Send controller data into form field in Yii

Question

UPDATED THE SOLUTION::: IT IS WORKING NOW

I have created a controller with join query. Now I want to include controller to actionCreate function and show data into text filed in form.

Controller

public function actionCreate()
{
    $model=new Grndetail;

    if (isset($_POST['Grndetail'])) {
        $model->attributes=$_POST['Grndetail'];
        if ($model->save())
            $this->redirect(array('view','id'=>$model->id));
    }
    
    $grndata = $this->GrnData1();
    $model->im_grnnumber = $grndata['im_grnnumber'];
    $this->render('create',array('model'=>$model, 'grndata'=>$grndata,));
}



public function GrnData1()
{               
    $sql = Yii::app()->db->createCommand()
              ->select('t.im_grnnumber, t.im_purordnum, r.pp_purordnum, r.cm_code, r.pp_quantity, r.pp_unit, r.pp_unitqty, r.pp_purchasrate, p.cm_description, p.cm_code')
              ->from('im_grnheader t')
              ->join('pp_purchaseorddt r', 't.im_purordnum = r.pp_purordnum')
              ->join('cm_productmaster p', 'p.cm_code = r.cm_code')
             //->where('id=:id', array(':id'=>$id))
              ->order('im_grnnumber DESC')
              ->queryRow();
                    
     return $sql;
}

form filed

<?php echo $form->textField($model,'im_grnnumber'); ?>

Is anything I am missing. How can I view the data as value in form when I am creating something.

UPDATED THE SOLUTION::: IT IS WORKING NOW

Muhammad Rizwan Kaim Khani · Accepted Answer · 2014-01-07 04:31:20Z

1

You should change your program as

public function GrnData1(){

$sql = Yii::app()->db->createCommand()->setFetchMode(PDO::FETCH_OBJ) // change it

->select('t.im_grnnumber, t.im_purordnum, r.pp_purordnum, r.cm_code, r.pp_quantity,  r.pp_unit, r.pp_unitqty, r.pp_purchasrate, p.cm_description, p.cm_code')

->from('im_grnheader t')
->join('pp_purchaseorddt r', 't.im_purordnum = r.pp_purordnum')
->join('cm_productmaster p', 'p.cm_code = r.cm_code')
->order('im_grnnumber DESC')
->queryRow();

Return $sql; // write here return statement

}

Now you can access your data like this

echo $form->textField($model,'im_grnnumber', array('value'=>$grndata->im_grnnumber));

Hope, It will help you.

Thanks

edited Jan 7, 2014 at 4:31

answered Dec 31, 2013 at 10:29

Muhammad Rizwan Kaim Khani

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1 Comment

Selim Reza Over a year ago

Thank you Dear. I forgot to return $sql. thanks for help. But in textfield is not working. I got the Solutions and Updated my POST/ CODE. Please Have a look. Thanks AGAIN

marc_s · Accepted Answer · 2023-02-16 19:30:56Z

1

Var $grndata is an array, not a model. So the correct way to call its values should be this:

<?php echo $form->textField($model,'im_grnnumber', array('value'=>$grndata['im_grnnumber']) ); ?>

Or something like that, I'm not sure of the format of the array returned by queryRow(). Maybe print_r($grndata) to see the style of the array.

edited Feb 16, 2023 at 19:30

marc_s

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answered Dec 31, 2013 at 9:20

Alfredo Castaneda Garcia

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3 Comments

Selim Reza Over a year ago

it's showing ERROR: GrndetailController and its behaviors do not have a method or closure named "GrnData1".

Selim Reza Over a year ago

I changed it. But still same error. I found something for static .. help me for dynamic. Like from controller :: $model->im_grnnumber ="1111"; is working. it is showing in the field. How can I do that dynamically. Please help –

Selim Reza Over a year ago

Thank you Dear. I forgot to return $sql. thanks for help. I got the Solutions and Updated my POST/ CODE. Please Have a look. Thanks AGAIN

yoshi · Accepted Answer · 2013-12-31 09:33:34Z

0

change actionGrnData1() to GrnData1()

answered Dec 31, 2013 at 9:33

yoshi

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3 Comments

Selim Reza Over a year ago

I changed it. But still same error. I found something for static .. help me for dynamic. Like from controller :: $model->im_grnnumber ="1111"; is working. it is showing in the field. How can I do that dynamically. Please help

Let me see Over a year ago

Do you have any idea what you are saying. he has created a function and why are you converting it to an action??

Selim Reza Over a year ago

Thank you Dear. I forgot to return $sql. thanks for help. I got the Solutions and Updated my POST/ CODE. Please Have a look. Thanks AGAIN

Let me see · Accepted Answer · 2013-12-31 11:12:24Z

0

You should complete your textField like this

    <?php echo $form->textField($model,'im_grnnumber',
array('value'=>$grndata->im_grnnumber
)); ?>

answered Dec 31, 2013 at 11:12

Let me see

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