hasMany through: my neverending problem?
I have two tables: attractions accessibilities. I want to display
which people can go on which rides so I invented a new, joined table called
accessibilities_attractions. Here's a visual representation of it:
https://lh6.googleusercontent.com/-nvhLQ93GqnY/Ut6PLdpucjI/ADU/ZHLBmBMJWB4/s1600/table.jpg
Data in table attraction (id - name) can be:
1 - Rollercoaster name
2 - Big Rollercoaster name
3 - Realy Big Rollercoaster name
4 - Biggest Rollercoaster ever seen name
Table accessibilities (id - name) has exactly 3 records:
1 - Forbidden
2 - Accessible with supervisor
3 - Accessible
Data in accessibilities_attractions (accessibility_id - attraction_id -
min - max) can be:
Forbidden - Big Rollercoaster name - null - 110
Accessible with supervisor - Big Rollercoaster name - 110 - 130
Accessibile - Big Rollercoaster name - 130 - null
Forbidden - Biggest Rollercoaster ever seen name - Forbidden - null - 140
Accessibile - Biggest Rollercoaster ever seen name - 140 - null
As I am developping a CMS system, I've made a plugin called CoasterCMS.
By this way, I implement a whole CMS system to CRUD the website elements of
my favourite theme park (e.g. attractions, animals, events, ...)
To go further, I've created three models:
*Accessibility.php*
?php
class Accessibility extends CoasterCmsAppModel
{
public $hasMany = array(
'CoasterCms.AccessibilityAttraction'
);
}
*Attraction.php*
class Attraction extends CoasterCmsAppModel
{
...
public $hasMany = array(
'CoasterCms.AccessibilityAttraction'
);
...
}
*AccessibilityAttraction.php*
?php
class AccessibilityAttraction extends CoasterCmsAppModel
{
public $belongsTo = array(
'CoasterCms.Accessibility', 'CoasterCms.Attraction'
);
}
Now, I've created a model so I can insert, change and delete records in
table accessibilities_attractions:
*AccessibilitiesAttractionsController.php*
?php
class AccessibilitiesAttractionsController extends CoasterCmsAppController
{
public $helpers = array('Html', 'Form', 'Session');
public $components = array('Session');
public function index()
{
debug($this-AccessibilityAttraction-find('all'));
}
...
}
And that's the point where my problem comes in. Every time I try to get the
records via the find('all') method in my index action, I'm not getting an
object at all:
Error: Call to a member function find() on a non-object
File: D:\Websites\BellewaerdeFun 2014\03 -
Online\app\Plugin\CoasterCms\Controller\AccessibilitiesAttractionsController.php
Line: 10
I realy don't understand what goes wrong, because I ordered myself to stick
on the CakePHP name conventions so nothing can go wrong...
Can anyone see what I'm doing wrong? After 2 days of searching, I cannot
povide the solution on my own...
Thanks in advance :)
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Little trick to return json
Sorry for the poor english.
I created a piece of code that solved my problem in returning a JSON object
and helped me not to re-write all views. I await your feedback pros and
cons.
class AppController extends Controller {
public $components = array('RequestHandler');
function beforeRender() {
if ($this-RequestHandler-*isAjax*()) {
$this-layout = '*ajax*';
$this-RequestHandler-setContent('*json*', '
*application/json*');
$this-*viewVars *= json_encode($this-*viewVars*);
*return false*;
}
}
}
If an Ajax request is performed automatically be returns a JSON object with
the query result.
$.ajax({
url: http://localhost/aplication/users;,
type: POST,
dataType: *json*
});
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cookbook
As I am studying CakePHP, the 2.0 cookbook is essential. I discovered many places in it where the level of knowledge needed to understand a topic is raised. In these conditions I think it's a heroic act to start learning Cake. I propose to you that the cookbook or any other introductory manual be organized as a wiki project. Something like the php manual in html form, with many comments at the base of the page: extremely useful. The learning curve could be steeper. The overall productivity and user base will increase. With a translation tool on each page. With many links to explain the terms. -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out.
Re: CakePHP bake command not working in CLI
In Ubuntu Linux I had to install it using the following command: apt-get install cakephp-scripts then it worked: cake bake luni, 30 septembrie 2013, 16:40:25 UTC+3, Victor Musvibe a scris: I have got an application that i am working on built using Cakephp 1.3 problem that i am having is that i am failing to bake using CLI. I am using Windows 7 and xampp as my localserver. This is how i have added my Environment Variables. C:\xampp\php;C:\xampp\htdocs\myappname\cake\console\ When i type cake command in the CLI i get an error 'cake' is not recognized as an internal or external command, operable program or batch file. Does anyone have an idea on how to fix this. I have also asked the same question on SO and i dont seem to get answer - http://stackoverflow.com/questions/19094349/cakephp-bake-command-not-working-in-cli -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out.
HMVC vs MVC
I kindly ask you all if HMVC (Hierarhical MVC) in cakePHP is actually MVC with controllers+actions. -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out.
RE: cookbook
No one is stopping you to do so. Dave Maharaj Freelance Designer | Developer Description: header_logo www.movepixels.com | mailto:d...@movepixels.com d...@movepixels.com | 709.800.0852 From: cake-php@googlegroups.com [mailto:cake-php@googlegroups.com] On Behalf Of MihaiTL Sent: Tuesday, January 21, 2014 8:10 PM To: cake-php@googlegroups.com Subject: cookbook As I am studying CakePHP, the 2.0 cookbook is essential. I discovered many places in it where the level of knowledge needed to understand a topic is raised. In these conditions I think it's a heroic act to start learning Cake. I propose to you that the cookbook or any other introductory manual be organized as a wiki project. Something like the php manual in html form, with many comments at the base of the page: extremely useful. The learning curve could be steeper. The overall productivity and user base will increase. With a translation tool on each page. With many links to explain the terms. -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out. -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out. image001.jpg
Re: cookbook
Hello Mihaitl, In order to learn cake or any other framework, you should first had knowledge of OO programming, i really dont feel like aa heroe...haha, and im pretty new to cake. Give it a try again, it is really simple to start and the learning curve is quite short. Have my best regards. 2014/1/21 MihaiTL maft...@gmail.com As I am studying CakePHP, the 2.0 cookbook is essential. I discovered many places in it where the level of knowledge needed to understand a topic is raised. In these conditions I think it's a heroic act to start learning Cake. I propose to you that the cookbook or any other introductory manual be organized as a wiki project. Something like the php manual in html form, with many comments at the base of the page: extremely useful. The learning curve could be steeper. The overall productivity and user base will increase. With a translation tool on each page. With many links to explain the terms. -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out. -- *Carlos Baeza Negroni* +56985644026 http://carlosbaeza.net @cjbaezilla http://cl.linkedin.com/in/cjbaeza http://cl.linkedin.com/in/cjbaeza -- Like Us on FaceBook https://www.facebook.com/CakePHP Find us on Twitter http://twitter.com/CakePHP --- You received this message because you are subscribed to the Google Groups CakePHP group. To unsubscribe from this group and stop receiving emails from it, send an email to cake-php+unsubscr...@googlegroups.com. To post to this group, send email to cake-php@googlegroups.com. Visit this group at http://groups.google.com/group/cake-php. For more options, visit https://groups.google.com/groups/opt_out.
Re: Router redirect pass all parameters to another domian
As Reuben said - you will be far better off with a Web Server based
redirect.
This way your application will not get called at all.
On Monday, 20 January 2014 23:49:09 UTC+2, gonzela2006 wrote:
Hello,
I want to use *Router::redirect* to redirect to another domain and pass
all parameters to this domain
Example:
Redirect this url
http://domain-*one*.com/posts/view/1
to
http://domain-*two*.com/posts/view/1
I tried the following
Router::redirect( '/*', 'http://domain-*two*.com/', array('persist' =
true) );
but it redirect to http://domain-*two*.com/ without parameters.
Please advise
Thanks
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RE: Little trick to return json
Where is the magic?
What's so special?
Simply spitting out the results in JSON……… and?
Dave Maharaj
Freelance Designer | Developer
Description: header_logo
www.movepixels.com | mailto:d...@movepixels.com d...@movepixels.com |
709.800.0852
From: cake-php@googlegroups.com [mailto:cake-php@googlegroups.com] On Behalf Of
Rubensvsx
Sent: Tuesday, January 21, 2014 12:29 PM
To: cake-php@googlegroups.com
Subject: Little trick to return json
Sorry for the poor english.
I created a piece of code that solved my problem in returning a JSON object and
helped me not to re-write all views. I await your feedback pros and cons.
class AppController extends Controller {
public $components = array('RequestHandler');
function beforeRender() {
if ($this-RequestHandler-isAjax()) {
$this-layout = 'ajax';
$this-RequestHandler-setContent('json', 'application/json');
$this-viewVars = json_encode($this-viewVars);
return false;
}
}
}
If an Ajax request is performed automatically be returns a JSON object with the
query result.
$.ajax({
url: http://localhost/aplication/users;,
type: POST,
dataType: json
});
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image001.jpg
Re: Little trick to return json
This is an excellent idea. Thx for sharing. Sent from my BlackBerry 10 smartphone on the TELUS network. From: RubensvsxSent: Wednesday, January 22, 2014 4:07 AMTo: cake-php@googlegroups.comReply To: cake-php@googlegroups.comSubject: Little trick to return jsonSorry for the poor english.I created a piece of code that solved my problem in returning a JSON object and helped me not to re-write all views. I await your feedback pros and cons.class AppController extends Controller { public $components = array('RequestHandler'); function beforeRender() {if ($this-RequestHandler-isAjax()) { $this-layout = 'ajax'; $this-RequestHandler-setContent('json', 'application/json'); $this-viewVars = json_encode($this-viewVars); return false;} }}If an Ajax request is performed automatically be returns a JSON object with the query result.$.ajax({ url: "http://localhost/aplication/users", type: "POST", dataType: "json"});
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Re: Little trick to return json
You could cut down your code by almost have and drop every second line when
following the documentation regarding JsonView:
http://book.cakephp.org/2.0/en/views/json-and-xml-views.html
No need to fight the framework and manually set response type and alike
Also, I highly recommend using /users.json as request url.
This way you also trigger a lot of Cake magic as well. And you are down to
none of your code above, as CakePHP would do all that for you right away.
See www.dereuromark.de/2014/01/09/ajax-and-cakephp
Am Dienstag, 21. Januar 2014 16:59:11 UTC+1 schrieb Rubensvsx:
Sorry for the poor english.
I created a piece of code that solved my problem in returning a JSON
object and helped me not to re-write all views. I await your feedback pros
and cons.
class AppController extends Controller {
public $components = array('RequestHandler');
function beforeRender() {
if ($this-RequestHandler-*isAjax*()) {
$this-layout = '*ajax*';
$this-RequestHandler-setContent('*json*', '
*application/json*');
$this-*viewVars *= json_encode($this-*viewVars*);
*return false*;
}
}
}
If an Ajax request is performed automatically be returns a JSON object
with the query result.
$.ajax({
url: http://localhost/aplication/users;,
type: POST,
dataType: *json*
});
--
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Find us on Twitter http://twitter.com/CakePHP
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