所有的名字我解析从http://hackage.haskell.org/package/xml-conduit-1.1.0.9/docs/Text-XML-Stream-Parse.html获得从XML的管道
这里修改了XML是什么样子:
<?xml version="1.0" encoding="utf-8"?>
<population xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://example.com">
<success>true</success>
<row_count>2</row_count>
<summary>
<bananas>0</bananas>
</summary>
<people>
<person>
<firstname>Michael</firstname>
<age>25</age>
</person>
<person>
<firstname>Eliezer</firstname>
<age>2</age>
</person>
</people>
</population>
如何获得的firstname
和age
每个人名单?
我的目标是使用HTTP的管道下载此XML,然后解析它,但是我正在寻找如何在没有属性分析的解决方案(使用tagNoAttrs?)
这里是我的”已经尝试过了,我已经将我的问题在哈斯克尔评论:
{-# LANGUAGE OverloadedStrings #-}
import Control.Monad.Trans.Resource
import Data.Conduit (($$))
import Data.Text (Text, unpack)
import Text.XML.Stream.Parse
import Control.Applicative ((<*))
data Person = Person Int Text
deriving Show
-- Do I need to change the lambda function \age to something else to get both name and age?
parsePerson = tagNoAttr "person" $ \age -> do
name <- content -- How do I get age from the content? "unpack" is for attributes
return $ Person age name
parsePeople = tagNoAttr "people" $ many parsePerson
-- This doesn't ignore the xmlns attributes
parsePopulation = tagName "population" (optionalAttr "xmlns" <* ignoreAttrs) $ parsePeople
main = do
people <- runResourceT $
parseFile def "people2.xml" $$ parsePopulation
print people
编辑添加我迄今试过的和评论 – Lionel