Php 从表单输入检索数据
我试图从使用symfony创建的表单输入中检索数据 这是我的FormUploaderType:Php 从表单输入检索数据,php,forms,symfony,Php,Forms,Symfony,我试图从使用symfony创建的表单输入中检索数据 这是我的FormUploaderType: class FormUploaderType extends AbstractType { public function buildForm(FormBuilderInterface $builder, array $options) { $builder ->add('service', ChoiceType::class, [
class FormUploaderType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('service', ChoiceType::class, [
'placeholder' => 'Wybierz serwis',
'choices' => [
'home' => 'home',
'az' => 'az',
'ionos.bg' => 'ionosbg',
'ionos.ro' => 'ionosro',
'ionos.hu' => 'ionoshu',
],
'mapped' => false,
])
->add('date', DateType::class)
->add('terms', FileType::class)
->add('button', SubmitType::class);
}
下面是我的表单在html中的外观:
我试图从form_uploader_date输入中获取数据,但这行代码:
$date = $form->get('form_uploader')->getData();
或:
似乎不起作用。。。我不知道该怎么办…你处理了这个请求吗?你应该这样做:
$form = $this->createForm(CLASS);
$form->handleRequest($request);` // you should inject Request from HttpFoundation namespace
if($form->isSubmitted() & $form->isValid()) {
$data = $form->getData();
// This array holds all your data.
// Use this instead of accesing each element as $form->get('property-name')->getData();
}
之后,验证您的请求是否是POST
,如下所示:
$form = $this->createForm(CLASS);
$form->handleRequest($request);` // you should inject Request from HttpFoundation namespace
if($form->isSubmitted() & $form->isValid()) {
$data = $form->getData();
// This array holds all your data.
// Use this instead of accesing each element as $form->get('property-name')->getData();
}
注意:如果表单类与实体/模型关联,则该表单将映射到该类上。否则,它将只是一个数组。你不需要那样访问它。此外,您还可以使用
dump()
或dd()
(对于Symfony 4.x+)函数以漂亮的格式转储内容。您是否尝试过$date=$form->getData()['date'];?顺便说一句,最好在表单中有一个DTO,并通过该DTO访问数据。谢谢,这一个很有帮助!:)