Question

目前我有3张桌子。

articles_categories

+-------------+---------+------+-----+---------+-------+
| Field       | Type    | Null | Key | Default | Extra |
+-------------+---------+------+-----+---------+-------+
| article_id  | int(11) | NO   | PRI | NULL    |       |
| category_id | int(11) | NO   | PRI | NULL    |       |
+-------------+---------+------+-----+---------+-------+

类别

+-----------------+--------------+------+-----+---------+----------------+
| Field           | Type         | Null | Key | Default | Extra          |
+-----------------+--------------+------+-----+---------+----------------+
| id              | int(11)      | NO   | PRI | NULL    | auto_increment |
| title           | varchar(255) | NO   |     | NULL    |                |
+-----------------+--------------+------+-----+---------+----------------+

制品

+-----------------+--------------+------+-----+---------+----------------+
| Field           | Type         | Null | Key | Default | Extra          |
+-----------------+--------------+------+-----+---------+----------------+
| id              | int(11)      | NO   | PRI | NULL    | auto_increment |
| title           | varchar(255) | NO   |     | NULL    |                |
+-----------------+--------------+------+-----+---------+----------------+

在类别实体

中

/**
     * @ORM\ManyToMany(targetEntity="Article", mappedBy="categories")
     * @ORM\OrderBy({"createdAt" = "DESC"})
     */
    protected $articles;

在文章实体中

/**
     * All categories this article belongs to.
     *
     * @ORM\ManyToMany(targetEntity="Category", inversedBy="articles", cascade={"persist"})
     * @ORM\JoinTable(name="articles_categories")
     */
    protected $categories;

大多数查询都运行正常。但我想获得属于某一类别的文章。或者具有特定类别或想要根据类别过滤文章。

为此，我写了以下查询。

$em = $this->getEntityManager();
        $qb = $em->createQueryBuilder('c');
        $qb->select('1')
            ->from('articles_categories', 'a_c')
            ->leftJoin('\\Chip\\Entity\\Article', 'a', 'WITH', 'a.id = a_c.article_id')
            ->leftJoin('\\Chip\\Entity\\Category', 'c', 'WITH', 'c.id = a_c.category_id')
        ;

        $result = $qb->getQuery()->getResult();

但它会引发跟随错误。

[Semantical Error] line 0, col 14 near 'articles_categories': Error: Class 'articles_categories' is not defined.
500 Internal Server Error - QueryException
1 linked Exception: QueryException »

任何帮助或提示或任何更好的查询方式都会很棒。

先谢谢。

Answer 1

您不应在查询构建器中使用表名，而应使用类名。

$em = $this->getEntityManager();
$qb = $em->createQueryBuilder('c');
$qb->select('c', 'a')
   ->from('Chip\Entity\Category', 'c') // this line is not necessary when performing this query in your category repository
   ->leftJoin('c.articles, a')
   ->where('c.id = 1');
$result = $qb->getQuery()->getResult();

教义查询多对多[语义错误]

1 个答案: