or unbindModel()/bindModel(). I'm definitely open to
  suggestions.

  What are post_start  post_end, anyway?

  Records in the menu_postings table define when each menu is visible on
  the website. We might have a particular menu visible on the website at
  various times throughout the year (defined by post_start and
  post_end).

  On Sat, Dec 6, 2008 at 10:06 PM, erturne [EMAIL PROTECTED] wrote:

   I've seen similar questions raised but haven't found a solution that
   works. Perhaps it's because I'm new to CakePHP and don't fully
   understand unbindModel(), bindModel(), container behaviors, etc.
   Anyway, I'll start simple and take suggestions.

   I have the following models in which a Menu hasAndBelongsToMany Entree
   and hasMany MenuPosting, and a MenuPosting belongsTo Menu:

   class Menu extends AppModel {
          var $name = 'Menu';
          var $hasAndBelongsToMany = 'Entree';
          var $hasMany = array(
                  'MenuPosting' = array( 'dependent' = true )
          );
    }

   class MenuPosting extends AppModel {
          var $name = 'MenuPosting';
          var $belongsTo = 'Menu';
   }

   class Entree extends AppModel {
          var $name = 'Entree';
    }

   In my MenusController I have a display_current() action to find all
   menus that are currently posted:

   class MenusController extends AppController {
          var $name = 'Menus';
          function display_current() {

                  // Limit the query results to the menu that is visible 
   (i.e. is
                  // currently posted)
                  $conditions = array(
                          'MenuPosting.post_start =' = date('Y-m-d 
   H:i:s', strtotime
   (now)),
                          'MenuPosting.post_end ' = date('Y-m-d H:i:s', 
   strtotime(now)),
                  );

                  // Find the menus that meets the conditions.
                  $this-set('menu', $this-Menu-find('all', 
   array('conditions' =
   $conditions)));
          }
    }

   The error I get from this is:
   Warning (512): SQL Error: 1054: Unknown column
   'MenuPosting.post_start' in 'where clause'

   $sql    =       SELECT `Menu`.`id`, `Menu`.`old_id`, `Menu`.`created`,
   `Menu`.`modified`, `Menu`.`title` FROM `menus` AS `Menu`   WHERE
   `MenuPosting`.`post_start` = '2008-12-06 22:02:57' AND
   `MenuPosting`.`post_end`  '2008-12-06 22:02:57'   

   If I remove the conditions from the find I am able to retrieve all
   Menu and associated Entree, so I believe I have set up my database
   tables correctly.

   Any suggestions?
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Auth-login() Not working correctly?

[gearvOsh]

So im trying to get the Auth component to work, here is my
AppController.

function beforeFilter() {
// Referer
$referer = $this-referer(null, true);
if (empty($referer)) {
$referer = array('controller' = 'dashboard', 'action' = 'index');
}

// Authenticate
if (isset($this-Auth)) {
$this-Auth-sessionKey = 'User';
//$this-Auth-authorize = 'controller';
$this-Auth-loginAction = array('controller' = 'users', 'action' =
'login');
$this-Auth-loginRedirect = $referer;
$this-Auth-logoutRedirect = array('controller' = 'site', 'action'
= 'index', 'home');
}
}

That all works fine, its just when I submit my login form (users/
login), the password field does not get encrypted, so it never finds
any rows. I read the docs and it said that it should auto-encrypt any
field named password, obviously its not.

Login action:

// Form processing
if (!empty($this-data)) {
$this-Auth-login($this-data['Form']);
}
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Re: Auth-login() Not working correctly?

[gearvOsh]

Also heres a debug of $this-data:

Array
(
[Form] = Array
(
[username] = test123
[password] = pass123
)

)

And the query being run by $this-Auth-login():

SELECT `User`.`id`, `User`.`username`, `User`.`password`  FROM
`users` AS `User` WHERE `User`.`username` = 'test123' AND
`User`.`password` = 'pass123' LIMIT 1
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Re: Auth-login() Not working correctly?

[Drew Pearson]

try not putting anything in the login action. It does what you are  
trying to do automatically. You also don't need to redirect to the  
referrer as the Auth component stores this automatically. The Auth  
component also looks for logout in a new action logout. Thats where to  
put your redirect.

