我必须解析一些 xml,我决定使用 xml-conduit 来完成该任务并使用它的流式传输部分。
xml 的结构由包含元素及其出现频率的 xsd 文件给出。但不是他们预期的顺序。
如何使用 xml 结构的子级解析所有可能的重新排序Text.XML.Stream.Parse
?
问题
假设我们有一个 xml 描述,例如
Root
/ \
A B
那么两者<Root><A>atext</A><B>btext</B></Root>
和 <Root><B>btext</B><A>atext</A></Root>
都是此 xml 结构的有效实例。但是在流设置中解析需要一个顺序才能成功。
我想过使用类似的东西,parseRoot1 <|> parseRoot2
但后来我必须实现Alternative
实例并手动编写所有可能性,我真的不想这样做。
这是一个最小的示例 haskell 程序。
Example.hs
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE Rank2Types #-}
module Main where
import Control.Exception
import Control.Monad.Trans.Resource (MonadThrow)
import Text.XML.Stream.Parse
import Data.Monoid ((<>))
import Data.Maybe
import Data.Text (Text)
import Data.XML.Types (Event)
import Data.Conduit (ConduitM, Consumer, yield, ($=), ($$))
data Root = Root {a :: A, b :: B} deriving (Show, Eq)
data A = A Text deriving (Show, Eq)
data B = B Text deriving (Show, Eq)
ex1, ex2 :: Text
ex1 = "<Root>"<>
"<A>Atest</A>"<>
"<B>Btest</B>"<>
"</Root>"
ex2 = "<Root>"<>
"<B>Btest</B>"<>
"<A>Atest</A>"<>
"</Root>"
ex :: Root
ex = Root {a = A "Atest", b = B "Btest"}
parseA :: MonadThrow m => ConduitM Event o m (Maybe A)
parseA = tagIgnoreAttrs "A"
$ do result <- content
return (A $ result)
parseB :: MonadThrow m => ConduitM Event o m (Maybe B)
parseB = tagIgnoreAttrs "B"
$ do result <- content
return (B result)
parseRoot1 :: MonadThrow m => ConduitM Event o m (Maybe Root)
parseRoot1 = tagIgnoreAttrs "Root" $ do
a' <- fromMaybe (error "error parsing A") <$> parseA
b' <- fromMaybe (error "error parsing B") <$> parseB
return $ Root{a = a', b = b'}
parseRoot2 :: MonadThrow m => ConduitM Event o m (Maybe Root)
parseRoot2 = tagIgnoreAttrs "Root" $ do
b' <- fromMaybe (error "error parsing B") <$> parseB
a' <- fromMaybe (error "error parsing A") <$> parseA
return $ Root{a = a', b = b'}
parseTxt :: Consumer Event (Either SomeException) (Maybe a)
-> Text
-> Either SomeException (Maybe a)
parseTxt p inTxt = yield inTxt
$= parseText' def
$$ p
main :: IO ()
main = do putStrLn "Poor Mans Test Suite"
putStrLn "===================="
putStrLn "test1 Root -> A - B " -- works
print $ parseTxt parseRoot1 ex1
putStrLn "test1 Root -> B - A " -- fails
print $ parseTxt parseRoot1 ex2
putStrLn "test2 Root -> A - B " -- fails
print $ parseTxt parseRoot2 ex1
putStrLn "test2 Root -> B - A " -- works again
print $ parseTxt parseRoot2 ex2
笔记
example.cabal
[...]
build-depends: base >=4.8 && <4.9
, conduit
, resourcet
, text
, xml-conduit
, xml-types
[...]