Symfony/Formbuilder。选择查询构建器上的特定字段

Question

如何在symfony上的表单上选择查询构建器上的特定字段？

我triying做到这一点：

 ->add('tienda', EntityType::class, array(
      'class' => 'AdminBundle:MyTable', 
      'query_builder' => function (EntityRepository $er) { 
       return $er->createQueryBuilder('select tienda') 
        ->distinct() 
        ->orderBy('u.tienda', 'ASC'); 
      }, 
      'required' => false, 
     )) 

Answer 1

使用choice_label选项它

->add('tienda', EntityType::class, array(
     'class' => 'AdminBundle:MyTable', 
     'choice_label' => 'xxx', 
     'query_builder' => function (EntityRepository $er) { 
      return $er->createQueryBuilder('u') 
       ->distinct() 
       ->orderBy('u.tienda', 'ASC'); 
     }, 
     'required' => false, 
    )) 

http://symfony.com/doc/current/reference/forms/types/entity.html#choice-label