Also make sure the password you stored in your database is the hashed  
version. You can view a hash result by putting  echo $this-Auth- 
 password(password string here); in an action.

On Dec 7, 2008, at 6:15 PM, gearvOsh wrote:


 So im trying to get the Auth component to work, here is my
 AppController.

 function beforeFilter() {
 // Referer
 $referer = $this-referer(null, true);
 if (empty($referer)) {
   $referer = array('controller' = 'dashboard', 'action' = 'index');
 }

 // Authenticate
 if (isset($this-Auth)) {
   $this-Auth-sessionKey = 'User';
   //$this-Auth-authorize = 'controller';
   $this-Auth-loginAction = array('controller' = 'users', 'action' =
 'login');
   $this-Auth-loginRedirect = $referer;
   $this-Auth-logoutRedirect = array('controller' = 'site', 'action'
 = 'index', 'home');
 }
 }

 That all works fine, its just when I submit my login form (users/
 login), the password field does not get encrypted, so it never finds
 any rows. I read the docs and it said that it should auto-encrypt any
 field named password, obviously its not.

 Login action:

 // Form processing
 if (!empty($this-data)) {
   $this-Auth-login($this-data['Form']);
 }
 


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Re: Well formed XML for RESTful web services ...

[Jay Reeder]

We use a debug datasource that routes the showLog output to firebug as
well.  You can enable this datasource while debugging (instead of using the
mysql driver) and when done, re-enable the mysql driver.

ex:

require_once (LIBS . 'model' . DS . 'datasources' . DS .
'dbo_source.php');
require_once (LIBS . 'model' . DS . 'datasources' . DS . 'dbo' . DS .
'dbo_mysql.php');

class DboMysqlExSource extends DboMysql {
var $description = MySQL DBO Driver - Extended;

function showLog($sorted = false) {

if ($sorted) {
$log = sortByKey($this-_queriesLog, 'took', 'desc',
SORT_NUMERIC);
} else {
$log = $this-_queriesLog;
   }

if ($this-_queriesCnt  1) {
$text = 'queries';
} else {
$text = 'query';
}
fb(({$this-configKeyName}) {$this-_queriesCnt} {$text} took
{$this-_queriesTime} ms\n);
foreach ($log as $k = $i) {
fb (($k +1) . . {$i['query']} {$i['error']}\n);
fb (\tAffected: {$i['affected']}\tTook:{$i['took']} ms\n);
}
}
}

On Sun, Dec 7, 2008 at 2:13 PM, lightglitch [EMAIL PROTECTED] wrote:


 Maybe this will help you:

 http://www.firephp.org/Wiki/Libraries/CakePHP

 On Dec 7, 4:32 am, Rob [EMAIL PROTECTED] wrote:
  No, I don't expect Cake to format the output at all since we are doing
  MVC.
 
  I want to use the debug output without it affecting my HTTP stream.
 
  The bug/problem is that if debug  1, the content type gets flipped to
  text/html, and the SQL debug gets appended to the XML (after the end
  of the HTML because it's being output directly from the dbo_source.php
 
  If we're truly doing MVC, there should never be debug information
  going directly to the view, and the debug information should be
  accessible in the model to the controller so we can decide how to deal
  with it.
 
  Say for instance I have debug set to 1 to catch errors. Now in my
  controller code, I have to turn off debug for XML, so I don't get any
  chance to catch those errors, which may or may not need to be dealt
  with.
 
  I know the workaround is to detect that I'm doing XML and set debug to
  0, but that doesn't help in development in any way.
 
  Formatting the debug data and spitting it into the view like it does
  is violating the MVC design (while it is nice when you're debugging
  HTML).
 
  On Dec 6, 6:31 pm, James K [EMAIL PROTECTED] wrote:
 
   Not sure I understand the problem. Do you expect Cake to format it's
   debug information based on the content type? It's debug information -
   it's meant for debugging.
 
   You can type Configure::write('debug', 0) at the top of any action in
   order to lower the debug level selectively per action without having
   to shut if off for the entire app.
 
