I am trying to use join with two table by using has many relation of CakePHP with condition my model code are here which am using
public $userid = 3;
public $name = 'Course';
public $hasMany = array(
'Enrollcourse' => array(
'className' => 'Enrollcourse',
'foreignKey' => 'course_id',
'conditions' => array('Enrollcourse.student_id' => $this->userid),
'dependent' => true
)
);
OR
public $userid = 3;
public $name = 'Course';
public $hasMany = array(
'Enrollcourse' => array(
'className' => 'Enrollcourse',
'foreignKey' => 'course_id',
'conditions' => array('Enrollcourse.student_id' => $userid),
'dependent' => true
)
);
here is $userid
is a variable which is used to check to retrieve data of selected user but am unable to get this & following are occur
Error: parse error
File: E:\wamp\www\simpleApp\app\Model\course.php
Line: 10
Any help?
You cannot use variables in the declaration of a variable within a class. This means that use of $userid
will cause the parse error you are seeing.
The best way to overcome this for dynamic information is to replace/overload the constructor for the model:
public function __construct($id = false, $table = null, $ds = null) {
$this->hasMany = array(
'Enrollcourse' => array(
'className' => 'Enrollcourse',
'foreignKey' => 'course_id',
'conditions' => array('Enrollcourse.student_id' => $this->userid),
'dependent' => true
)
);
parent::__construct($id, $table, $ds);
}
This overcomes the use of a variable during a class variable declaration.