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 book中获取了这些代码,并尝试将其清理干净,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
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.
同
main = do
updateDB "Frank Silver" 40
updateDB“Frank Silver”40的类型被推断为IO(),因为这是main的默认类型(对于某些a,它必须具有类型IO a).但是从定义来看,它的类型被推断为MonadRescource m => m a for some a(可能是a =(),但我不确定),并且没有实例MonadResource IO.因此,您需要将updateDB转换为IO操作,正常的方法是runResourceT,它将ResourceT m a转换为m a(此处为m = IO),因此
main = runResourceT $updateDB "Frank Silver" 40
作品.
