mysql - Symfony2 多对多与 COUNT

Question

在我的一个存储库中获得了这个 QueryBuilder。

        $query = $em->createQueryBuilder('d')
                ->select('d, i, u, SUM(CASE WHEN t.user = :userId THEN 1 ELSE 0 END) as myTickets')
                ->leftJoin('d.item','i')
                ->leftJoin('d.users','u')
                ->leftJoin('d.tickets','t')
                ->where('d.active = 1')
                ->andWhere('d.state = 1')
                ->setParameter('userId',$user->getId())
                ->orderBy('d.dateFinish', 'ASC');

当我执行代码时，MySQL 抛出这个错误。

Key "premium" for array with keys "0, myTickets" does not exist

“premium”是“d”的一个字段。

如何使用自定义 SUM 接收字段？

Answer 1

由于您在查询中使用聚合函数，因此您会得到所谓的混合结果。混合结果通常会将使用 FROM 子句获取的对象返回为零索引 [0]。结果的其余部分将根据您为自定义字段设置的别名进行填充。

$result[0] 将返回您要访问的对象。

$result['myTickets'] 将返回聚合函数的结果。在本例中，它是一个 SUM。

引用自文档:

SELECT u, UPPER(u.name) nameUpper FROM MyProject\Model\User u

This query makes use of the UPPER DQL function that returns a scalar value and because there is now a scalar value in the SELECT clause, we get a mixed result.

Conventions for mixed results are as follows:

The object fetched in the FROM clause is always positioned with the key ‘0’.

Every scalar without a name is numbered in the order given in the query, starting with 1.

Every aliased scalar is given with its alias-name as the key. The case of the name is kept.

If several objects are fetched from the FROM clause they alternate every row.

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