here is my table association code:
class UserMastersTable extends Table {
public function initialize(array $config) {
parent::initialize($config);
$this->table('user_masters');
$this->hasOne('person_masters', [
'className' => 'person_masters',
'foreign_key'=>'user_master_id',
'dependent' => true
]);
}
}
when in controller i am using:
$this->UserMasters->get($id);
it results only user_masters table data..
so how can i also get Associated tables data??
Use contain()
Copy-Paste from the manual:
You should use contain() when you want to load the primary model, and
its associated data. While contain() will let you apply additional
conditions to the loaded associations, you cannot constrain the
primary model based on the associations. For more details on the
contain(), look at Eager Loading Associations.
// In a controller or table method.
// As an option to find()
$query = $articles->find('all', ['contain' => ['Authors', 'Comments']]);
// As a method on the query object
$query = $articles->find('all');
$query->contain(['Authors', 'Comments']);
Read the manual before jumping into trial and error driven development! If you would have done one of the tutorials in the manual before this would be clear. So do them now, they'll cover a lot more of the basics.
Related
I have created method in the Yii2 model Users to get all the replies for the current user
public function getAllRepliesForUsers() { return $this->hasMany(Replies::class, ['user_id' => 'id'])->viaTable('replies_links', ['replies_id' => 'id'])->where(['entity'=>'user']); }
My replies table
My users table
and the final table that links these two tables
Is my method is correct?
Here's the relationship of Users to the Replies. You can use the Model generator of Gii module so you won't get confused by manually typing them.
public function getReplies()
{
return $this->hasMany(Replies::className(), ['id' => 'reply_id'])->viaTable('rply_links', ['user_id' => 'id']);
}
(May I know what do you intend to do with the condition ->where(['entity'=>'user'])?).
Since this mornong i am facing a very big problem. I am using CodeIgniter to develop a website, and GAS ORM for the database.
I have basically two tables. One named "pool", and one named "partners". I am having two associations between these two tables, so I have two foreign keys in my table Partners referencing the table pool.
Pool(#id:integer, name:varchar)
Partners(#id:integer, associated_pool_id=>Pool, futur_associated_pool_id=>Pool).
As I have two references to the same table, I can't name the foreign keys "pool_id". So in my relationships with Gas ORM, I have to specify the names of the columns. I do it, but it doesn't work...
Here is what I do:
class Partner extends ORM {
public $primary_key = 'id';
public $foreign_key = array('\\Model\\Pool' => 'associated_pool_id', '\\Model\\Pool' => 'future_associated_pool_id');
function _init()
{
// Relationship definition
self::$relationships = array(
'associated_pool' => ORM::belongs_to('\\Model\\Pool'),
'future_association_pool' => ORM::belongs_to('\\Model\\Pool'),
);
self::$fields = array(
'id' => ORM::field('auto[11]'),
'name' => ORM::field('char[255]'),
'associated_pool_id' => ORM::field('int[11]'),
'future_associated_pool_id' => ORM::field('int[11]')
);
}
and in my Pool class :
class Pool extends ORM {
public $primary_key = 'id';
function _init()
{
// Relationship definition
self::$relationships = array(
'associated_partner' => ORM::has_many('\\Model\\Partner'),
'future_associated_partner' => ORM::has_many('\\Model\\Partner'),
);
self::$fields = array(
'id' => ORM::field('auto[11]'),
'name' => ORM::field('char[50]'),
);
}
I have a test controller testing if everything is okay:
class Welcome extends CI_Controller {
public function index()
{
$pool = \Model\Pool::find(1);
echo $pool->name;
$partners = $pool->associated_partner();
var_dump($partners);
}
But I have an error saying:
Error Number: 1054
Champ 'partner.pool_id' inconnu dans where clause
SELECT * FROM partner WHERE partner.pool_id IN (1)
I don't know how to specify to Gas ORM that it shouldn't take "pool_id" but "associated_pool_id"....
Thank you for your help!!!!!!!!!!!!
I don't know, if this topic is still up to date and interesting to some of you, but in general, I had the exact same problem.
I decided Gas ORM to be my mapper in combination with CodeIgniter. As my database structure was given and it was not following the table_pk convention of Gas, I had to define a foreign key by myself which shall refer to my custom database foreign key. However, the definition of it had no impact on anything. Like your error above, the mapper was not able to build the right SQL-statement. The statement looked similar to yours:
SELECT * FROM partner WHERE partner.pool_id IN (1)
Well, it seems like Gas ignores the self-defined foreign keys and tries to use the default table_pk convention. This means, it takes the table (in your case: pool) and the primary key (id) by merging it with a underscore character.
