Symfony CollectionType -> EntryType -> 带参数的表单
Symfony CollectionType -> EntryType -> Form with params
我有两个表单,第一个允许我添加项目 ProjectType,第二个是允许向项目添加贡献者的表单 ProjectContactType。
在第一种形式中,我得到了带有解析器的实体公司
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',
]);
}
我希望能够在我的第二种形式中接收到相同的实体。
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,
]);
}
我的目标是能够按公司实体过滤用户。
谢谢。
您应该能够像这样将 ProjectType 的公司传递给 ProjectContactType:
在 ProjectContactType 中创建所需的公司:
public function configureOptions(OptionsResolver $resolver): void
{
$resolver
->setDefaults([
'data_class' => ProjectContact::class,
])
->setRequired([
'company',
]);
}
然后(在 ProjectType 中)将公司从 ProjectType 的选项传递给 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'],
],
]);
}
我有两个表单,第一个允许我添加项目 ProjectType,第二个是允许向项目添加贡献者的表单 ProjectContactType。
在第一种形式中,我得到了带有解析器的实体公司
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',
]);
}
我希望能够在我的第二种形式中接收到相同的实体。
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,
]);
}
我的目标是能够按公司实体过滤用户。
谢谢。
您应该能够像这样将 ProjectType 的公司传递给 ProjectContactType:
在 ProjectContactType 中创建所需的公司:
public function configureOptions(OptionsResolver $resolver): void
{
$resolver
->setDefaults([
'data_class' => ProjectContact::class,
])
->setRequired([
'company',
]);
}
然后(在 ProjectType 中)将公司从 ProjectType 的选项传递给 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'],
],
]);
}