Question

我对Doctrine还很陌生，我正在尝试检索要关注的用户建议列表。

基本上，鉴于用户A，我需要选择用户A关注的用户所遵循的所有用户，不包括用户A已经关注的用户。

我如何使用Doctrine查询构建器执行此操作？

class User
{

...

/**
 * @ORM\ManyToMany(targetEntity="User", inversedBy="followees")
 * @ORM\JoinTable(name="user_followers",
 *      joinColumns={@ORM\JoinColumn(name="user_id", referencedColumnName="id", onDelete="CASCADE")},
 *      inverseJoinColumns={@ORM\JoinColumn(name="follower_id", referencedColumnName="id", onDelete="CASCADE")}
 *      )
 */
private $followers;

/**
 * @ORM\ManyToMany(targetEntity="User", mappedBy="followers")
 */
private $followees;
}

编辑：根据slaur4解决方案，我试过了

    $qb = $this->createQueryBuilder('u');

    $qb->select('suggestions')
    ->join('u.followees', 'followees')
    ->join('followees.followers', 'suggestions')
    ->where('u.id = :id')
    ->andWhere($qb->expr()->notIn('suggestions.id', 'u.followees'))
    ->setParameter('id', $user->getId());

但它给了我以下例外：

QueryException: [Syntax Error] line 0, col 171: Error: Expected Literal, got 'u'

Answer 1

这是一个自引用查询。我会试试这个：

QueryBuilder（用户Symfony2存储库）

<?php

//Subquery to exclude user A followees from suggestions
$notsQb = $this->createQueryBuilder('user')
    ->select('followees_excluded.id')
    ->join('user.followees', 'followees_excluded')
    ->where('user.id = :id');

$qb = $this->createQueryBuilder('suggestions');
$qb->join('suggestions.followers', 'suggestions_followers')
    ->join('suggestions_followers.followers', 'users')
    ->where('users.id = :id')
    ->andWhere('suggestions.id != :id') //Exclude user A from suggestions
    ->andWhere($qb->expr()->notIn('suggestions.id', $notsQb->getDql()))
    ->setParameter('id', $userA->getId());
$query = $qb->getQuery();
$users = $query->getResult(); // array of User

Doctrine 2自加入查询

1 个答案: