使用 SUM(a.id=1) 作为 `ìdentifier` 时的教义错误：预期的 Doctrine\ORM\Query\Lexer::T_CLOSE_PARENTHESIS，得到 '='

Question

我正在尝试在学说中执行包含类似内容的查询

SUM(a.id = 1) as `1`

由于某些原因，它总是给我以下错误：

[Syntax Error] line 0, col 15: Error: Expected Doctrine\ORM\Query\Lexer::T_CLOSE_PARENTHESIS, got '='

这是我正在使用的代码

 $result = $em->getRepository('MyBundle:PlayerAction')
            ->createQueryBuilder('pa')
            ->select(array(
                'SUM(a.id=1) as `1`,
                SUM(a.id=2) as `2`,
                SUM(a.id=3) as `3`,
                p.playerName,
                pa.timestamp'
            ))
            ->innerJoin('pa.action', 'a')
            ->innerJoin('pa.player', 'p')
            ->where('pa.timestamp > ?1')
            ->groupBy('p')
            ->setParameter(1, time() - $time)
            ->orderBy('p.playerName', 'ASC');

Answer 1

考虑到错误消息，Doctrine 似乎对您的 MySQL 表达式进行了一些解析。正如 nietonfir 已经建议的那样，这个解析器可能不完全符合 MySQL 的。您可以尝试一些语法变化，例如：

SUM(CASE a.id WHEN 1 THEN 1 ELSE 0 END) as `1`,
SUM(CASE a.id WHEN 2 THEN 1 ELSE 0 END) as `2`,
SUM(CASE a.id WHEN 3 THEN 1 ELSE 0 END) as `3`

或：

SUM(IF(a.id=1, 1, 0)) as `1`,
SUM(IF(a.id=2, 1, 0)) as `2`,
SUM(IF(a.id=3, 1, 0)) as `3`

如果失败，重构查询；沿着这条线的东西（只是在这里大声思考；我将 QueryBuilder 等价留给你）：

SELECT
   (SELECT COUNT(*) FROM PlayerAction AS pa WHERE pa.player = p.id AND pa.timestamp > ?1 AND pa.action = 1) AS `1`,
   (SELECT COUNT(*) FROM PlayerAction AS pa WHERE pa.player = p.id AND pa.timestamp > ?1 AND pa.action = 2) AS `2`,
   (SELECT COUNT(*) FROM PlayerAction AS pa WHERE pa.player = p.id AND pa.timestamp > ?1 AND pa.action = 3) AS `3`,
   p.playerName
FROM Player p
ORDER BY p.playerName ASC

这可能会对查询性能产生负面影响。

Answer 2

Doctrine 似乎不支持这种语法。您可以重写您的查询（可能会想到子查询或某些连接）或使用原生 MySQL 查询。