   I have lots of views in a particular application that return XML or
   JSON, and of course having debug info at the bottom of that response
   will break those requests. Once I confirm the response is returning
   correctly formatted XML or JSON, I turn the debugging off for that
   action and only turn it back on in that action in the event I need
   to debug :P
 
   Good luck,
   James
 
   On Dec 6, 6:24 pm, Rob [EMAIL PROTECTED] wrote:
 
I'll probably get no response or flames telling me to just set debug
to 1, but 
 
I've set up my VolunteerCake code to be a RESTful web service by
 using
the Router::parseExtensions() and Router::mapResources() magic. This
works fine as long as debug is not greater than 1.
 
When debug is greater than 1 however, the XML that gets returned is
broken due to problems with the way the SQL debug is spit out.
 
I traced this down to a call In the close() function of
dbo_source.php, to showLog() which spits out the SQL in an HTML table
 
I call the web service and get the XML as expected. The closing XML
 is
followed by the dump of the SQL like:
 
  /VolunteerCake!-- 0.9699s --table class=cake-sql-log
id=cakeSqlLog_1228604311493b0397a19568_61229664 summary=Cake SQL
Log cellspacing=0 border = 0
  caption(default) 36 queries took 127 ms/caption
 
Other web service tools I've worked with operate by giving you access
to the debug data so you can spit it out as part of the XML if you
want to, so you end up with something like:
 
VolunteerCakemyDatanon-debug data/myDatadebugdebug data
(usually wrapped in a CDATA)/debug/myData/VolunteerCake
 
Ideally Cake shouldn't be changing the content-type and spitting out
HTML after the well-formed document has been created anyway.
 
 
 


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Re: ORM, fetching related rows of corresponding model: undefined index..

[Tanay]

i have done that..
here is my full questions_controller.php

?php
class QuestionsController extends AppController {
   var $name = 'Questions';
   function home() {
  $this-Question-recursive = 'all';
  $this-set('questions', $this-Question-find('all'));
   }
   function show( $id = null) {
  if (!$id) {
 $this-Session-setFlash('Invalid Question.');
 $this-redirect(array('action'='home'));
  }
  $this-set('question', $this-Question-read(null, $id));
   }
}
?

===


On Dec 8, 12:10 am, Rob [EMAIL PROTECTED] wrote:
 Well based on your post, you are not getting the related model, and it
 appears you've only done a find on the Questions without setting
 recursive in your find('all'), try this:

  $this-set('questions', $this-Question-find('all', array
 ('recursive' = 1)));

 On Dec 7, 9:55 am, Tanay [EMAIL PROTECTED] wrote:



  MODEL answers.php

      ?php
      class Answer extends AppModel {
         var $name = 'Answer';
         var $belongsTo = array(
            'Question' = array(
               'className'    = 'Question'
            )
         );
      }
      ?

  Model questions.php

      ?php
      class Question extends AppModel {
         var $name = 'Question';
         var $hasMany = array(
            'Answer' = array(
               'className'     = 'Answer'
            )
         );
      }
      ?

  ===

  On Dec 7, 11:00 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

   Post your model code so we can see if you are doing your associations
   correctly, or if they are spelled correctly.

   On Dec 6, 9:50 pm, Tanay [EMAIL PROTECTED] wrote:

it seems no relationships are working with my cakephp.
has it got anything to do with my version of php5?

i have got struck at this for a long time.
help will be very greatly appreciated.

here is the var_dump of $questions as seen in home.ctp
array(2) { [0]= array(1) { [Question]= array(5) { [id]= string
(1) 1 [question]= string(22) Why do you use
CakePHP [questioner]= string(5) Ahsan [created]= string(19)
2008-02-11 22:19:04 [modified]= string(19) 2008-02-11
22:19:04 } } [1]= array(1) { [Question]= array(5) { [id]=
string(1) 2 [question]= string(25) Why won't you use
CakePHP [questioner]= string(5) Ahsan [created]= string(19)
2008-02-11 22:22:06 [modified]= string(19) 2008-02-11
22:22:06 } } }

coresponding queries sql dump is
DESCRIBE `questions`
SELECT `Question`.`id`, `Question`.`question`,
`Question`.`questioner`, `Question`.`created`, `Question`.`modified`
FROM `questions` AS `Question` WHERE 1 = 1

On Dec 7, 1:47 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

 I see you are using the Packt Publishing book, ensure that your model
 relations aren't misspelled or anything else it won't load the answers
 and thus won't exist, in the controller do a debug on the find, that
 or in the view do a pr($questions) before the foreach this will
 display the array and see what exactly what is in the array.

