php - FosMessageBundle sendId 不能为空

Question

我用我的消息包实现了 fosMessageBundle ，但是如果我想发送一条新消息使用 fos_message.composer ，我得到了这个错误：

    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

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我的问题是，为什么消息包没有设置发送 ID？

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注意：我正在调试 setSender 数据。每个数据似乎都可以。

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代码：

$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

问题解决了！

message.orm.yml 中的 manyToOne.sender.joinColumn.name 写错了。

错误的 ：

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

真的：

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