5
我有保存實體回我這個錯誤的一個問題:Symfony2的__toString()錯誤
Catchable Fatal Error: Method My\BusinessBundle\Entity\Type::__toString()
must return a string value in
/var/www/MyBusiness0_1/vendor/doctrine/orm/lib/Doctrine/ORM/ORMInvalidArgumentException.php line 113
奇怪的是,該方法__的toString()是實體型!
class Type
{
//..
/**
* @var string
*
* @ORM\Column(name="type", type="string", length=100)
*/
private $type;
/**
* @ORM\OneToMany(targetEntity="MailTelCont", mappedBy="type")
*/
protected $mailTelContacts;
public function __construct()
{
$this->mailTelContacts = new \Doctrine\Common\Collections\ArrayCollection();
}
public function __toString()
{
return $this->getType();
}
//...
另一個奇怪的事情是,如果我把級聯= {「堅持」}類MailTelCont上多對一的關係「型」不顯示我這個錯誤,但在保存類型的新領域..
類MailTelCont
class MailTelCont
{
//..
/**
* @var string
*
* @ORM\Column(name="contact", type="string", length=100)
*/
private $contact;
/**
* @ORM\ManyToOne(targetEntity="Type", inversedBy="mailTelContacts")
* @ORM\JoinColumn(name="type_id", referencedColumnName="id")
*/
private $type;
/**
* @ORM\ManyToOne(targetEntity="Anagrafica", inversedBy="mailTelContacts", cascade={"persist"})
* @ORM\JoinColumn(name="anagrafica_id", referencedColumnName="id")
* @Assert\Type(type="My\BusinessBundle\Entity\Anagrafica")
*/
private $anagrafica;
public function __toString()
{
return $this->getContact();
}
呼叫以這種方式嵌套 「AnagraficType」 的形式:
class TypeType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('type', 'entity', array(
'class' => 'My\BusinessBundle\Entity\Type',
'attr' => array('class' => 'conct'),
'property' => 'type',
'label' => 'Tipologia',
))
;
}
*****
class MailTelContType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('type', new TypeType())
->add('contact', 'text', array('label' => 'Contatto'))
;
}
*****
class AnagraficaType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('mailTelContacts', 'collection', array('type' => new MailTelContType(),
'allow_add' => true,
'allow_delete' => true,
'prototype' => true,
'by_reference' => false
))
我在哪裏做錯了?
謝謝你的回覆, 但現在這個錯誤: '一個新的實體通過中沒有配置級聯的關係,「我的\ BusinessBundle \實體\ MailTelCont#型」找到持續經營的實體:Telefono。要解決這個問題:要麼顯式調用EntityManager#persist()在這個未知的實體上,要麼配置cascade在映射中堅持這個關聯,例如@ManyToOne(..,cascade = {「persist」})' 如果我把'cascade = {「persist」}'然而,在類型中保存一個新的字段,你不必因爲類型只是一個容器的聯繫類型的選擇(例如電子郵件,傳真,電話..) – Lughino 2013-02-17 17:01:20
它適用於我,謝謝;) – 2014-01-03 17:39:29