[英]Composing Database.Esqueleto queries, conditional joins and counting
[英]Counting rows with esqueleto
我正在嘗試計算Esqueleto(版本2.1.2.1)的內部聯接的行。 不幸的是,我的代碼無法編譯,我也不明白為什么。 我查看了以下示例,了解如何執行此操作,但無法弄清自己做錯了什么: example1 , example2 。
我的架構如下所示(簡化):
User
Game
state
Player
user UserId Maybe
game GameId
用戶可以在該網站上注冊玩游戲。 您也可以不注冊而玩。 因此,有一個單獨的表Player
。 游戲具有狀態。 它可以是Ongoing
中或某種形式的游戲結束。 我現在要計算用戶正在玩的所有正在進行的游戲。
下面的SQL查詢可以很好地做到這一點(對於固定的用戶標識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查詢無法編譯:
[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
錯誤消息如下所示:
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,如果刪除countRows
,則完全相同的查詢有效。 即下面的代碼可以編譯並執行我想要的操作。
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)
我究竟做錯了什么?
事實證明,上面的Esqueleto代碼是正確的。 錯誤發生在代碼的另一部分,其中缺少限制導致類型歧義。
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