Need Help With Inheritance In Yii

Yii 1.1 General Discussion
1

Hi,

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

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!

2

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

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

3

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…