 On Dec 6, 9:40 am, Tanay [EMAIL PROTECTED] wrote:

  Tanay wrote:
   i have two dbs questions , and answers..

   corresponding models

   questions has many answers
   answers belogns to questions

   in questions_controller, i have this home function
    function home() {
             $this-Question-recursive = 1;
             $this-set('questions', $this-Question-find('all'));
          }

   in my home.ctp, i tried to show the corresponding answers using 
   the
   code below
   i took a question from an array of questions using foreach
   foreach question, $question['Answer'] is expected to give array of
   answers, but it says  Undefined index:  Answer
   THAT LIne is marked in code below

   home.ctp:

       ?php if(empty($questions)): ?
          p class=no_answerNo Questions yet. Be the first one to
                    post a Question!/p
       ?php else: ?
          dl
          ?php foreach ($questions as $question): ?
             dtspan?php e($question['Question']['questioner']);
              ?/span/dt
                        dd
                ?php e($html-link($question['Question']['question']
       .'?', array('action' = 'show', 
   $question['Question']['id']))); ?
                ?php
                   $answer_count = count($question
   ['Answer']);   //  -ERROR HERE
                   if(!$answer_count)
                      e((no answers yet));
                   else if($answer_count == 1)
                      e((1 answer));
                   else
                      e((.$answer_count. answers));
                 ?
             /dd
          ?php endforeach; ?
          /dl
       

Re: How can I view the screencasts on Linux

[Marcelo Andrade]

On Sun, Dec 7, 2008 at 10:09 PM, Ed Howland [EMAIL PROTECTED] wrote:

 I am having trouble playing the screencasts on the cakephp.org site. I
 am using Ubuntu 8.10 w/FF.

 Does anyone have any suggestions?

What kind of problems?

--
MARCELO DE F. ANDRADE (aka eleKtron)
Belem, PA, Amazonia, Brazil
Linux User #221105

[EMAIL PROTECTED] ~]# links http://pa.slackwarebrasil.org/

For Libby's backstory be told onLost
http://www.petitiononline.com/libby423/petition.html

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Re: ORM, fetching related rows of corresponding model: undefined index..

[thatsgreat2345]

recursive = 'all'? Should be set to an integer 
http://book.cakephp.org/view/439/recursive

On Dec 7, 7:19 pm, Tanay [EMAIL PROTECTED] wrote:
 i have done that..
 here is my full questions_controller.php

     ?php
     class QuestionsController extends AppController {
        var $name = 'Questions';
        function home() {
           $this-Question-recursive = 'all';
           $this-set('questions', $this-Question-find('all'));
        }
        function show( $id = null) {
           if (!$id) {
              $this-Session-setFlash('Invalid Question.');
              $this-redirect(array('action'='home'));
           }
           $this-set('question', $this-Question-read(null, $id));
        }
     }
     ?

 ===

 On Dec 8, 12:10 am, Rob [EMAIL PROTECTED] wrote:

  Well based on your post, you are not getting the related model, and it
  appears you've only done a find on the Questions without setting
  recursive in your find('all'), try this:

   $this-set('questions', $this-Question-find('all', array
  ('recursive' = 1)));

  On Dec 7, 9:55 am, Tanay [EMAIL PROTECTED] wrote:

   MODEL answers.php

       ?php
       class Answer extends AppModel {
          var $name = 'Answer';
          var $belongsTo = array(
             'Question' = array(
                'className'    = 'Question'
             )
          );
       }
       ?

   Model questions.php

       ?php
       class Question extends AppModel {
          var $name = 'Question';
          var $hasMany = array(
             'Answer' = array(
                'className'     = 'Answer'
             )
          );
       }
       ?

   ===

   On Dec 7, 11:00 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

Post your model code so we can see if you are doing your associations
correctly, or if they are spelled correctly.

