Question

在Symfony 1.4中，Doctrine我有来自schema.yml的这个片段

Attendance:
  columns:
    id: { type: integer(4), primary: true, autoincrement: true }
    user_id: { type: integer(4) }

  relations:
    JoomlaUser: { class: JosUser, local: user_id, foreignAlias: AttendanceList }

和

JosUser:
  tableName: jos_users
  columns:
    id: { type: integer(4), primary: true, autoincrement: true }
  relations:
    AttendanceList: { class: Attendance, local: id, foreign: user_id }

从JosUser确定没有出勤记录的最有效方法是什么？我尝试过model / doctrine / JosUser.class.php

count($this->getAttendanceList())

但是这会返回一个Doctrine Record，所有字段都为空但user_id

Answer 1

架构：

JosUser:
    tableName: jos_users
    columns:
        id: { type: integer(4), primary: true, autoincrement: true }
    relations:
    AttendanceList: { type: many, class: Attendance, local: id, foreign: user_id }

使用type: many，getAttendanceList()应该返回一个Doctrine_Collection（参见lib / model / doctrine / base / BaseJosUser.class.php中的JosUser基类），然后你可以使用

$this->getAttendanceList()->count()

Symfony学说确定外来Alias List是空的

1 个答案: