我希望能够访问
XML文件的数据
<?xml version="1.0"?>
<MY>
<Foo id="1" name="test">
<Argument name="a" />
</Foo>
<Foo id="2" name="test2">
<Argument name="a" />
<Argument name="b" />
</Foo>
<Other id="2" name="someOther"/>
</MY>
我想要,例如用它的Arguments读出每个Foo,我怎么能用Haskell做到这一点？ (我想使用HaXml模块)
我不知道从哪里开始.
我找不到haXml的最新文档和示例.
原文链接：https://www.f2er.com/xml/292517.html但是有一些HXT的文档可用.
我知道这可能对你的例子来说太过分了,但无论如何.
如果你想使用tagsoup,也许以下答案可能会有所帮助：
xml-tree parser (Haskell) for graph-library
In Haskell how do you extract strings from an XML document?
以下是HXT的文档示例：
http://www.haskell.org/haskellwiki/HXT/Conversion_of_Haskell_data_from/to_XML
http://www.haskell.org/haskellwiki/HXT
http://www.haskell.org/haskellwiki/HXT/Practical
http://en.wikibooks.org/wiki/Haskell/XML
现在代码使用HXT.
(警告我不确定这是否正确)
我按照教程：
http://www.haskell.org/haskellwiki/HXT/Conversion_of_Haskell_data_from/to_XML
你需要你的xml文件为“data.xml”
import Data.Map (Map,fromList,toList)
import Text.XML.HXT.Core
type Foos = Map String [Foo]
data Foo = Foo
{
fooId :: String,fooName :: String,arguments :: [Argument]
}
deriving (Show,Eq)
data Argument = Argument
{ argName :: String
}
deriving (Show,Eq)
instance XmlPickler Foo where
xpickle = xpFoo
instance XmlPickler Argument where
xpickle = xpArgument
-- WHY do we need this?? no clue
instance XmlPickler Char where
xpickle = xpPrim
-- this could be wrong
xpFoos :: PU Foos
xpFoos
= xpWrap (fromList,toList
) $
xpList $
xpElem "MY" $
xpickle
xpFoo :: PU Foo
xpFoo
= xpElem "Foo" $
xpWrap ( uncurry3 Foo,\ f -> (fooId f,fooName f,arguments f
)
) $
xpTriple (xpAttr "id" xpText)
(xpAttr "name" xpText)
(xpList xpickle)
xpArgument :: PU Argument
xpArgument
= xpElem "Argument" $
xpWrap ( \ ((a)) -> Argument a,\ t -> (argName t)
) $
(xpAttr "name" xpText )
main :: IO ()
main
= do
runX ( xunpickleDocument xpFoos
[ withValidate no,withTrace 1,withRemoveWS yes,withPreserveComment no
] "data.xml"
>>>
arrIO ( \ x -> do {print x ; return x})
)
return ()
结果(您需要xml示例为“data.xml”)：
-- (1) getXmlContents
-- (1) readDocument: "data.xml" (mime type: "text/xml" ) will be processed
-- (1) readDocument: "data.xml" processed
fromList [("",[Foo {fooId = "1",fooName = "test",arguments = [Argument {argName = "a"}]},Foo {fooId = "2",fooName = "test2",arguments = [Argument {argName = "a"},Argument {argName = "b"}]}])]
