Haskell/Persistent-Sqlite:"没有(Control.Monad.Trans.Resource.MonadResource IO)的实例"

Question

我一直试图想出一种简单直观的方法来使用Haskell数据库.我从Yesod书中获取了这些代码并试图将其清理干净,以便更容易理解和使用.

{-# LANGUAGE QuasiQuotes, TemplateHaskell, TypeFamilies, OverloadedStrings #-}
{-# LANGUAGE GADTs, FlexibleContexts #-}

import Database.Persist
import Database.Persist.Sqlite (withSqliteConn, runSqlConn, runMigration)
import Database.Persist.TH (share, mkPersist, mkMigrate, sqlSettings, persist)

share [mkPersist sqlSettings, mkMigrate "migrateAll"] [persist|
Person                              -- Table name
    name String                     -- String value
    age Int Maybe                   -- Numerical value
|]

updateDB x y = withSqliteConn "data.db" $ runSqlConn $ do
    runMigration migrateAll         -- Creates "Person" table if one doesn't exist
    insert $ Person x $ Just y      -- Inserts values into .db file

main = do
    updateDB "Frank Silver" 40      -- adds name "Frank Silver" and age "40" to data.db file

这段代码几乎可以工作,但我得到以下错误,我无法解决.

No instance for (Control.Monad.Trans.Resource.MonadResource IO)
      arising from a use of `updateDB'
    Possible fix:
      add an instance declaration for
      (Control.Monad.Trans.Resource.MonadResource IO)
    In a stmt of a 'do' block: updateDB "Frank Silver" 40
    In the expression: do { updateDB "Frank Silver" 40 }
    In an equation for `main': main = do { updateDB "Frank Silver" 40 }

任何建议指向我正确的方向将不胜感激.

Answer 1

同

main = do
    updateDB "Frank Silver" 40

updateDB "Frank Silver" 40推断的类型是IO (),因为它是默认类型main(它必须具有IO a某些类型a).但是根据定义,它的类型被推断MonadRescource m => m a为某些a(可能a = (),但我不确定),并且没有instance MonadResource IO.所以你需要一些东西来转换updateDB成一个IO动作,正常的方法就是runResourceT把它ResourceT m a变成一个m a(这里m = IO),所以

main = runResourceT $ updateDB "Frank Silver" 40

作品.