CODEWITHSUNDEEP
validationrestsymfony
user1930254
user1930254

Reputation: 1321

REST form validation in symfony 2 - how to test post

The form is never valid, but $form->getErrors() gives no error. Since its gonna be a REST API test it with DEV HTTP client

test data is

Header: Content-Type:application/json

Body: {"username":"sad","password":"123","email":"[email protected]"}

I don't have any validation.yml file

Is there any any method to find out whats going wrong (error Message)? 

public function postUserAction(Request $request)
{

    return $this->processForm(new User(),$request);
}

private function processForm(User $user, Request $request )
{


    $form = $this->createForm(new UserType(), $user);
    $form->handleRequest($request);

    if ($form->isValid()) {

         return array('form' => 'valid');
    }

    return \FOS\RestBundle\View\View::create($form->getErrors(),400);
}

Upvotes: 5

Views: 4873

Answers (4)

George Mylonas
George Mylonas

Reputation: 722

I am late to reply to this one but, all answers are wrong. submit should be avoided and use handleRequest instead. Read this. That form was not valid with handleRequest because UserType was missing one of the next:

  1. A form field was missing
  2. A form field was not in appropriate format

  3. In UserType class getName function should return empty string and csrf_protection should be disabled (IF you are using API approach where form was not generated from your Symfony2 project):

    public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults(array(
        'data_class' => 'UserApiBundle\Entity\User',
        'csrf_protection'   => false,
    ));
}

public function getName()
{
    return '';
}

    submit would not check for that and it is a bad approach for POST. You could use it for PUT request to update a User where password field would not be submitted for xyz reasons (ex: permissions).

    $form->submit($request, false);

It is hard to dig for form errors in Symfony2. You can use this function to get errors:

public function getErrors($form)
{
    $errors = array();
    foreach ($form as $fieldName => $formField) {
        $currentError = $formField->getErrors();

        if ($currentError->current()) {
            $current = $currentError->current();
            $errors[$fieldName] = $current->getMessage();
        }
    }

    return $errors;
}

Upvotes: 0

electronix384128
electronix384128

Reputation: 6733

This worked for me:

$form = $this->createForm(new UserType(), $user);
$jsonData = json_decode($request->getContent(), true); // "true" to get an associative array

if ('POST' === $request->getMethod()) {
    $form->bind($jsonData);
    ...
}

Upvotes: 1

user1930254
user1930254

Reputation: 1321

After a little debug: isValid() checks form several things eg if the form has been submitted. It wasnt, so i changed to

...
        $form = $this->createForm(new UserType(), $user);
        //$form->handleRequest($request);
        $form->submit($request);

        if ($form->isValid()) {
...

now its valid.

Upvotes: 3

xdazz
xdazz

Reputation: 160903

When debugging a form validation,

use $form->getErrorsAsString() instead of $form->getErrors().

(which will run in to deep level including the form children.)

Upvotes: 1

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