有一种形式与两个实体相关联.我需要使一个实体可选-symfony 2

Question

我有一份员工登记表，如下所示，

namespace HrmEmployeeBundleForm;
use SymfonyComponentFormAbstractType;
use SymfonyComponentFormFormBuilder;
use HrmUserBundleFormUserUserForm;
class EmpForm extends AbstractType
{
public function buildForm(FormBuilder $builder, array $options)
{   
    $builder->add('id','hidden');
    $builder->add('firs_name', 'text');
    $builder->add('middle_name', 'text');
    $builder->add('last_name', 'text');     
    $builder->add('user', new UserForm(), array('required' => false));
}
public function getName()
{
    return 'EmpForm';
}
}

我还有一个用户登录表，如下所示，

namespace HrmUserBundleFormUser;
use SymfonyComponentFormAbstractType;
use SymfonyComponentFormFormBuilder;
class UserForm extends AbstractType
{
  public function buildForm(FormBuilder $builder, array $options)
  {   
    $builder->add("id","hidden",array("required"=>false));
    $builder->add("userName","text",array("label"=>"Username"));
    $builder->add("password","password",array("label"=>"Password"));
    $builder->add("confirmPassword","password",array("property_path" => 
                   false,"label"=>"Confirm Password"));
    $builder->add("email","email",array("label"=>"Email"));
  }
  public function getName()
  {
    return 'user';
  }
}

员工YML文件如下，

HrmEmployeeBundleEntityEmployee:
type: entity
table: employee
fields:
  id:
    type: integer
    id: true
    generator:
      strategy: AUTO
  firs_name:
    type: string
    length: '255'
    default: null
  middle_name:
    type: string
    length: '255'
    default: null
  last_name:
    type: string
    length: '255'
    default: null  
  oneToOne:
    User:
      targetEntity: HrmUserBundleEntityUser
      cascade: ["remove", "persist"]
      joinColumn:
        name: login_id        
        referencedColumnName: id        
        nullable: true                   
 lifecycleCallbacks: {  }

员工管理员是这样的，

public function addEmployeeAction(Request $request) {
    $employee = new Employee();
    $user = new User();
    $employee->setUser($user);
    $form = $this->createForm(new EmpForm(), $employee);
    if ($request->getMethod() == 'POST') {
        $form->bindRequest($request);
        if ($form->isValid()) {
            $em = $this->getDoctrine()->getEntityManager();
            $em->persist($employee);
            $em->flush();
            return $this->redirect($this->generateUrl('HrmEmployeeBundle_homepage'));
        }
    }
    return $this->render('HrmEmployeeBundle:Pages:add.html.twig', array(
                'form' => $form->createView(), 'status' => 'addEmployee',
            ));
}

所以我需要做的是以下选项应该可用，

1. 填写员工注册详细信息并创建登录名（已完成）
2. 只填写员工注册详细信息，不需要为此创建登录名

所以第二个选项对我不起作用，它说列"用户名"不能为空我该如何解决这个问题？

Answer 1

要回答第二个问题，请在addEmployeeAction中，不要在新员工上添加用户，删除以下行：

$user = new User();
$employee->setUser($user);

目前，$employee是用一个没有用户名的空用户保存的，所以MySQL很不高兴！通过删除这些行，只有在填写了表单的情况下才会创建用户！

Answer 2

如果要设置SQL字段optionnal，请添加可为null的选项true:

oneToOne:
    User:
      targetEntity: SysUserBundleEntityUser
      cascade: ["persist"]
      joinColumn:
        name: login_id
        referencedColumnName: id
        nullable: true

但在您的情况下，似乎您注册了一个没有用户名的用户，这是不合乎逻辑的。。。也许用户名字段工作不正常，请给我们更多的细节。