CODEWITHSUNDEEP
phpyii
Dharmesh patel
Dharmesh patel

Reputation: 654

how to get data from other tables in grid view in yii

I have a module named attendance but value in it is comes from other tables with some conditions. so how can i get id of other tables. my code is like below:

models are:

    if($type == 2)
    {
     $model = new Supplier(); 
    }
    elseif($type==3)
    {
     $model = new TruckingCompany();
    }
    elseif($type==4)
    {
     $model = new ServiceProvider();
    }
    elseif($type==5)
    {
     $model = new Landowner();
    }
    elseif($type==6)
    {
     $model = new Refiner();
    }
    elseif($type==8)
    {
     $model = new Worker();
    }
    elseif($type==9)
    {
     $model = new Trainer();
    }

now i want to update this record

    array(
        'header'=>'Action',         
        'class' => 'CButtonColumn',
        'template'=>'{update}',
        'updateButtonUrl'=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))',
   );

now i want to get id of these diffent table for update record and these are different in all tables. i can get it by below line:

 'updateButtonUrl'=> 'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))';

but it is not a proper way to define by these for all tables so what to do in this case... any idea... i am new in yii.

Upvotes: 1

Views: 97

Answers (2)

Jelle de Fries
Jelle de Fries

Reputation: 875

In addition to let me see's answer. I would put it in array form:

$types = array(
    1 => array("modelName"=>"Supplier","url"=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))'),
    2 => array("modelName"=>"TruckingCompany","url"=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))'),
);

Of course you can improve on this by dividing the "url" key for example like this:

"url"=>array(
     "route"=>"attendance/update",
     "params"=>'"id"=>$data->id',
 )

This way you can pass it like this:

if(array_key_exists($type,$types)){ //Checks if you've defined the type.   
    $this->render('view',array(
            'updateUrl'=>$types[$type]['url'],
            'model'=>new $types[$type]['modelName'](), //Creates your object.
         )
    );
    return; //just to be sure.
}
throw new CHTTPException(404, "Type not Found!");

Your view would look something like this with the improved "url" key:

array(
    'header'=>'Action',         
    'class' => 'CButtonColumn',
    'template'=>'{update}',
    'updateButtonUrl'=>'Yii::app()->createUrl(' . $updateUrl['route'] . ', array(' . $updateUrl['params'] . '))',
);

This clears up your code a bit and gets rid of the huge if statement/switch.

The createUrl string is still a bit ugly, so you can improve on that further if you would like.

Upvotes: 1

Let me see
Let me see

Reputation: 5094

Why do not you create the URL along with the models like

if($type == 2)
    {
     $model = new Supplier(); 
     $updateUrl = 'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))';
    }
    elseif($type==3)
    {
     $model = new TruckingCompany();
     $updateUrl = 'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))';
    }

Pass it to the view while rendering

$this->render('view',array('updateUrl'=>$updateUrl));

And then finally use it in view file like

array(
        'header'=>'Action',         
        'class' => 'CButtonColumn',
        'template'=>'{update}',
        'updateButtonUrl'=>$updateUrl,
   );

Upvotes: 2

Related Questions