On Dec 6, 9:50 pm, Tanay [EMAIL PROTECTED] wrote:

 it seems no relationships are working with my cakephp.
 has it got anything to do with my version of php5?

 i have got struck at this for a long time.
 help will be very greatly appreciated.

 here is the var_dump of $questions as seen in home.ctp
 array(2) { [0]= array(1) { [Question]= array(5) { [id]= string
 (1) 1 [question]= string(22) Why do you use
 CakePHP [questioner]= string(5) Ahsan [created]= string(19)
 2008-02-11 22:19:04 [modified]= string(19) 2008-02-11
 22:19:04 } } [1]= array(1) { [Question]= array(5) { [id]=
 string(1) 2 [question]= string(25) Why won't you use
 CakePHP [questioner]= string(5) Ahsan [created]= string(19)
 2008-02-11 22:22:06 [modified]= string(19) 2008-02-11
 22:22:06 } } }

 coresponding queries sql dump is
 DESCRIBE `questions`
 SELECT `Question`.`id`, `Question`.`question`,
 `Question`.`questioner`, `Question`.`created`, `Question`.`modified`
 FROM `questions` AS `Question` WHERE 1 = 1

 On Dec 7, 1:47 am, thatsgreat2345 [EMAIL PROTECTED] wrote:

  I see you are using the Packt Publishing book, ensure that your 
  model
  relations aren't misspelled or anything else it won't load the 
  answers
  and thus won't exist, in the controller do a debug on the find, that
  or in the view do a pr($questions) before the foreach this will
  display the array and see what exactly what is in the array.

  On Dec 6, 9:40 am, Tanay [EMAIL PROTECTED] wrote:

   Tanay wrote:
i have two dbs questions , and answers..

corresponding models

questions has many answers
answers belogns to questions

in questions_controller, i have this home function
 function home() {
          $this-Question-recursive = 1;
          $this-set('questions', $this-Question-find('all'));
       }

in my home.ctp, i tried to show the corresponding answers using 
the
code below
i took a question from an array of questions using foreach
foreach question, $question['Answer'] is expected to give array 
of
answers, but it says  Undefined index:  Answer
THAT LIne is marked in code below

home.ctp:

    ?php if(empty($questions)): ?
       p class=no_answerNo Questions yet. Be the first one 
to
                 post a Question!/p
    ?php else: ?
       dl
       ?php foreach ($questions as $question): ?
          dtspan?php 
e($question['Question']['questioner']);
           ?/span/dt
                     dd
             ?php 
e($html-link($question['Question']['question']
    .'?', array('action' = 'show', 
$question['Question']['id']))); ?
             ?php
                $answer_count = count($question
['Answer']);   //  -ERROR HERE
                if(!$answer_count)
                  

Re: How can I view the screencasts on Linux

[Ed Howland]

On Sun, Dec 7, 2008 at 9:32 PM, Marcelo Andrade [EMAIL PROTECTED] wrote:
 On Sun, Dec 7, 2008 at 10:09 PM, Ed Howland [EMAIL PROTECTED] wrote:

 I am having trouble playing the screencasts on the cakephp.org site. I
 am using Ubuntu 8.10 w/FF.

 Does anyone have any suggestions?

 What kind of problems?

It just says video loading and then quits. Do you know what kind of
video format it is?

Thanks
Ed


 --
 MARCELO DE F. ANDRADE (aka eleKtron)
 Belem, PA, Amazonia, Brazil
 Linux User #221105

 [EMAIL PROTECTED] ~]# links http://pa.slackwarebrasil.org/

 For Libby's backstory be told onLost
 http://www.petitiononline.com/libby423/petition.html

 




-- 
Ed Howland
http://greenprogrammer.blogspot.com
http://twitter.com/ed_howland

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Re: How can I view the screencasts on Linux

[thatsgreat2345]

It is quicktime.

On Dec 7, 8:47 pm, Ed Howland [EMAIL PROTECTED] wrote:
 On Sun, Dec 7, 2008 at 9:32 PM, Marcelo Andrade [EMAIL PROTECTED] wrote:
  On Sun, Dec 7, 2008 at 10:09 PM, Ed Howland [EMAIL PROTECTED] wrote:

  I am having trouble playing the screencasts on the cakephp.org site. I
  am using Ubuntu 8.10 w/FF.

