User Model has relation:
public $hasMany = array(
'MyRecipe' => array(
'className' => 'Recipe',
)
);
I want to select all users who have recipes with ID: 1,2
How I can use that conditions in select:
$this->User->find('all', array(
'conditions' => array(
'Recipe.Id' => [1,2]
)
));
But in this example I will get also Users without recipes, how to prevent that ?
please give this relation in User model
class User extends AppModel
{
var $name = 'User';
var $belongsTo = array("Recipe");
}
and in user controller your query as
$list = $this->User->find('all',array("conditions"=>array("recipe_id IN"=> [1,2] )));
its gives output which you want..
Related
My Category Model:
class Category extends AppModel {
public $displayField = 'name';
// public $actsAs = array('Containable');
public $hasAndBelongsToMany = array(
'Post' => array(
'className' => 'Post',
'joinTable' => 'categories_postss',
'foreignKey' => 'category_id',
'associationForeignKey' => 'post_id',
'unique' => 'keepExisting'
)
);
}
$params['contain'] = array('Post' => array(
'limit'=> 3));
pr($this->Category->find('first',$params)); exit;
It is fetching all Posts, irrespective of limit.
What I want to do:
I have this page where I ma listing all the categories and latest 5 posts related to it.
I want to limit the associated model to only 5 rows.
Any ideas?
Containable behavior is not in use
The most likely reason for this problem is that the containable behavior is not being used at all.
Compare, for the below code example:
$results = $this->Category->find('first', array(
'contain' => array(
'Post' => array(
'limit' => 3
)
)
));
Without containable behavior, it'll generate the following queries:
SELECT ... FROM `crud`.`categories` AS `Category` WHERE 1 = 1 LIMIT
SELECT ... FROM `crud`.`posts` AS `Post`
JOIN `crud`.`categories_posts` AS `CategoriesPost` ON (...)
With containable behavior, it'll generate the following queries:
SELECT ... FROM `crud`.`categories` AS `Category` WHERE 1 = 1 LIMIT
SELECT ... FROM `crud`.`posts` AS `Post`
JOIN `crud`.`categories_posts` AS `CategoriesPost` ON (...) LIMIT 3
Given this (and the code in the question) check that the AppModel has the containable behavior in $actsAs:
<?php
// app/Model/AppModel.php
class AppModel extends Model {
public $actsAs = array('Containable');
}
Limit always required?
Alternatively, or possibly in addition, you may prefer to put a limit in the association definition - To do so just define the 'limit' key:
class Category extends AppModel {
public $hasAndBelongsToMany = array(
'Post' => array(
'limit' => 100, // default to a high but usable number of results
)
);
}
the hasAndBelongsToMany relationship seems unnecessary to me. I think you only need Category hasMany Post and Post belongsTo Category relationships. Add category_id to the posts table. Make both models actAs containable.
Post Model
class Post extends AppModel {
public $actsAs = array('Containable');
var $belongsTo = array(
'Category' => array(
'className' => 'Category',
'foreignKey' => 'category_id'
),
// ... more relationships
);
Category Model
class Category extends AppModel {
public $actsAs = array('Containable');
var $hasMany = array(
'Post' => array(
'className' => 'Post',
'foreignKey' => 'category_id'
),
// ... more relationships
);
Categories Controller
class CategoriesController extends AppController {
public $paginate = array(
'Category' => array(
'contain' => array(
'Post' => array(
'limit' => 3
), // end Post
) // end Category contain
) // end Category pagination
); // end pagination
public function index() {
// for paginated results
$this->set('categories', $this->paginate());
// for find results
$this->Category->contain(array(
'Post' => array(
'limit' => 3
)
));
$this->set('categories', $this->Category->find('all'));
}
I'm trying to retrieve some data through two model relationships with CakePHP. The models and their associations are as follows:
User hasOne Profile HABTM Skill
I would like the user's skills to be returned when I do a find() operation on the User model, and right now it isn't returned. Here's my find call which is being executed against the User model:
$this->paginate = array(
'conditions' => array(
'OR' => array(
'Profile.firstname LIKE' => "%$q%",
'Profile.lastname LIKE' => "%$q%"
)
)
);
It's returning user data and profile data, but not any skill data. I tried setting recursive to 2 or 3, but that doesn't help either. The only way I can get skill data is if I run find() against the Profile model, but I can't do that. For clarification here's the relevant models and their relationships:
// app/Model/User.php
class User extends AppModel {
public $hasOne = 'Profile';
}
// app/Model/Profile.php
class Profile extends AppModel {
public $belongsTo = 'User';
public $hasAndBelongsToMany = 'Skill';
// app/Model/Skill.php
class Skill extends AppModel {
public $hasAndBelongsToMany = 'Profile';
Can anyone help me get the users skills when retrieving user data? Thanks.
