我的目标是用haskell编写一个程序,该程序采用json文件的名称,并将其余参数解释为导航json文件的路径,并打印导航到的值。问题在于JSON可以包含多个值类型,我不知道如何让Haskell的类型系统理解我想要的下面是haskell代码,其中包含我无法正确实现的“navigate”函数:

import qualified Data.Aeson as A
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import Data.List
import Data.Maybe
import System.Environment

parse :: String -> A.Value
parse = fromJust . A.decode . BSL.pack

isInteger xs = case reads xs :: [(Integer, String)] of
    [(_, "")] -> True
    _ -> False

navigate :: A.Value -> String -> String
navigate value [] = value
navigate value [x:xs]
    | isInteger x = ??? -- value is an array, get the xth element of it.
    | otherwise = ??? -- value is an map, x is a key in it.

main :: IO ()
main = do
     [filename:path] <- getArgs
     contents <- readFile filename
     let d = parse contents
     putStrLn (show (navigate d path))

作为参考,下面是用Python编写同一程序的方法:
from json import load
from sys import argv
def navigate(obj, path):
    if not path:
        return obj
    head, tail = path[0], path[1:]
    return navigate(obj[int(head) if head.isdigit() else head], tail)
if __name__ == '__main__':
    fname, path = argv[1], argv[2:]
    obj = load(open(fname))
    print navigate(obj, path)

程序将如下运行:
$ cat data.json
{"foo" : [[1, 2, 3, {"bar" : "barf"}]]}
$ python showjson.py data.json foo 0 3 bar
barf

最佳答案

您可以简单地对A.Value的构造函数进行模式匹配,以确定您正在处理的是哪种JSON对象:

import qualified Data.HashMap.Strict as M
import qualified Data.Vector as V
import qualified Data.Text as T

-- ... rest of the code more or less as before ...

navigate :: A.Value -> [String] -> BSL.ByteString
navigate value        []       = A.encode value
navigate (A.Array vs) (x : xs) = navigate (vs V.! read   x) xs
navigate (A.Object o) (x : xs) = navigate (o  M.! T.pack x) xs

注意A.Value的定义如下:
data Value
  = Object !(HashMap Text Value)
  | Array  !(Vector Value)
  | ...  -- other constructors

因此,navigate的代码在向量和散列映射上使用查找函数(在两种情况下都称为!)。函数read用于在需要时将命令行参数解释为一个数字(如果不是,它将失败得可怕),而T.pack则将字符串重新解释为Text类型的值。

09-10 04:02
查看更多