Haskell Groundhog-Db 函数类型

Question

我是 haskell 的新手，我正在摆弄 postgreSQL 数据库上的 Groundhog-ORM 库。如果我将所有数据库内容放在一个函数中，一切都可以正常工作。但是，现在我想在不同的功能中分离不同的主题。

{-# LANGUAGE GADTs, TypeFamilies, TemplateHaskell, QuasiQuotes, FlexibleInstances, StandaloneDeriving #-}

module Main where

import System.IO
import Database.Groundhog.TH
import Database.Groundhog.Postgresql
import Data.DateTime
import Data.List.Split (splitOn)
import Control.Monad (mapM_, when, unless)
import Control.Monad.IO.Class (liftIO)

postgres_connection_string = "host=localhost port=5432 user=itsme password=unknown dbname=stocks"

data Symbol = Symbol {
    name :: String
} deriving (Show, Eq)

mkPersist defaultCodegenConfig [groundhog|
definitions:
  - entity: Symbol
    dbName: symbols
    schema: stocks
    keys:
      - name: unique_symbol
    constructors:
      - name: Symbol
        uniques:
          - name: unique_symbol
            fields: [name]
|]


ensureSymbol :: (PersistBackend m) => String -> m (Symbol)
ensureSymbol sym = do
    loadedSymbols <- select (NameField ==. sym)
    return (loadedSymbols !! 0)

main = do
    contents <- readFile "src\inputTable.csv"
    putStrLn ("Read " ++ show (length contents) ++ " Bytes input data.")
    withPostgresqlConn postgres_connection_string . runDbConn $ do
        liftIO $ putStrLn "About to ensure table structures."
        runMigration $ do
          migrate (undefined :: Symbol)
        liftIO $ putStrLn "Checking for the symbol."
        symbol <- ensureSymbol "EURUSD"
        ... [ Some more stuff ] ...

如您所见，我已经分离了一个从数据库中读取符号实体的小函数。这个例子编译正常。

但是，该函数被称为 ensureSymbol 而不是 readSymbol，因此该函数应显示为

ensureSymbol :: (PersistBackend m) => String -> m (Symbol)
ensureSymbol sym = do
    loadedSymbols <- select (NameField ==. sym)
    when (null loadedSymbols) (return (Symbol sym))
    unless (null loadedSymbols) (return (loadedSymbols !! 0))

但现在我收到一堆错误消息，甚至在

行

loadedSymbols <- select (NameField ==. sym)

之前编译正确。我假设我需要一个额外的类型说明符来使 when 和 unless 对函数可见，但我不确定。

src\Main.hs:117:40:
    Couldn't match type `Symbol' with `()'
    In the first argument of `select', namely `(NameField ==. sym)'
    In a stmt of a 'do' block:
      loadedSymbols <- select (NameField ==. sym)
    In the expression:
      do { loadedSymbols <- select (NameField ==. sym);
           when (null loadedSymbols) (return (Symbol sym));
           unless (null loadedSymbols) (return (loadedSymbols !! 0)) }

 src\Main.hs:118:40:
    Couldn't match expected type `()' with actual type `Symbol'
    In the first argument of `return', namely `(Symbol sym)'
    In the second argument of `when', namely `(return (Symbol sym))'

src\Main.hs:119:5:
    Couldn't match type `()' with `Symbol'
    Expected type: m Symbol
      Actual type: m ()
    In a stmt of a 'do' block:
      unless (null loadedSymbols) (return (loadedSymbols !! 0))
    In the expression:
      do { loadedSymbols <- select (NameField ==. sym);
           when (null loadedSymbols) (return (Symbol sym));
           unless (null loadedSymbols) (return (loadedSymbols !! 0)) }

也许有更优雅的方法来做到这一点？

Answer 1

when 和 unless 都具有相同的类型：

when, unless :: Monad m => Bool -> m () -> m ()

因此作为第二个参数传递的值必须具有类型 m ()。在这两种情况下，您都通过了 m Symbol。看起来你只想使用 if:

ensureSymbol sym = do
    loadedSymbols <- select (NameField ==. sym)
    if (null loadedSymbols) then (return (Symbol sym)) else (return (loadedSymbols !! 0))

或 case:

ensureSymbol sym = do
        loadedSymbols <- select (NameField ==. sym)
        case loadedSymbols of
          [] -> return (Symbol sym)
          s:ss -> return s

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