CODEWITHSUNDEEP
phptemplatesyii
ajaybc
ajaybc

Reputation: 4059

Yii display and validation of fields not present in table

I am a newbie in Yii. I have created a page where user can change their password.

So in my changePassword view I have :

<div class="row">
    <?php echo $form->labelEx($model,'oldpwd'); ?>
    <?php echo $form->textField($model,'oldpwd'); ?>
    <?php echo $form->error($model,'oldpwd'); ?>
</div>

<div class="row">
    <?php echo $form->labelEx($model,'pwd'); ?>
    <?php echo $form->textField($model,'pwd'); ?>
    <?php echo $form->error($model,'pwd'); ?>
</div>

<div class="row">
    <?php echo $form->labelEx($model,'pwd_repeat'); ?>
    <?php echo $form->passwordField($model,'pwd_repeat'); ?>
    <?php echo $form->error($model,'pwd_repeat'); ?>
</div>

Now obviously I am getting an error as only the field 'pwd' is in the table and thereby in the model. I am new to MVC frameworks and can use some help here. Thanks

Upvotes: 1

Views: 1643

Answers (2)

Suriyan Suresh
Suriyan Suresh

Reputation: 3024

add the following line to your model 

public $old_pwd;  
public $pwd_repeat;

we instruct Yii to use this field as virtual field instead of searching in database field

Upvotes: 0

Rajat Singhal
Rajat Singhal

Reputation: 11264

Declare them in your model as property of model First..

public $old_pwd;
public $pwd;
public $pwd_repeat;

As you are asking model Labels of these attributes..define them in your attributeLabels function in model..

public function attributeLabels()
{
    return array(
        'old_pwd'=>'Old Passw....',
                     '.....same way for all those who are not already there..'
    );
}

Declare them safe in rules if required...

Upvotes: 2

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