[NeMewSys]

Hi,

I'm trying to use yii to model this kind of relation:

h**p://i45.tinypic.com/28rh5dk.png

So I made the following tables:

Animal {
int Id (Pk)
varchar Name
}

Dog {
int Id (Pk)
varchar Hair
foreign key, relates this Id with Animal.Id
}

What's the best way to integrate this with yii? I've tried myself and i also read alot of documentation, but could find the answer.
So far i had something like this:

$animal = new Animal;
$dog = new Dog;
$animal->Name = ...;
$animal->save();
$animal->refresh() // Tried with and without this line
$dog->Hair = ...;
$dog->Id = $animal->Id;
$dog->save(); // I get and error here because he build this query: INSERT INTO Dog (`Hair`) VALUES(:0ypb) instead of INSERT INTO Dog (`Id`,`Hair`) VALUES("the same Id has the Animal that I just created","this is a valid hair")

But probably even if this worked, there must be a better way to manage this relationship in yii I guess.

Thanks!

[alirz23]

Hi you first of all you have to change your table struct make sure you have your table structure as following

Quote

Animal
ID (Pk)
Name


Dog
Id (Pk)
Hair
Animal_Id

and then in your dog model under relations add the following line

'animal' => array(self::BELONGS_TO, 'Animal', 'animal_id'),

and your animal model under relations add following line

'dogs' => array(self::HAS_MANY, 'Dog', 'animal_id'),

for creating dogs you first have to create animals and then in your dog form you can have a select menu to select an animal for your dog and then and then in your dogs controller you can do some thing as following

$dog = new Dog;
$dog->attributes = $_POST['Dog'];
$dog->save(); 

there you have it

[NeMewSys]

Thanks for helping!

I'm still getting an error, quite strange one...

I have this code in my DogController's create:

/**
	 * Creates a new model.
	 * If creation is successful, the browser will be redirected to the 'view' page.
	 */
	public function actionCreate()
	{
		$animal = new Animal;
		$dog=new Dog;

		// Uncomment the following line if AJAX validation is needed
		// $this->performAjaxValidation($model);

		if(isset($_POST['Dog']))
		{
			$animal->nome=$_POST['Dog']['nome'];
			$dog->CadeiaDeADN=$_POST['Dog']['CadeiaDeADN'];
			if($animal->save()
			&& $animal->refresh()
			&& $dog->animal_Id = $animal->Id
			&& $dog->save())
				$this->redirect(array('view','id'=>$dog->Id));
		}

		$this->render('create',array(
			'model'=>$dog,
		));
	}

With this now I get this:

CDbCommand failed to execute the SQL statement: SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails (`myDB`.`dog`, CONSTRAINT `NewForeignKey2` FOREIGN KEY (`animal_Id`) REFERENCES `animal` (`Id`)). The SQL statement executed was: INSERT INTO `dog` (`animal_Id`, `CadeiaDeADN`) VALUES (:yp0, :yp1)

What should I do? I've confirmed that he can create animals and retrieve their Id by $animal->Id right after saving them, but still I got this error...