haskell - 持久性 selectList 导致“无法将类型 'BaseBackend backend0' 与 'SqlBackend' 匹配”的错误

Question

我遇到了以下编译错误：

• Couldn't match type ‘BaseBackend backend0’ with ‘SqlBackend’
    arising from a use of ‘runSqlite’
  The type variable ‘backend0’ is ambiguous
• In the expression: runSqlite ":memory:"
  In the expression:
    runSqlite ":memory:"
    $ do { records <- selectList [UserUsername ==. "abc"] [LimitTo 10];
           liftIO $ print (records :: [Entity User]) }
  In an equation for ‘selectAll’:
      selectAll
        = runSqlite ":memory:"
          $ do { records <- selectList [UserUsername ==. "abc"] [LimitTo 10];
                 liftIO $ print (records :: [Entity User]) }

代码：

selectAll :: IO ()
selectAll = runSqlite ":memory:" $ do
  records <- selectList [UserUsername ==. "abc"] [LimitTo 10]
  liftIO $ print (records :: [Entity User])

看看 runSqlite 的类型签名：

runSqlite    
:: (MonadBaseControl IO m, MonadIO m, IsSqlBackend backend)  
=> Text 
-> ReaderT backend (NoLoggingT (ResourceT m)) a 
-> m a

我假设我需要为 runSqlite 指定一个显式类型，尽管我不太确定我在 in 中设置了backend什么ReaderT backend (NoLoggingT (ResourceT m)) a？

Answer 1

你可以专攻它SqlBackend。

asSqlBackendReader :: ReaderT SqlBackend m a -> ReaderT SqlBackend m a
asSqlBackendReader = id

selectAll :: IO ()
selectAll = runSqlite ":memory:" . asSqlBackendReader $ do
  records <- selectList [UserUsername ==. "abc"] [LimitTo 10]
  liftIO $ print (records :: [Entity User])

---

查看 的类型，需要满足runSqlite一个约束条件。IsSqlBackend backend

的定义IsSqlBackend是：

type IsSqlBackend backend =
  (IsPersistBackend backend, BaseBackend backend ~ SqlBackend)

然后抬头看IsPersistBackend。

就在类的定义下方，我们看到它具有三个实例：

instance IsPersistBackend SqlWriteBackend
instance IsPersistBackend SqlReadBackend
instance IsPersistBackend SqlBackend

这三种类型指定了具有各种功能的后端，其中SqlBackend最通用的一种（未知功能）。如果这就是您所需要的，请随意使用更受限制的。