  Does anyone have any suggestions?

  What kind of problems?

 It just says video loading and then quits. Do you know what kind of
 video format it is?

 Thanks
 Ed



  --
  MARCELO DE F. ANDRADE (aka eleKtron)
  Belem, PA, Amazonia, Brazil
  Linux User #221105

  [EMAIL PROTECTED] ~]# linkshttp://pa.slackwarebrasil.org/

  For Libby's backstory be told onLost
 http://www.petitiononline.com/libby423/petition.html

 --
 Ed Howlandhttp://greenprogrammer.blogspot.comhttp://twitter.com/ed_howland
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Error when submitting form to a controller using different model name

[Tony]

I'm in the process of creating a website that is for my affiliates and
their information is stored into a model called 'User' and I have a
controller called businesses with two actions;  index and thank_you.

When a prospect lands onto the page they are given a sales letter and
a form to fill out. This form will use the User model but I want it to
submit to the businesses thank_you action.

I'm putting the prospects information into the Users model because
they could be converted into an affiliate.

Where I'm having the problem is that currently my form code looks like
this:

===CODE===
echo $form-create('User', array('url' = '/businesses/thank_you'));
echo $form-hidden('parent_id', array('value' = $member['User']
['affiliate_number']));
echo 'span class=required*/span spanName:/span';
echo $form-input('first_name', array('label' = false, 'size' =
20));
echo 'span class=required*/span spanEmail:/span';
echo $form-input('email', array('label' = false, 'size' = 20));
echo 'br /';
echo 'spanstrongPhone: em(optional, for those who really want
change)/em/strong/spanbr /';
echo $form-input('phone', array('label' = false));
echo 'br /';
echo $form-end(array('label' = 'I Want Change in My Life'));
===CODE===

The problem is that when I submit this form the url at the top is
correct:
http://example.com/businesses/thank_you

But I get this error:

Not Found

Error: The requested address '/businesses/thank_you' was not found on
this server.


What am I doing wrong here?

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Re: Error when submitting form to a controller using different model name

[thatsgreat2345]

Do you have a controller called businesses_controller.php and in that
a function called thank_you?

On Dec 7, 9:37 pm, Tony [EMAIL PROTECTED] wrote:
 I'm in the process of creating a website that is for my affiliates and
 their information is stored into a model called 'User' and I have a
 controller called businesses with two actions;  index and thank_you.

 When a prospect lands onto the page they are given a sales letter and
 a form to fill out. This form will use the User model but I want it to
 submit to the businesses thank_you action.

 I'm putting the prospects information into the Users model because
 they could be converted into an affiliate.

 Where I'm having the problem is that currently my form code looks like
 this:

 ===CODE===
 echo $form-create('User', array('url' = '/businesses/thank_you'));
 echo $form-hidden('parent_id', array('value' = $member['User']
 ['affiliate_number']));
 echo 'span class=required*/span spanName:/span';
 echo $form-input('first_name', array('label' = false, 'size' =
 20));
 echo 'span class=required*/span spanEmail:/span';
 echo $form-input('email', array('label' = false, 'size' = 20));
 echo 'br /';
 echo 'spanstrongPhone: em(optional, for those who really want
 change)/em/strong/spanbr /';
 echo $form-input('phone', array('label' = false));
 echo 'br /';
 echo $form-end(array('label' = 'I Want Change in My Life'));
 ===CODE===

 The problem is that when I submit this form the url at the top is
 correct:http://example.com/businesses/thank_you

 But I get this error:

 Not Found

 Error: The requested address '/businesses/thank_you' was not found on
 this server.

 What am I doing wrong here?
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Re: Auth-login() Not working correctly?

[gearvOsh]

I tried removing the $this-Auth-login() and now it doesnt do
anything. No query is fired.
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Re: Auth-login() Not working correctly?

[gearvOsh]

Actually nevermind. The $this-data had to be under User, not Form.
Now the login is submitting and the query is being fired... BUT now
everytime I hit login the password field goes empty and it never
works.

