FosMessageBundle sendId不能為null

Question

我用我的消息包實現了fosMessageBundle，但是如果我想發送新消息，請使用fos_message.composer，我收到此錯誤：

    An exception occurred while executing 'INSERT INTO Message (body, created_at, threadId,   sendId) VALUES (?, ?, ?, ?)' with params ["Test mesaj\u0131", "2013-12-23 12:15:48", 32, null]:

SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'sendId' cannot be null

---

我的問題是，為什么未設置消息束發送ID？

---

注意：我正在調試setSender數據。 每個數據似乎都可以。

---

代碼：

$sender     = $this->getUser();
        $threadBuilder = $this->get('fos_message.composer')->newThread();
        $threadBuilder->
            addRecipient($clinicOwner)
            ->setSender($sender)
            ->setSubject($form['subject'])
            ->setBody($form['message']);
        $sender = $this->get('fos_message.sender');
        $sender->send($threadBuilder->getMessage());

Answer 1

問題解決了！

message.orm.yml中的manyToOne.sender.joinColumn.name寫錯。

錯誤的：

manyToOne:
thread:
  targetEntity: ATL\MessageBundle\Entity\Thread
  inversedBy: messages
  joinColumn:
    name: threadId
    referencedColumnName: id
sender:
  targetEntity: ATL\UserBundle\Entity\User
  joinColumn:
    name: **threadId**
    referencedColumnName: id

真正：

manyToOne:
thread:
  targetEntity: ATL\MessageBundle\Entity\Thread
  inversedBy: messages
  joinColumn:
    name: threadId
    referencedColumnName: id
sender:
  targetEntity: ATL\UserBundle\Entity\User
  joinColumn:
    name: sendId
    referencedColumnName: id