Question

我正在尝试使用Esqueleto（版本2.1.2.1）计算内部联接的行数。不幸的是，我的代码没有编译，我不明白为什么。我查看了以下如何执行此操作的示例，但无法弄清楚我做错了什么：example1，example2。

我的架构看起来像这样（简化）：

User
Game
  state
Player
  user UserId Maybe
  game GameId

用户可以在网站上注册玩游戏。您也可以在不注册的情况下玩游戏。因此，有一个单独的表Player。游戏有州。它可以是Ongoing，也可以是某种形式的游戏结束。我现在想要计算用户正在玩的所有正在进行的游戏。

以下SQL查询就可以了（对于固定的userId为1）：

SELECT COUNT (*)
FROM Player INNER JOIN Game
ON Player.game = Game.id
WHERE Player.user = 1 AND game.state = "Ongoing"

但是，以下Esqueleto查询无法编译：

[count1] <- runDB $ E.select                                    -- Line 25
                  $ E.from $ \(player `E.InnerJoin` game) -> do
                      E.on $ player^.PlayerGame E.==. game^.GameId
                      E.where_ $
                          player^.PlayerUser E.==. E.just (E.val userId) E.&&.
                          game^.GameState E.==. E.val MyGame.Ongoing
                      return (game, player)
                      return E.countRows                        -- Line 32

错误消息如下所示：

Handler/ListUserGames.hs:25:23:
    No instance for (E.SqlSelect (expr0 (E.Value a0)) r0)
      arising from a use of ‘E.select’
    The type variables ‘r0’, ‘expr0’, ‘a0’ are ambiguous
    Note: there are several potential instances:
      instance (E.SqlSelect a ra, E.SqlSelect b rb) =>
               E.SqlSelect (a, b) (ra, rb)
        -- Defined in ‘Database.Esqueleto.Internal.Sql’
      instance (E.SqlSelect a ra, E.SqlSelect b rb, E.SqlSelect c rc) =>
               E.SqlSelect (a, b, c) (ra, rb, rc)
        -- Defined in ‘Database.Esqueleto.Internal.Sql’
      instance (E.SqlSelect a ra, E.SqlSelect b rb, E.SqlSelect c rc,
                E.SqlSelect d rd) =>
               E.SqlSelect (a, b, c, d) (ra, rb, rc, rd)
        -- Defined in ‘Database.Esqueleto.Internal.Sql’
      ...plus 13 others
    In the expression: E.select
    In the second argument of ‘($)’, namely
      ‘E.select
       $ E.from
         $ \ (player `E.InnerJoin` game)
             -> do { E.on $ player ^. PlayerGame E.==. game ^. GameId;
                     E.where_
                     $ player ^. PlayerUser E.==. E.just (E.val userId)
                       E.&&. game ^. GameState E.==. E.val MyGame.Ongoing;
                     .... }’
    In a stmt of a 'do' block:
      [count1] <- runDB
                  $ E.select
                    $ E.from
                      $ \ (player `E.InnerJoin` game)
                          -> do { E.on $ player ^. PlayerGame E.==. game ^. GameId;
                                  E.where_
                                  $ player ^. PlayerUser E.==. E.just (E.val userId)
                                    E.&&. game ^. GameState E.==. E.val MyGame.Ongoing;
                                  .... }

Handler/ListUserGames.hs:32:32:
    No instance for (E.Esqueleto query0 expr0 backend0)
      arising from a use of ‘E.countRows’
    The type variables ‘query0’, ‘expr0’, ‘backend0’ are ambiguous
    Note: there is a potential instance available:
      instance E.Esqueleto E.SqlQuery E.SqlExpr SqlBackend
        -- Defined in ‘Database.Esqueleto.Internal.Sql’
    In the first argument of ‘return’, namely ‘E.countRows’
    In a stmt of a 'do' block: return E.countRows
    In the expression:
      do { E.on $ player ^. PlayerGame E.==. game ^. GameId;
           E.where_
           $ player ^. PlayerUser E.==. E.just (E.val userId)
             E.&&. game ^. GameState E.==. E.val MyGame.Ongoing;
           return (game, player);
           return E.countRows }

但是，如果删除countRows，则完全相同的查询有效。即以下代码编译并执行我希望它执行的操作。

ongoing <- runDB $ E.select
                 $ E.from $ \(player `E.InnerJoin` game) -> do
                     E.on $ player^.PlayerGame E.==. game^.GameId
                     E.where_ $
                       player^.PlayerUser E.==. E.just (E.val userId) E.&&.
                       game^.GameState E.==. E.val MyGame.Ongoing
                     E.orderBy [E.desc $ game^.GameLastActionTime]
                     E.limit pagelen
                     E.offset $ max 0 $ (page1 - 1) * pagelen
                     return (game, player)

我做错了什么？

Answer 1

事实证明上面的Esqueleto代码是正确的。错误发生在代码的另一部分，缺乏限制导致类型模糊。

用esqueleto计数行

1 个答案: