I am trying to implement cakedc search plugin into my cakephp 2 application. I had the plugin working correctly at one point and something else in the application has knocked it off. However I would like to check that I am going about using the search pluging the right way as it could be the method I am using that is causing a conflict or something similar.
The search is only searching for one field which is order_id from the order model within the orders controller.
In my model I have :
// Search Filters
public $filterArgs = array(
array('name' => 'order_id', 'type' => 'like')
);
Within my controller I have:
public $presetVars = true;
public $components = array('Search.Prg', 'RequestHandler');
public $uses = array('order', 'product');
public function find () {
$this->Prg->commonProcess();
//debug($this->Order->parseCriteria($this->passedArgs));
$this->paginate = array('conditions' => $this->Order->parseCriteria($this->passedArgs));
$this->set('orders', $this->paginate());
}
EDIT this is happening because I am using the $uses variable in my class to define the controller models. Does anyone know how to define the cakedc search model. I have tried search, searchable and searchablebehavior
Try to stick to conventions. Note the casing:
public $uses = array('Order', 'Product');.
Also note: The first one will be your primary model here.
Related
I have taken over a project for cakePHP, I am new and I find it not easy, Magento should be difficult but I find cakePHP more difficult, but maybe I have not reach the moment I know it ...
I have the next model (for table postcodes):
public $belongsTo = array(
'Postcode' => array(
'className' => 'Postcode',
'foreignKey' => 'postcode_id',
'conditions' => '',
'fields' => '',
'order' => ''
),
Everything works, but with the tool AgentRansack I can't find the model Postcode. And besides the name of the table is postcodes, I can't even find a relation for Postcode and postcodes.
How can such a model setup in a different way than with a class Postcode.php ?
When requesting models for which no concrete model class exists (or cannot be found for whatever reason), dynamic model objects will be generated from the AppModel class.
From the CakePHP Cookbook:
CakePHP will dynamically create a model object for you if it cannot find a corresponding file in /app/Model. This also means that if your model file isn’t named correctly (for instance, if it is named ingredient.php or Ingredients.php rather than Ingredient.php), CakePHP will use an instance of AppModel rather than your model file (which CakePHP assumes is missing). If you’re trying to use a method you’ve defined in your model, or a behavior attached to your model, and you’re getting SQL errors that are the name of the method you’re calling, it’s a sure sign that CakePHP can’t find your model and you need to check the file names, your application cache, or both.
http://book.cakephp.org/2.0/en/models.html#understanding-models
By the sounds of it you need to create a class for your 'postcodes',
Read the following: http://book.cakephp.org/2.0/en/models.html
But it should look something like this:
// app/Model/Postcode.php
App::uses('AppModel', 'Model');
class PostCode extends AppModel {
public $validate = array(
// Validation
);
// Replace 'ModelName' with the name of the class where the code sites you mentioned above
public $hasMany = array(
'ModelName' => array(
'className' => 'ModelName',
'foreignKey' => 'postcode_id'
)
);
}
Can I use another Model inside one model?
Eg.
<?php
class Form extends AppModel
{
var $name='Form';
var $helpers=array('Html','Ajax','Javascript','Form');
var $components = array( 'RequestHandler','Email');
function saveFormName($data)
{
$this->data['Form']['formname']=$data['Form']['formname'];
$this->saveField('name',$this->data['Form']['formname']);
}
function saveFieldname($data)
{
$this->data['Attribute']['fieldname']=$data['Attribute']['fieldname'];
}
}
?>
Old thread but I'm going to chime in because I believe the answers to be incomplete and lacking in "why". CakePHP has three ways to load models. Though only two methods work outside of a Controller, I'll mention all three. I'm not sure about version availability but this is core stuff so I believe they'll work.
App::import() only finds and require()s the file and you'll need to instantiate the class to use it. You can tell import() the type of class, the name and file path details.
ClassRegistry::init() loads the file, adds the instance to the object map and returns the instance. This is the better way to load something because it sets up "Cake" things as would happen if you loaded the class through normal means. You can also set an alias for the class name which I've found useful.
Controller::loadModel() uses ClassRegistry::init() as well as adds the Model as a property of the controller. It also allows $persistModel for model caching on future requests. This only works in a Controller and, if that's your situation, I'd use this method before the others.
You can create instances of other models from within any model/controller using one of these two methods.
If you're using Cake 1.2:
App::import('model','Attribute');
$attr = new Attribute();
$attr->save($dataYouWantToSavetoAttribute);
If you're using Cake 1.1:
loadModel('Attribute');
$attr = new Attribute();
$attr->save($dataYouWantToSavetoAttribute);
An obvious solution everyone missed is to create an association between two models, if appropriate. You can use it to be able to reference one model from inside another.
class Creation extends AppModel {
public $belongsTo = array(
'Inventor' => array(
'className' => 'Inventor',
'foreignKey' => 'inventor_id',
)
);
public function whoIsMyMaker() {
$this->Inventor->id = $this->field('inventor_id');
return $this->Inventor->field('name');
}
}
In CakePHP 1.2, it's better to use:
ClassRegistry::init('Attribute')->save($data);
This will do simply
<?php
class Form extends AppModel
{
//...
$another_model = ClassRegistry::init('AnotherModel');
//...
}
?>
In CakePHP 3 we may use TableRegistry::get(modelName)
use Cake\ORM\TableRegistry;
$itemsOb = TableRegistry::get('Items');
$items = $itemsOb->find("all");
debug($items);
If you want to use Model_B inside Model_A, add this line at the beginning of Model_A file:
App::uses('Model_B_ClassName', 'Model');
and then you will be able to use it inside Model_A. For example:
$Model_B = new Model_B_ClassName();
$result = $Model_B->findById($some_id);
var $uses = array('ModeloneName','ModeltwoName');
By using $uses property, you can use multiple models in controller instead of using loadModel('Model Name').
App::import('model','Attribute');
is way to use one model into other model. Best way will be to used association.
As an example lets imagine we have a simple tv show database. Show and Episode as Model.
An Episode belongsTo one Show and one Show hasMany Episodes.
In the episodes/index view we are just echoing all episodes, the same goes for the shows/index view. But I also want to echo lets say the first 5 episodes of each show (just the title). I could simply limit the episodes by setting the limit attribute for the hasMany association.
In shows/episode/x(id) view I want to echo all episodes. And therefore I can't simply use the limit attribute for the hasMany association since it is view dependent.
What solution should I choose to implement that? I could only archive that by using some "dirty workarounds/hacks" but I feel like this is an usual problem and there might be some actual solution.
I believe what you are looking for is the containable behaviour.
Read the doc:
http://book.cakephp.org/2.0/en/core-libraries/behaviors/containable.html
Then remove any limit to your associations. Below there is a way of how you can use containable behavior in your example.
class Shows extends AppModel {
public $actsAs = array('Containable');
}
class ShowsController extends AppController {
//Bring all the shows and 5 episodes
public function index(){
$this->Show->find('all', array('contain' => array(
'Episode' => array('limit' => 5)
)));
}
public function view($id){
//Bring the show
$this->Show->findById($id);
//Then bring the episodes of the show
$this->Show->Episode->findByShowId($id);
//Or you can use
$this->Show->find('all', array(
'contain' => array('Episode')),
'conditions' => array('id' => $id)
);
}
}
So I have a company based system and I am keeping all data within the one database and I separate the data by using a site_id field which is present in all tables and in the users table.
Now at the moment I am doing a condition on every single find('all'). Is there a more global way to do this. Maybe in the AppController? Even for saving data I am having to set the site_id every single save.
Something like
public function beforeFilter() {
parent::beforeFilter();
$this->set('site_id', $this->Auth->user('site_id'));
}
Any direction would only help. Thanks
TLDR:
Create a Behavior with a beforeFind() method that appends a site_id condition to all queries.
Example/Details:
Create a behavior something along the lines of the below. Note, I'm getting the siteId from a Configure variable, but feel free to get that however you want.
<?php
class SiteSpecificBehavior extends ModelBehavior {
public function setup(Model $Model, $settings = array()) {
$Model->siteId = Configure::read('Site.current.id');
}
public function beforeFind(Model $Model, $query = array()) {
$siteId = $Model->siteId;
$query['conditions'][$Model->alias . '.site_id'] = $siteId ;
return $query;
}
}
Then on any/all models where you want to make sure it's site-specific, add:
public $actsAs = array('SiteSpecific');
This can be tweaked and improved up certainly, but it should give you a good idea.
Can I use another Model inside one model?
Eg.
<?php
class Form extends AppModel
{
var $name='Form';
var $helpers=array('Html','Ajax','Javascript','Form');
var $components = array( 'RequestHandler','Email');
function saveFormName($data)
{
$this->data['Form']['formname']=$data['Form']['formname'];
$this->saveField('name',$this->data['Form']['formname']);
}
function saveFieldname($data)
{
$this->data['Attribute']['fieldname']=$data['Attribute']['fieldname'];
}
}
?>
Old thread but I'm going to chime in because I believe the answers to be incomplete and lacking in "why". CakePHP has three ways to load models. Though only two methods work outside of a Controller, I'll mention all three. I'm not sure about version availability but this is core stuff so I believe they'll work.
App::import() only finds and require()s the file and you'll need to instantiate the class to use it. You can tell import() the type of class, the name and file path details.
ClassRegistry::init() loads the file, adds the instance to the object map and returns the instance. This is the better way to load something because it sets up "Cake" things as would happen if you loaded the class through normal means. You can also set an alias for the class name which I've found useful.
Controller::loadModel() uses ClassRegistry::init() as well as adds the Model as a property of the controller. It also allows $persistModel for model caching on future requests. This only works in a Controller and, if that's your situation, I'd use this method before the others.
You can create instances of other models from within any model/controller using one of these two methods.
If you're using Cake 1.2:
App::import('model','Attribute');
$attr = new Attribute();
$attr->save($dataYouWantToSavetoAttribute);
If you're using Cake 1.1:
loadModel('Attribute');
$attr = new Attribute();
$attr->save($dataYouWantToSavetoAttribute);
An obvious solution everyone missed is to create an association between two models, if appropriate. You can use it to be able to reference one model from inside another.
class Creation extends AppModel {
public $belongsTo = array(
'Inventor' => array(
'className' => 'Inventor',
'foreignKey' => 'inventor_id',
)
);
public function whoIsMyMaker() {
$this->Inventor->id = $this->field('inventor_id');
return $this->Inventor->field('name');
}
}
In CakePHP 1.2, it's better to use:
ClassRegistry::init('Attribute')->save($data);
This will do simply
<?php
class Form extends AppModel
{
//...
$another_model = ClassRegistry::init('AnotherModel');
//...
}
?>
In CakePHP 3 we may use TableRegistry::get(modelName)
use Cake\ORM\TableRegistry;
$itemsOb = TableRegistry::get('Items');
$items = $itemsOb->find("all");
debug($items);
If you want to use Model_B inside Model_A, add this line at the beginning of Model_A file:
App::uses('Model_B_ClassName', 'Model');
and then you will be able to use it inside Model_A. For example:
$Model_B = new Model_B_ClassName();
$result = $Model_B->findById($some_id);
var $uses = array('ModeloneName','ModeltwoName');
By using $uses property, you can use multiple models in controller instead of using loadModel('Model Name').
App::import('model','Attribute');
is way to use one model into other model. Best way will be to used association.