Add dropdown to symfony 3 form - php

Question

I want to add a dropdow field for category in my form in symfony version 3, I have tried to solutions but each one have their own problem
First got all categories and pass them to my view and show them:
Action:
/**
* Creates a new News entity.
*
* #Route("/new", name="news_new")
* #Method({"GET", "POST"})
*/
public function newAction(Request $request)
{
$news = new News();
$form = $this->createForm('AppBundle\Form\NewsType', $news);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$em = $this->getDoctrine()->getManager();
$em->persist($news);
$em->flush();
return $this->redirectToRoute('news_show', array('id' => $news->getId()));
}
$em = $this->getDoctrine()->getManager();
$categories = $em->getRepository('AppBundle:Category')->findAll();
return $this->render('news/new.html.twig', array(
'news' => $news,
'form' => $form->createView(),
'categories' => $categories,
));
}
View:
{% extends 'base.html.twig' %}
{% block body %}
<h1>News creation</h1>
{{ form_start(form) }}
<label for="news_content" class="required">Category</label>
<select name="news[categoryId]">
{% for category in categories %}
<option value="{{ category.id }}">{{ category.title }}</option>
{% endfor %}
</select>
{{ form_widget(form) }}
<input class="btn btn-sm btn-success" type="submit" value="Create" />
{{ form_end(form) }}
<ul>
<li>
<a class="label label-sm label-info" href="{{ path('news_index') }}">Back to the list</a>
</li>
</ul>
{% endblock %}
The form is created as I expected but when i want to submit it, it show an validation error as bellow:
This form should not contain extra fields.
second solution that I have tried is to generate the dropdown from my Type, so in NewsType I changed the buildForm function as bellow:
/**
* #param FormBuilderInterface $builder
* #param array $options
*/
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('categoryId', EntityType::class, [
'class' => 'AppBundle:Category',
'choice_label' => 'title',
])
->add('title')
->add('content')
;
}
It this way, the form also have been created nicely but after submit, an database exception returned and said:
An exception occurred while executing 'INSERT INTO news (category_id, title, content) VALUES (?, ?, ?)' with params [{}, "asdf", "asdf"]:
Catchable Fatal Error: Object of class AppBundle\Entity\Category could not be converted to string
It mean that my category_id passed as an object !
What should I do?
BTW, my english is a little weak, please do not put and minus on my post, I had been ban multiple times.
Thanks in advance.

Answer 1

all symfony is trying to do is to find a string representation of the Category object, so it can populate the option fields.
you can solve this in a couple of ways:
In the Category object, you can make a __toString method.
public function __toString() {
return $this->name; // or whatever field you want displayed
}
or
you can tell symfony which field to use as the label for the field. From docs
$builder->add('attending', ChoiceType::class, array(
'choices' => array(
new Status(Status::YES),
new Status(Status::NO),
new Status(Status::MAYBE),
),
'choice_label' => 'displayName', // <-- where this is getDisplayName() on the object.
));