使用SUM(a.id = 1)作为"标识符"时的学说错误:预期的Doctrine\ORM\Query\Lexer :: T_CLOSE_PARENTHESIS,得到'='

Question

我试图在包含这样的东西的学说中执行查询

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,似乎Doctrine在MySQL表达式上对它自己进行了一些解析.正如nietonfir已经提出的那样,这个解析器可能不完全符合MySQL的要求.您可以尝试一些语法变体,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`

要么:

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;我将保留QueryBuilder等同于你):

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.