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selectDistinct似乎对我不起作用，这可能是一个简单的错误。查询：

 info <- runDB $ 
        E.selectDistinct $ 
        E.from $ \(tp `E.InnerJoin` rnd `E.InnerJoin` h) -> do
        E.on (rnd E.^. RoundId E.==. h E.^. HoleRound)
        E.on (tp E.^. TpartTournament E.==. rnd E.^. RoundTourn)
        E.where_ ((tp E.^. TpartTournament E.==. E.val tId ))
        E.orderBy [E.asc (tp E.^. TpartId)]
        return (tp, rnd, h)

我很确定这代表了有效的 sql 查询：

SELECT DISTINCT tpart.id, round.name, hole.hole_num, hole.score
from tpart
inner join round on round.tourn = tpart.tournament
inner join hole on hole.round = round.id
where tpart.tournament = 1;

要查看结果，我有一个测试处理程序来打印结果表。请注意，对于 tpart 1，第 1 轮，有多个洞 1 和洞 2。在 postgresqlSELECT DISTINICT中删除了这些重复项。

     TpartId, RoundName, holeNum, HoleScore

Key {unKey = PersistInt64 1}, round 1, 1, 6
Key {unKey = PersistInt64 1}, round 1, 2, 4
Key {unKey = PersistInt64 1}, round 1, 1, 6
Key {unKey = PersistInt64 1}, round 1, 2, 4
Key {unKey = PersistInt64 1}, round 1, 1, 6
Key {unKey = PersistInt64 1}, round 1, 2, 4
Key {unKey = PersistInt64 1}, round 2, 1, 3
Key {unKey = PersistInt64 1}, round 2, 2, 5
Key {unKey = PersistInt64 1}, round 2, 1, 3
Key {unKey = PersistInt64 1}, round 2, 2, 5
Key {unKey = PersistInt64 1}, round 2, 1, 3
Key {unKey = PersistInt64 1}, round 2, 2, 5
Key {unKey = PersistInt64 3}, round 1, 1, 6
Key {unKey = PersistInt64 3}, round 1, 2, 4
Key {unKey = PersistInt64 3}, round 1, 1, 6
Key {unKey = PersistInt64 3}, round 1, 2, 4
Key {unKey = PersistInt64 3}, round 1, 1, 6
Key {unKey = PersistInt64 3}, round 1, 2, 4
Key {unKey = PersistInt64 3}, round 2, 1, 3
Key {unKey = PersistInt64 3}, round 2, 2, 5
Key {unKey = PersistInt64 3}, round 2, 1, 3
Key {unKey = PersistInt64 3}, round 2, 2, 5
Key {unKey = PersistInt64 3}, round 2, 1, 3
Key {unKey = PersistInt64 3}, round 2, 2, 5

对不起，难以辨认。任何帮助，将不胜感激！

4

1 回答 1

0

错误是对于某个洞，洞的round和洞的part必须相等，它们是各自的部分。此外，在这种情况下，内部连接是多余的。

于 2013-11-24T15:35:27.817 回答