I figured out, that the constructor of orm.php handles every primary and foreign key defined within the entities. In line 191, the code calls an if clause combined with the empty function of php. As the primary key is defined always and there is no negation in the statement, it skips the inner part of the clause every time. However, the inner part takes care of the self-defined foreign keys.
Long story short, I added a negation (!) in line 191 of orm.php which leads me to the following code:
if ( ! empty($this->primary_key))
{
if ( ! empty($this->foreign_key))
{
// Validate foreign keys for consistency naming convention recognizer
$foreign_key = array();
foreach($this->foreign_key as $namespace => $fk)
{
$foreign_key[strtolower($namespace)] = $fk;
}
$this->foreign_key = $foreign_key;
}
else
{
// If so far we didnt have any keys yet,
// then hopefully someone is really follow Gas convention
// while he define his entity relationship (yes, YOU!)
foreach ($this->meta->get('entities') as $name => $entity)
{
if ($entity['type'] == 'belongs_to')
{
$child_name = $entity['child'];
$child_instance = new $child_name;
$child_table = $child_instance->table;
$child_key = $child_instance->primary_key;
$this->foreign_key[strtolower($child_name)] = $child_table.'_'.$child_key;
}
}
}
}
Well, this little fix helped me out a lot and I hope some of you can take advantage of this hint as well.
How can I add where condition to my Articles model so that slug(From category model) is equal to $slug?
And this is a function that Gii generated:
public function getCategory()
{
return $this->hasOne(Categories::className(), ['id' => 'category_id']);
}
Here's my code:
public function specificItems($slug)
{
$query = Articles::find()->with('category');
$countQuery = clone $query;
$pages = new Pagination(['totalCount' => $countQuery->count(),'pageSize' => 12]);
$articles = $query->offset($pages->offset)
->limit($pages->limit)
->all();
return ['articles' => $articles,'pages' => $pages];
}
Your SQL query should contain columns from both article and category table. For that you need to use joinWith().
$result = Articles::find()
->joinWith('category')
->andWhere(['category.slug' => $slug])
->all();
Where 'category' is then name of your category table.
However, in your code you deviate from certain best practices. I would recommend the following:
Have both table name and model class in singular (Article and article). A relation can be in plural, like getCategories if an article has multiple categories.
Avoid functions that return result sets. Better return ActiveQuery class. If you have a query object, all you need to get the actual models is ->all(). However, you can further manipulate this object, add more conditions, change result format (->asArray()) and other useful stuff. Returning array of results does not allow that.
Consider extending ActiveQuery class into ArticleQuery and implementing conditions there. You'll then be able to do things like Article::find()->joinWith('category')->byCategorySlug('foo')->all().
Table.linkedIndex is related to LinkedIndex.ID. The value of the field LinkedIndex.TableName is either Linked1 or Linked2 and defines which of these tables is related to a row in Table.
Now i want to make a dynamical link with Yii models so that i can easily get from a Table row to the corresponding Linked1 or Linked2 row:
Table.linkedID = [LinkedIndex.TableName].ID
Example
Table values:
LinkedIndex values:
Now I should get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;
Model
In the model Table, I tried to make the relation to the table with the name of the value of linkedIndex.TableName:
public function relations()
{
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
'linked' => array(
self::HAS_ONE,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
)
}
But then I get the error:
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
Is there any way to make a dynamic relation Table.linkedID -> [LinkedIndex.TableName].ID with Yii Models?
Per the Yii docs here:
http://www.yiiframework.com/doc/api/1.1/CActiveRecord#relations-detail
I'd suggest using self::HAS_ONE instead (unless there can be multiple rows in LinkedIndex with the same ID - although from the looks of above, I doubt that's the case).
You can link tables together that have different keys by following the schema:
foreign_key => primary_key
In case you need to specify custom PK->FK association you can define it as array('fk'=>'pk'). For composite keys it will be array('fk_c1'=>'pk_с1','fk_c2'=>'pk_c2').
so in your case:
public function relations(){
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
);
}
where LinkedIndex is the class name for the LinkedIndex model (relative to your Table model - i.e. same folder. You could change that, of course) and array('ID' => 'linkedIndex') specifies the relationship as LinkedIndex.ID = Table.linkedIndex.
Edit
Looking at your updated example, I think you're misunderstanding how the relations function works. You're getting the error
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
because you're trying to create another relation here:
'linked' => array(
self::BELONGS_TO,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
This part: linkedIndex.TableName refers to a new model class linkedIndex.TableName, so Yii attempts to load that class' file linkedIndex.TableName.php and throws an error since it doesn't exist.
