[英]Symfony CollectionType -> EntryType -> Form with params
我有兩個 forms,第一個允許我添加項目ProjectType ,第二個是允許向項目ProjectContactType添加貢獻者的表單。
在第一種形式中,我得到帶有解析器的實體公司
public function buildForm(FormBuilderInterface $builder, array $options): void
{
$builder
->add('name', TextType::class)
->add('projectContacts', CollectionType::class, [
'label' => false,
'entry_type' => ProjectContactType::class,
'by_reference' => false,
'allow_add' => true,
'allow_delete' => true,
]);
}
public function configureOptions(OptionsResolver $resolver): void
{
$resolver
->setDefaults([
'data_class' => Project::class,
])
->setRequired([
'company',
]);
}
我希望能夠以我的第二種形式接收同樣的實體。
public function buildForm(FormBuilderInterface $builder, array $options): void
{
$builder
->add('user', EntityType::class, [
'placeholder' => 'Choisir un client',
'required' => true,
'class' => User::class,
'choice_label' => function (User $user) {
return $user->getFirstName() . " " . $user->getLastName();
}
]);
}
public function configureOptions(OptionsResolver $resolver): void
{
$resolver
->setDefaults([
'data_class' => ProjectContact::class,
]);
}
我的目標是能夠按公司實體過濾用戶。
謝謝你。
您應該能夠像這樣將 ProjectType 的公司傳遞給 ProjectContactType:
使 ProjectContactType 中所需的公司:
public function configureOptions(OptionsResolver $resolver): void
{
$resolver
->setDefaults([
'data_class' => ProjectContact::class,
])
->setRequired([
'company',
]);
}
然后(在 ProjectType 中)將公司從 ProjectType 的選項傳遞給 ProjectContactType:
public function buildForm(FormBuilderInterface $builder, array $options): void
{
$builder
->add('name', TextType::class)
->add('projectContacts', CollectionType::class, [
'label' => false,
'entry_type' => ProjectContactType::class,
'by_reference' => false,
'allow_add' => true,
'allow_delete' => true,
'entry_options' => [
'company' => $options['company'],
],
]);
}
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