On Dec 7, 9:50 pm, gearvOsh [EMAIL PROTECTED] wrote:
 I tried removing the $this-Auth-login() and now it doesnt do
 anything. No query is fired.
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Re: How can I view the screencasts on Linux

[Alexandru Ciobanu]

Ed Howland wrote:
 I am having trouble playing the screencasts on the cakephp.org site. I
 am using Ubuntu 8.10 w/FF.

 Does anyone have any suggestions?

 Thanks
 Ed

   
I'm using the Totem mozplugin to do this on Fedora 10.

On my notebook I'm using mplayer and mozplugger.

If you can't get the videos to play in your browser,look at the HTML 
source, search for the .mov files and play them with your favorite player.

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Ajax and Sessions

[JermWorm]

I'm trying to upgrade and app from an older beta of cake to RC3.

In my app_controller, I have a before filter that checks whether the
user is logged in via a session variable, but in the course of
debugging I've noticed that the session variable isn't there(i.e.
session not started?) when ajax requests are being processed, it is
there when normal requests are being processed, I set security in
core.php to low (it was on medium) with no difference, $helpers/
$components definetly includes 'Session'.

Am I missing something in the update I need to do I have looked
through the latest api/manuals but can't see why the session start is
being bypassed.

Regards
Jeremy
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Re: How to add line break in text mail layout?

[Hipnotik]

On 6 Gru, 17:32, majna [EMAIL PROTECTED] wrote:
 Just write \n
 Examples:
 ?php echo \n;?
 ?php echo $name . \n;?

Wow! It works!
I don't know why, but it works correct. The only thing I had to do was
moving \n into inside of echo function.

Does not work directly in template:
Something: ?php e($Model['field']); ?\n

but this code works correct:
Something: ?php e($Model['field']).\n; ?\n

Thanks all for help ;)

 On Dec 5, 1:44 pm, dr. Hannibal Lecter [EMAIL PROTECTED] wrote:



  Well...that's not supposed to happen. Can you check the mail source
  and mime headers? Maybe you're not sending your email in plain text..?

  On Dec 5, 1:11 pm, Hipnotik [EMAIL PROTECTED] wrote:

   On 5 Gru, 12:35, dr. Hannibal Lecter [EMAIL PROTECTED] wrote:

What about simply hitting the enter key? :)

   Doesn't work... displays all lines in one :(

On Dec 5, 12:14 pm, Hipnotik [EMAIL PROTECTED] wrote:

 How to add line break in text mail layout?

 It's in elements/email/text/default.ctp file.
 I tried to use:
 \n
 \r\n
 \r
 and still it doesn't work correct.
 It breaks the lines but new line character (i.e. \n) appears in the
 received message too.

 How to do that?
 Thanks for help ;)
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Re: How to add line break in text mail layout?

[Hipnotik]

On 6 Gru, 17:32, majna [EMAIL PROTECTED] wrote:
 Just write \n
 Examples:
 ?php echo \n;?
 ?php echo $name . \n;?

Wow! It works!
I don't know why, but it works correct. The only thing I had to do
was
moving \n into inside of echo function.

Does not work directly in template:
Something: ?php e($Model['field']); ?\n

but this code works correct:
Something: ?php e($Model['field']).\n; ?

Thanks all for help ;)

 On Dec 5, 1:44 pm, dr. Hannibal Lecter [EMAIL PROTECTED] wrote:



  Well...that's not supposed to happen. Can you check the mail source
  and mime headers? Maybe you're not sending your email in plain text..?

  On Dec 5, 1:11 pm, Hipnotik [EMAIL PROTECTED] wrote:

   On 5 Gru, 12:35, dr. Hannibal Lecter [EMAIL PROTECTED] wrote:

What about simply hitting the enter key? :)

   Doesn't work... displays all lines in one :(

On Dec 5, 12:14 pm, Hipnotik [EMAIL PROTECTED] wrote:

 How to add line break in text mail layout?

 It's in elements/email/text/default.ctp file.
 I tried to use:
 \n
 \r\n
 \r
 and still it doesn't work correct.
 It breaks the lines but new line character (i.e. \n) appears in the
 received message too.

 How to do that?
 Thanks for help ;)
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