I think what you're looking for is to be able to access the value TableName within the table LinkedIndex, correct? If so, that's accessible from within the Table model via:
$this->linkedIndex->TableName
This is made possible by the relation we set up above. $this refers to the Table model, linkedIndex refers to the LinkedIndex relation we made above, and TableName is an attribute of that LinkedIndex model.
Edit 2
Per your comments, it looks like you're trying to make a more complex relationship. I'll be honest that this isn't really the way you should be using linking tables (ideally you should have a linking table between two tables, not a linking table that says which 3rd table to link to) but I'll try and answer your question as best as possible within Yii.
Ideally, this relationship should be made from within the LinkedIndex model, since that's where the relationship lies.
Since you're using the table name as the linking factor, you'll need to create a way to dynamically pass in the table you want to use after the record is found.
You can use the LinkedIndex model's afterFind function to create the secondary link after the model is created within Yii, and instantiate the new linked model there.
Something like this for your LinkedIndex model:
class LinkedIndex extends CActiveRecord{
public $linked;
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return 'LinkedIndex';
}
public function afterFind(){
$this->linked = new Linked($this->TableName);
parent::afterFind();
}
//...etc.
}
The afterFind instantiates a new Linked model, and passes in the table name to use. That allows us to do something like this from within the Linked model:
class Linked extends CActiveRecord{
private $table_name;
public function __construct($table_name){
$this->table_name = $table_name;
}
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return $this->table_name;
}
//...etc.
}
which is how we dynamically create a class with interchangeable table names. Of course, this fails of the classes need to have separate operations done per-method, but you could check what the table_name is and act accordingly (that's pretty janky, but would work).
All of this would result in being to access a property of the linked table via (from within the Table model):
$this->linkedIndex->linked->foo;
Because the value of LinkedIndex.TableName and Table.linkedID is needed to get the values, I moved the afterFind, suggested by M Sost, directly into the Table-Class and changed its content accordingly. No more need for a virtual model.
class Table extends CActiveRecord {
public $linked; // Needs to be public, to be accessible
// ...etc.
public function afterFind() {
$model = new $this->linkedIndex->TableName;
$this->linked = $model::model()->findByPk( $this->linkedID );
parent::afterFind();
}
// ...
}
Now I get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;
I'm new to cakephp. I'm trying to search through mysql tables. I want to use nested query.
class TableController extends AppController{
.
.
public function show(){
$this->set('discouns', $this->DiscounsController->query("SELECT * FROM discoun as Discoun WHERE gcil_id = 1"));//(SELECT id FROM gcils WHERE genre = 'Shoes' AND company_name = 'Adidas')"));
}
}
Error:
Error: Call to a member function query() on a non-object
I've also tried
public function show(){
$this->DiscounsController->query("SELECT * FROM count as Count WHERE ctr_id = (SELECT id FROM ctrs WHERE genre = 'Shoes' AND company_name = 'Adidas')");
}
Error:
Error: Call to a member function query() on a non-object
File: C:\xampp\htdocs\cakephppro\myapp\Controller\CountsController.php
Please help. I've been trying this for last few hours. :/
As mentioned in the comments there are a few problems with your code.
Firstly, you are trying to call the query() method on a Controller, whereas you should be executing it on a Model, as it is models that handle database queries and the controller should simply be used to call these methods to get the data and pass them to the view.
The second thing is that you are executing a very simple SQL query raw instead of using CakePHPs built in functions <- Be sure to read this page in full.
Now for your problem, as long as you have setup your model relationships correctly and followed the correct naming conventions, this should be your code to run your SQL query from that controller:
public function show(){
$this->set('discouns', $this->Discouns->find('all', array(
'conditions' => array(
'gcil_id' => 1,
'genre' => 'shoes',
'company_name' => 'Adidas'
)
));
}
query() is not a Controller, but a Model method. That's what the error (Call to a member function on a non-object) is trying to tell you.
So the correct call would be:
$this->Discount->query()
But you are calling this in a TableController, so unless Table and Discount have some type of relationship, you won't be able to call query().
If the Table does have a relationship defined you will be able to call:
$this->Table->Discount->query()
Please not that query() is only used when performing complex SQL queries in scenarios where the standard methods (find, save, delete, etc.) are less practical.
$this->Counts->find('all',array(
'conditions' => array(
'ctrs.genre' => 'Shoes',
'ctrs.company_name' => 'Adidas'
), 'recursive' => 1
));
The above is with tables named counts and ctrs.
This is assuming you have the model set up to have some sort of relationship between the counts table and the ctrs table. It's kind of hard to tell in your code exactly what you tables are.
The CakePHP book should have all the answers you need. One of the reasons to run CakePHP over regular PHP is the FIND statement. Once you have your models set up correctly, using the find statement should be really easy.
http://book.cakephp.org/2.0/en/models.html