带有ManyToMany映射的Symfony形式
带有ManyToMany映射的Symfony形式
提问于 2018-12-11 15:06:11
Symfony 3.4原则1.9
我找到了很多关于这方面的文章。但没什么能解决我的问题。
我有两个实体User和Company，它们被映射为Many2Many。两个实体都应填写提交的表格。为此，我有以下FormTypes：
CustomerRegistrationType
这包括CollectionType with entry_type of CompanyFormType，它返回Company:class
class CustomerRegistrationType extends AbstractType
{
private $translator;
private $session;
public function __construct(TranslatorInterface $translator,Session $session)
{
$this->session = $session;
$this->translator = $translator;
}
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('mandant',HiddenType::class,array('data'=>
->add('newsletter',CheckboxType::class,array('label' => false,'required' => false))
->add('company', CollectionType::class, array('entry_type' => CompanyFormType::class, 'entry_options' => array('label' => false)));
}CompanyFormType
class CompanyFormType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
parent::buildForm($builder, $options);
$builder->add('telephone', TextType::class,array('required' => false))
->add('company_name', TextType::class,array('label' => false,'required' => false))
->add('company_supplement', TextType::class,array('label' =>
->add('send_info', CheckboxType::class,array('label' => false,'required' => false))
->add('send_invoice', CheckboxType::class,array('label' => false,'required' => false))
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults(array(
'data_class' => 'AppBundle\Entity\Company'));
}
}Many2Many映射
class User extends BaseUser
{
public function __construct()
{
parent::__construct();
$this->company = new ArrayCollection();
}
/**
* @ORM\ManyToMany(targetEntity="AppBundle\Entity\Company",inversedBy="users")
* @ORM\JoinTable(name="user_comp_comp_user",
* joinColumns={@ORM\JoinColumn(name="user_id", referencedColumnName="id")},
* inverseJoinColumns={@ORM\JoinColumn(name="company_id", referencedColumnName="id")}
* )
*/
protected $company;
public function getCompany()
{
return $this->company;
}
public function setCompany($company)
{
$this->company = $company;
return $this;
}
public function addCompany(Company $company)
{
if ($this->company->contains($company)) {
return;
}
$this->company[] = $company;
return $this;
}
public function removeCompany(Company $company)
{
if (!$this->company->contains($company)) {
return;
}
$this->company->removeElement($company);
return $this;
}
class Company
{
public function __construct()
{
$this->users = new ArrayCollection();
}
/**
* @var \Doctrine\Common\Collections\Collection|Company[]
* @ORM\ManyToMany(targetEntity="User",mappedBy="company")
*/
protected $users;
}在Controller中，我希望将所有数据保存到不同的实体中。为此，我要：
public function registerAction(Request $request)
{
$user = $this->userManager->createUser();
$form = $this->createForm(CustomerRegistrationType::class,$user);
$form->handleRequest($request);
$company = $user->getCompany();
$company->setSendInvoice(true);
$company->setSendInfo(true);
if ($form->isSubmitted()){
$user->addCompany($company);
$this->userManager->updateUser($user);
....
}
return $this->render('Registration/register.html.twig',[
'form' => $form->createView()
]);
}我不重要，不管我怎么做，每次我都会出错。在本例中，公司数据为空。其他方式与ArrayCollection() of $this->company in User()发生冲突。
User {#907 ▼
#id: null
#company: ArrayCollection {#926 ▼
-elements: []
}
#groups: null
#mandant: null
#salutation: null
#first_name: null
#last_name: null
#fullname: null
#created: DateTime @1544539763 {#901 ▶}
#telephone: null更新:我无法访问的Twig表单中的
{{ form_widget(form.company.company_name, {'attr': {'class': 'form-control'}})}}属性"company_name“和方法"company_name()”、company_name或"__call()“都不存在，并且在类”Symfony\Component\FormView“中具有公共访问权。
这是在表单提交之前出现的。
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-12-13 10:14:03
我在这里找到了解决方案：https://groups.google.com/forum/#!topic/symfony2/DjwwzOfUIuQ
因此，我在控制器中使用了一个基于数组的表单。
$form = $this->createFormBuilder()
->add('user', CustomerRegistrationType::class, array(
'data_class' => 'AppBundle\Entity\User'
))
->add('company', CompanyFormType::class, array(
'data_class' => 'AppBundle\Entity\Company'
))
->getForm();不同的对象可以像这样捕捉到：
$form->handleRequest($request);
$user = $form->get('user')->getData();
$company = $form->get('company')->getData();然后
if ($form->isSubmitted()){
...
$user->addCompany($company);
$this->userManager->updateUser($user);
}就这样
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53726939
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