Use CakePHP's Containable Behavior. Your find will then look something like this:
$this->User->find('all',array(
'conditions' => array(
'OR' => array(
'Profile.firstname LIKE' => "%$q%",
'Profile.lastname LIKE' => "%$q%"
)
),
'contain' => array(
'Profile' => array(
'Skill'
)
)
));
MUCH simpler, easier to read, and voila - you get the data you want without needing to use the dreaded 'recursive'.
I'm using an existing database (I can't change it and its table names are not like cake conventions want it), and I'd like to do some left joins but can't do it properly :/
I've already defined my tables, giving them primary keys and the relations in the models.
Here is my problem :
Table Wysipage can have 0 to n wysipage_content, and 0 to n wysipage_menu.
an element from wysipage_content corresponds to 1 and only 1 Wysipage.
an element from wysipage_menu corresponds to 0 or 1 Wysipage.
I'd like to make a request who would give me a list of all the elements from Wysipages, with their eventuals contents and menus, all that in a single table, and by only one request.
Here are my tables definitions (I'm avoiding you the entire schema, just be aware there is a wp_id and a wp_name column) :
class Wysipage extends AppModel {
var $actsAs = array('Containable');
public $useTable = 'wysipage';
public $primaryKey = 'wp_id';
public $displayField = 'wp_name';
var $hasMany = array(
'un' => array(
'Wysipage_contenu' => array(
'className' => 'Wysipage_contenu',
'foreignKey' => 'wpc_wp_id',
)),
'deux' => array(
'Wysipage_menu' => array(
'className' => 'Wysipage_menu',
'foreignKey' => 'wpm_wp_id',
))
);
class Wysipage_contenu extends AppModel {
var $actsAs = array('Containable');
public $useTable = 'wysipage_contenu';
public $primaryKey = 'wpc_id';
public $displayField = 'wpc_h1';
public $belongsTo = array(
'Wysipage' => array(
'className' => 'Wysipage',
'foreignKey' => 'wp_id'
)
);
class Wysipage_menu extends AppModel {
var $actsAs = array('Containable');
public $useTable = 'wysipage_menu';
public $primaryKey = 'wm_id';
public $displayField = 'wm_lien';
public $belongsTo = array(
'Wysipage' => array(
'className' => 'Wysipage',
'foreignKey' => 'wm_wp_id'
)
);
And here is my code to try request (but failed) :
$this->loadModel('Wysipage_contenu');
$this->loadModel('Wysipage_menu');
$this->Wysipage->contain();
$mes_wysipages = $this->Wysipage->find('all', array('joins' => array(
array(
'table' => 'wysipage_contenu',
'alias' => 'wpc',
'type' => 'LEFT',
'conditions'=> array('wpc.wpc_wp_id = Wysipage.wp_id')
),
array(
'table' => 'wysipage_menu',
'alias' => 'wpm',
'type' => 'LEFT',
'conditions'=> array('wpm.wm_wp_id = Wysipage.wp_id')
)
)));
$this->set('wysipages', $mes_wysipages);
$this->render();
What have I done wrong? Is the problem in my model declarations? Or do I use a wrong request type? :(
The request I'd like to make is simply :
SELECT wp_id, wp_name, wpc_id, wpc_name
FROM wysipage
LEFT JOIN wysipage_contenu ON wysipage.wp_id = wysipage_contenu.wpc_wp_id
Just this :(
I'm not even sure I want a LEFT join or a RIGHT join, but anyway the problem remains the same, this code gives me bad answers with multiple occurrences of the same lines :/
Thanks :/
PS : Sorry for my bad English, it's not my native language.
You can not join content and menu tables on wp_id in same query because they both have many-to-one relationship to wp_id, and there by for each row from content there are all rows from menu with same wp_id in result of such join. You need to do 2 queries: one for content and one for menu. Or if list columns you needed identical for both tables you can union both results from this two queries.
I have a site develop in cakephp 2.0
I want to make a HABTM relation to the same model: A product can has more products.
