哈斯克尔：字符串转换为现场MongoDB的

Question

什么是字符串"[\"id\"=:5977, \"brandId\"=:87]"转换为场MongoDB的["id"=:5977, "brandId"=:87]

，甚至更好，如果我有一个数据类型

data Body = Body 
{ body_id :: int 
    brandId :: Int } deriving (Show) 

给予最好的办法

Body {body_id = 5977, brandId = 87}

什么是将其转换为

的最佳方法

["id"=:5977, "brandId"=:87]

谢谢。

Answer 1

如果我理解正确，您正在使用hackage: mongoDB与数据库交互。

所以，我们要做的唯一事情是一个Haskell值转换为BSON价值这是hackage: bson提供，特别是在类型类Val

class (Typeable a, Show a, Eq a) => Val a where 

Haskell types of this class correspond to BSON value types 

Minimal complete definition 
val, cast' 

Methods 
val :: a -> Value 
... 
cast' :: Value -> Maybe a 
... 

所以它仍然实现val和cast'

但首先我们需要推导出比您拥有的更多属性：

Body.hs

{-# LANGUAGE DeriveDataTypeable #-} 
{-# LANGUAGE OverloadedStrings #-} 
module Body where 

import Data.Bson 
import Data.Typeable 

data Body = Body { bodyID :: Int 
       , brandID :: Int } 
       deriving (Show,Typeable,Eq) 


instance Val Body where 
    val (Body body brand) = Doc [ "bodyID" := (Int64 $ fromIntegral body) 
           , "brandID" := (Int64 $ fromIntegral brand)] 

    cast' (Doc bson) = case (bson !? "bodyID") of 
          Just (Int64 x) -> 
           case (bson !? "brandID") of 
             Just (Int64 y) -> Just (Body (fromIntegral x) 
                     (fromIntegral y)) 
             Just _ -> Nothing 
             Nothing -> Nothing 
          Just _ -> Nothing 
          Nothing -> Nothing 
    cast' (Array x) = undefined 
    cast' _ = Nothing 

这是一个有点集群而是简单。

---

下面我就给这个东西多一点思考 - 以前cast'真的集群和Haskell有很好的抽象应该做的更好

val (Body body brand) = Doc [ "bodyID" := (val body) 
           , "brandID" := (val brand)] 
    cast' (Doc bson) = do body <- (bson !? "bodyID") 
         brand <- (bson !? "brandID") 
         Body <$> cast' body <*> cast' brand 

    cast' _ = Nothing 

所以，你真的可以写这个有点短接受这样<$> （叫做fmap）和<*>（叫做ap做“正确”的事

另外我用do符号来简化FY工作，现在用这个

> cast' $ val (Body 30 4) :: Maybe Body 
Just (Body {bodyID = 30, brandID = 4}) 

注在你真正需要的注释:: Maybe Body最后一个例子，否则编译器无法弄清楚投什么，你会得到一个Nothing。