cakephp use another model inside current model

Question

I know I can use another model inside a controller by doing $this->loadModel("MyModel"), but how do I do this inside another Model? I tried to use loadModel but it didn't work.

Any idea?

Thank you

Possible duplicate of [Can I use one model inside of a different model in CakePHP?](https://stackoverflow.com/questions/980556/can-i-use-one-model-inside-of-a-different-model-in-cakephp) — icc97, Jul 06 '17 at 12:54

score 27 · Accepted Answer · edited May 23 '17 at 12:24

27

Easier is:

$my_model = ClassRegistry::init('MyModel');

More details: Can I use one model inside of a different model in CakePHP?

edited May 23 '17 at 12:24

Community

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answered Mar 13 '12 at 21:51

Costa

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This is the best way. If you use `App::import` or `App::uses` with model constructor then will be problem e.g. with TranslateBehavior. Reason: `DboSource` class use `ClassRegistry::getObject` to get model and read virtual fields. So to prevent using two instances of the same model use this solution. – dmajka Apr 28 '21 at 09:55

score 4 · Answer 2 · edited Dec 27 '15 at 22:18

4

You can use the following code to export a model that is not associated with the current model in any way:

App::import('Model', 'MyModel');
$my_model = new MyModel();

If MyModel is associated with current model you could use the chaining e.g. $this->SomeModel->MyModel

edited Dec 27 '15 at 22:18

ᴍᴇʜᴏᴠ

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answered Mar 13 '12 at 20:36

Ehtesham

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score 0 · Answer 3 · edited Mar 08 '15 at 03:15

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You don't need to import anything. Just do this:

$my_model = new MyModel();
//Then
$my_model->read(null,$id);

edited Mar 08 '15 at 03:15

Jeffrey Bosboom

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answered Mar 08 '15 at 02:51

Paulo Peres Junior

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cakephp use another model inside current model

3 Answers3