Question

我的目标是在Haskell中编写一个程序，它接受一个json文件的名称，并将其余参数解释为导航该json文件的路径并打印导航到的值。问题是因为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东西。这是具有“导航”功能的Haskell代码，我无法正确实现：

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))

供参考，以下是用Python编写相同程序的方法：

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)

程序将按如下方式运行：

$ cat data.json
{"foo" : [[1, 2, 3, {"bar" : "barf"}]]} 
$ python showjson.py data.json foo 0 3 bar
barf

Answer 1

您可以简单地在A.Value的构造函数上进行模式匹配，以便找出您正在处理的JSON对象类型：

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

请注意A.Value的定义如下：

data Value
  = Object !(HashMap Text Value)
  | Array  !(Vector Value)
  | ...  -- other constructors

navigate的代码因此在向量和哈希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数字（如果不是则会失败），而T.pack将字符串重新解释为类型{{1}的值}}

在Haskell中以通用方式导航JSON对象

1 个答案: