学说 通过多对多自引用关系的标准进行计数

Question

我有一个可以有很多朋友的用户。表示关系是自引用的。我现在需要根据性别计算我的朋友朋友的数量。到目前为止，我想出了以下解决方案。

我的用户实体：

<?php
namespace AcmeUserBundleEntity;
use DoctrineCommonCollectionsArrayCollection;
use DoctrineORMMapping as ORM;
/**
 * @ORMEntity(repositoryClass="AcmeUserBundleEntityRepositoryUserRepository")
 * @ORMTable(name="users")
 */
class User extends BaseUser
{
    /**
     * @ORMManyToMany(targetEntity="AcmeUserBundleEntityUser", inversedBy="friendsOf")
     * @ORMJoinTable(name="map_user_friend",
     *      joinColumns={@ORMJoinColumn(name="user_id", referencedColumnName="id")},
     *      inverseJoinColumns={@ORMJoinColumn(name="friend_id", referencedColumnName="id")}
     *      )
     */
    protected $friends;
    /**
     * @ORMManyToMany(targetEntity="AcmeUserBundleEntityUser", mappedBy="friends")
     */
    private $friendsOf;

    /**
     * @ORMColumn(type="string", length=1, nullable=true)
     */
    protected $gender;
    // other fields ...
}

我的用户存储库具有以下方法：

public function getFriendsQuery(User $user)
{
    return $this->getEntityManager()->createQueryBuilder()
        ->select('u as user, COUNT(DISTINCT mf.id) as male_friends, COUNT(DISTINCT ff.id) as female_friends')
        ->from('OneupUserBundleEntityUser', 'u')
        ->leftJoin('u.friendsOf', 'f')
        ->leftJoin('u.friendsOf', 'mf', 'with', 'mf.gender = :gender')->setParameter('gender', 'm')
        ->leftJoin('u.friendsOf', 'ff', 'with', 'ff.gender = :gender')->setParameter('gender', 'w')
        ->where('f = :user')->setParameter('user', $user)
        ->groupBy('u.id')
    ;
}

在控制器中，我现在首先通过返回DoctrineCommonUtilDebug::dump($qb->getQuery()->getSQL());转储 SQL

SELECT u0_.username AS username0,
       u0_.username_canonical AS username_canonical1,
       -- ... --
       u0_.gender AS gender19,
       -- ... --
       COUNT(DISTINCT u1_.id) AS sclr32,
       COUNT(DISTINCT u2_.id) AS sclr33
FROM users u0_
LEFT JOIN map_user_friend m4_ ON u0_.id = m4_.friend_id
LEFT JOIN users u3_ ON u3_.id = m4_.user_id
LEFT JOIN map_user_friend m5_ ON u0_.id = m5_.friend_id
LEFT JOIN users u1_ ON u1_.id = m5_.user_id
AND (u1_.gender = ?)
LEFT JOIN map_user_friend m6_ ON u0_.id = m6_.friend_id
LEFT JOIN users u2_ ON u2_.id = m6_.user_id
AND (u2_.gender = ?)
WHERE u3_.id = ?
GROUP BY u0_.id

m为男性，f为女性，21为当前用户。现在遵循奇怪的行为。 DoctrineCommonUtilDebug::dump($qb->getQuery()->execute());回报我

array (size=2)
  0 =>
    array (size=3)
      'user' => string 'OneupUserBundleEntityUser' (length=28)
      'male_friends' => string '2' (length=1)
      'female_friends' => string '2' (length=1)
  1 =>
    array (size=3)
      'user' => string 'OneupUserBundleEntityUser' (length=28)
      'male_friends' => string '0' (length=1)
      'female_friends' => string '0' (length=1)

如果我直接在 mysql 中执行查询，我会得到另一个计数：

+--------------+-----------------+-----------------+
| username0    | sclr32 (male)   | sclr33 (female) |
+--------------+-----------------+-----------------+
| foobar12     | 1               | 2               |
+--------------+-----------------+-----------------+
| foobar2      | 2               | 0               |
+--------------+-----------------+-----------------+

这是正确的。这是一个教义错误，还是我只是做错了？

Answer 1

为了回答我自己的问题，错误出在存储库中的参数上

   ->leftJoin('u.friendsOf', 'mf', 'with', 'mf.gender = :gender')->setParameter('gender', 'm')
   ->leftJoin('u.friendsOf', 'ff', 'with', 'ff.gender = :gender')->setParameter('gender', 'w')

错了。学说对这两个条目都采用第二个值。

   ->leftJoin('u.friendsOf', 'mf', 'with', 'mf.gender = :male')->setParameter('male', 'm')
   ->leftJoin('u.friendsOf', 'ff', 'with', 'ff.gender = :female')->setParameter('female', 'w')