Question

我有一个上课的帖子：

/**
 * @ORM\Entity(repositoryClass="App\Repository\PostRepository")
 */
class Post
{
    const TYPE_TEXT   = 1;
    const TYPE_PHOTOS = 2;
    const TYPE_VIDEO  = 3;

    /**
     * @ORM\OneToMany(targetEntity="Photo", mappedBy="post")
     */
    private $photos;

    and other properties, methods, etc...

我只想带照片的帖子带回来。

我有一个DQL查询，例如：

    $qb = $this->createQueryBuilder('p');
    $qb->select('p, postPhotos, etc...')
        ->leftJoin('p.photos', 'postPhotos')
        ->leftJoin('p.videos', 'postVideos')
        etc...

    if ($mediaType != null)
    {
        switch ($mediaType) {
            case Post::TYPE_PHOTOS:
                $qb->andWhere('postPhotos != :null')
                    ->setParameter('null', null);

“！=：null”无效，COUNT（postPhotos）也无效（显然是出于聚合原因）。

有没有一种方法我只能指定带回1张或更多照片的帖子？

Answer 1

快速解答：如果您仅使用join（或innerJoin）替换leftJoin的用法，那么您将得到想要的：只有具有至少1张照片的帖子。

详细信息

如果您查看此有用的SO问答：

Different MySQL Join Methods

...您会发现一些出色的维恩图，显示了左连接和内部连接之间的差异。然后，如果查看Doctrine的Doctrine \ ORM \ QueryBuilder类，您会发现它们具有三种连接方法：

join（仅调用innerJoin）
innerJoin
leftJoin

    /**
     * Creates and adds a join over an entity association to the query.
     *
     * The entities in the joined association will be fetched as part of the query
     * result if the alias used for the joined association is placed in the select
     * expressions.
     *
     * <code>
     *     $qb = $em->createQueryBuilder()
     *         ->select('u')
     *         ->from('User', 'u')
     *         ->join('u.Phonenumbers', 'p', Expr\Join::WITH, 'p.is_primary = 1');
     * </code>
     *
     * @param string      $join          The relationship to join.
     * @param string      $alias         The alias of the join.
     * @param string|null $conditionType The condition type constant. Either ON or WITH.
     * @param string|null $condition     The condition for the join.
     * @param string|null $indexBy       The index for the join.
     *
     * @return self
     */
    public function join($join, $alias, $conditionType = null, $condition = null, $indexBy = null)
    {
        return $this->innerJoin($join, $alias, $conditionType, $condition, $indexBy);
    }

    /**
     * Creates and adds a join over an entity association to the query.
     *
     * The entities in the joined association will be fetched as part of the query
     * result if the alias used for the joined association is placed in the select
     * expressions.
     *
     *     [php]
     *     $qb = $em->createQueryBuilder()
     *         ->select('u')
     *         ->from('User', 'u')
     *         ->innerJoin('u.Phonenumbers', 'p', Expr\Join::WITH, 'p.is_primary = 1');
     *
     * @param string      $join          The relationship to join.
     * @param string      $alias         The alias of the join.
     * @param string|null $conditionType The condition type constant. Either ON or WITH.
     * @param string|null $condition     The condition for the join.
     * @param string|null $indexBy       The index for the join.
     *
     * @return self
     */
    public function innerJoin($join, $alias, $conditionType = null, $condition = null, $indexBy = null)
    {
        $parentAlias = substr($join, 0, strpos($join, '.'));

        $rootAlias = $this->findRootAlias($alias, $parentAlias);

        $join = new Expr\Join(
            Expr\Join::INNER_JOIN, $join, $alias, $conditionType, $condition, $indexBy
        );

        return $this->add('join', [$rootAlias => $join], true);
    }

    /**
     * Creates and adds a left join over an entity association to the query.
     *
     * The entities in the joined association will be fetched as part of the query
     * result if the alias used for the joined association is placed in the select
     * expressions.
     *
     * <code>
     *     $qb = $em->createQueryBuilder()
     *         ->select('u')
     *         ->from('User', 'u')
     *         ->leftJoin('u.Phonenumbers', 'p', Expr\Join::WITH, 'p.is_primary = 1');
     * </code>
     *
     * @param string      $join          The relationship to join.
     * @param string      $alias         The alias of the join.
     * @param string|null $conditionType The condition type constant. Either ON or WITH.
     * @param string|null $condition     The condition for the join.
     * @param string|null $indexBy       The index for the join.
     *
     * @return self
     */
    public function leftJoin($join, $alias, $conditionType = null, $condition = null, $indexBy = null)
    {
        $parentAlias = substr($join, 0, strpos($join, '.'));

        $rootAlias = $this->findRootAlias($alias, $parentAlias);

        $join = new Expr\Join(
            Expr\Join::LEFT_JOIN, $join, $alias, $conditionType, $condition, $indexBy
        );

        return $this->add('join', [$rootAlias => $join], true);
    }

更改代码以使用innerJoin（或仅联接）将导致Doctrine在生成的SQL中发出INNER JOIN，该联接仅返回联接两侧都存在“某物”的记录，因此，任何没有照片不会包含在结果中。

（原则）选择数组集合不为空的地方？

1 个答案: