Simple question: Can get I get model data, using a variable instead of a model name?

Yii 1.1 General Discussion
1

Hi,

I wanted to know how I could return a model, if the name of the model was in a variable…like this:




$model = $ModelName::model()->findAll();

Does anyone know if this is possible?

Thanks,

Matt

2

That will throw an error, what about:




$model=ModelName::model();

$model->findAll();
3

Yeah, just came up with that myself…I’m building a lookup table in an array, of models that I want to be part of the API I’m building.

I’m ok with that solution for now, unless anyone else has something better!

Thanks,

Matt

4

You could always store the Model names in the array and do smth like:




$models=array('Model1','Model2','Model3');

foreach($models AS $model){

  if(some condition){

    $m=new $model();

    $m->findAll();

  }

}

This has the advantage that you don’t keep an array of objects therefore should be less resource intensive if you’d have, say 1000 models to load (i doubt that anyway)

5

Unless ive misunderstood the question, I use this when the model class is dynamic…




$variable = 'Post';

$model = CActiveRecord::model($variable)->find();
6

I use it too, but as of PHP 5.3 the following syntax will also work:




$model = $variable::model()->findAll();
7

Awesome solutions guys…I guess I’m not on 5.3 yet…LOL.

For now I just have a lookup array of key/values, instantiated upon load…I’m sure this will be horrible later in production, so I’ll look at upgrading to php 5.3.

Cheers,

Matt

8

I have a similarly problem…

I defined in my class a variable $table_name, then a try to do :

$model = $this->table_name::model()->findByPk($id);

but it doesn’t work… I tried ($this->table_name)::model()->findByPk($id) but doesn’t work too…

do you have an idea ?

error : unexpected T_PAAMAYIM_NEKUDOTAYIM

9

Zugluk, you might try:

$model = {$this->table_name}::model()->findByPk($id);