I thinked to do in this mode into my model:
class Product extends AppModel {
public $name = 'Product';
public $useTable = 'products';
public $belongsTo = 'User';
public $actsAs = array('Containable');
public $hasAndBelongsToMany = array(
'Ingredient' => array(
'className' => 'Product',
'joinTable' => 'ingredients_products',
'foreignKey' => 'product_id',
'associationForeignKey' => 'ingredient_id',
'unique' => true
)
);
}
Is correct my relation? But have I to insert into my table products the field product_id and ingredient_id?
And how can I save my data with a form? I know how to save data with HABTM but I never done an HABTM to the same table.
Your relation is fine. What you have written will create a Product Model that can have any number of Ingredients and allows an Ingredient to belong to any number of Products.
When saving, you must simply treat the Ingredient as if it were another Model. The CakePHP example on saving HABTM works just as well as for associating the same model as with 2 different models: http://book.cakephp.org/2.0/en/models/saving-your-data.html.
So, if you're saving multiple Ingredients to a Product, your Array structure will look like this:
Array(
[0] => Array(
Product => Array(
id => 1
),
Ingredient => Array(
id => 18
)
),
1 => Array(
Product => Array(
id => 1
),
Ingredient => Array(
id => 23
)
)
// ...
)
It is up to you how you capture this in a form, but the form example used in the link provided above should manage this properly.
I have the following tables for my CakePHP app that allows friendships between users:
**Users**
id
username
password
**Profiles**
id
firstname
lastname
user_id
**Friends**
id
user_id_to
user_id_from
status
So basically a user has a profile and a user can be friends with another user and this is recorded in the database table called friends with a simple status of confirmed or not using either 0 or 1 (it's an int). So friends is the join between two users.
I'm trying to list the friends for a user so for example if I get a url like:
/people/cameron/friends it will list the friends for the user Cameron.
However I'm struggling with the find statement to pass the user and find them (notice I contain the profile data) and then list friends that are related to that user. Can anyone help?
These are the Friend, User and Profile models:
class Friend extends AppModel
{
public $name = 'Friend';
public $belongsTo = array('User');
public $actsAs = array('Containable');
}
User.php
class User extends AppModel
{
public $name = 'User';
public $hasOne = 'Profile';
public $hasMany = array(
'Post',
'Answer',
'Friend' => array(
'foreignKey' => 'user_id_to'
)
);
public $belongsTo = array(
'Friend' => array(
'foreignKey' => 'user_id_from'
)
);
public $actsAs = array('Containable');
public function getFriends($username)
{
return $this->find('all',
array('conditions' => array('User.username' => $username, 'Friend.status'=>1),
'contain' => array('Friend' => array('User'))
));
}
}
Profile.php
class Profile extends AppModel
{
public $name = 'Profile';
public $belongsTo = 'User';
public $actsAs = array('Containable');
}
and this is my method for showing the friend list for a user:
public function index( $username )
{
$friends = $this->User->getFriends($username);
$this->set('friends', $this->paginate());
}
I'm currently getting this error:
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'User.user_id_from' in 'on clause'
SQL Query: SELECT `User`.`id`, `User`.`username`, `User`.`password`, `User`.`email`, `User`.`status`, `User`.`code`, `User`.`lastlogin`, `Friend`.`id`, `Friend`.`user_id_from`, `Friend`.`user_id_to`, `Friend`.`datetime`, `Friend`.`status` FROM `db52704_favorr`.`users` AS `User` LEFT JOIN `db52704_favorr`.`friends` AS `Friend` ON (`User`.`user_id_from` = `Friend`.`id`) WHERE `User`.`username` = 'cameron' AND `Friend`.`status` = 1
It looks as though the app thinks the foreign keys are in the User table rather than the friend table even though they called within the Friend association... Any ideas what the problem is?
Any time you specify foreign keys for $belongsTo, remember that the name you specify is the name of the field in the current table, not the other table.
So, for example, your $belongsTo references to 'foreignKey' => 'user_id_to' should be in the Friends model, not the Users model.
Re-read Cake's docs, as it does get confusing (even after years of Cake apps, I still need to refresh when I start a new project): http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html
Cake appears to be getting confused by those foreignKey assignments when constructing the join query.
you could try replacing each relation with the following to force the right join statement:
public $hasMany = array(
'Friend' => array(
'foreignKey' => null,
'conditions' => array('Friend.user_id_to = User.id')
)
);
public $belongsTo = array(
'Friend' => array(
'foreignKey' => null,
'conditions' => array('Friend.user_id_from = User.id')
)
);