I'm am not really familiar with XML parsing in Haskell, but I am
wondering why, if you have an xml file, you not simply name the
element after the type of contents:
<book>
<title>Some Story</title>
<description>This story about..</description>
<author>Tom Smith</author>
</book>
Alternatively you could use the class or some type attribute to
indicate the type.
Regards,
Wilfried van Asten
2012/3/15 Никитин Лев
Hello, haskellers.
Suppose we have this xml doc (maybe, little stupid):
<div> <span>Some story</span> <span>Description</span>: This story about... <span>Author</span>: Tom Smith </div>
In the end I whant to get list: [("Title", "Some story"), ("Description","This story about..."), ("Author", "Tom Smith")], or, maybe this: Book "Some story" [("description","This story about..."), ("Author", "Tom Smith")] (Book = Book String [(String, String)].
First span is a special case then others and I undestand how to process it:
===============
import Text.XML.HXT.Core import Text.XML.HXT.Curl import Text.XML.HXT.HTTP
pageURL = "http://localhost/test.xml"
main = do r <- runX (configSysVars [withCanonicalize no, withValidate no, withTrace 0, withParseHTML no] >>> readDocument [withErrors no, withWarnings no, withHTTP []] pageURL >>> getChildren >>> isElem >>> hasName "div" >>> listA (getChildren >>> hasName "span") >>> getTitle <+> getSections) putStrLn "Статьи:" putStr "<" mapM_ putStr $ map (\i -> (fst i) ++ ": " ++ (snd i) ++ "| ") r putStrLn ">"
getTitle = arr head >>> getChildren >>> getText >>> arr trim >>> arr ("Title",)
getSections = arr tail >>> unlistA >>> ((getChildren >>> getText >>> arr trim) &&& (getChildren >>> getText >>> arr trim))
ltrim [] = [] ltrim (' ':x) = ltrim x ltrim ('\n':x) = ltrim x ltrim ('\r':x) = ltrim x ltrim ('\t':x) = ltrim x ltrim x = x
rtrim = reverse . ltrim . reverse
trim = ltrim . rtrim
===================
And I' get list: [("Title", "Some story"), ("Description","Description"), ("Author", "Author")]
(Maybe, there is a better way to get this list?)
But I cannot find a way to get text that followes some span.
I suppose that I have to use function from Data.Tree.NavigatableTree.XPathAxis, but I don't "puzzle out" how to do it.
Please, help me.
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