Question

我有以下xml：

<list>
    <recipient>
        <name></name>
        <lastname></lastname>
        <email></email>
        <phone></phone>
        <home></home>
    </recipient>
    <recipient>
    </recipient>
</list>

以及以下数据类型：

data Recipient = Recipient { name     :: String
                           , lastname :: String
                           , email    :: String
                           , phone    :: Maybe String
                           , home     :: Maybe String }

我想要做的是阅读xml并获取收件人列表：[收件人]
为此，我写了以下内容：

import Text.XML.HXT.Core

readMyXml :: FilePath -> IO [Recipient]
readMyXml path = do
   -- lets read the doc
   fe   <- readFile path
   let
      -- parse it
      doc = readString [withValidate no] fe

   -- get all recipient nodes
   reps <- getAllRep

   -- the part I don't have
   -- the only thing wrong in the following is the map function
   -- I need an equivalent that runs on hxt trees
   return $ map frmRep2Dat reps
   --        ^
   --        |
   --       here
   -- end of part I don't have

 where
   getAllRep = runX $ doc
            >>> deep (hasName "list")
            >>> multi (hasName "recipient")

   frmRep2Dat branch = do
      let
         -- gets the recipient of a recipient node child
         getV name = runX $
                     branch
                 >>> deep (hasName name)
                 >>> removeAllWhiteSpace
                 >>> deep getText

         -- normaly there is no need to check because not maybe fields are
         -- mandatory and should not be empty
         getVal name = do
            val <- getV name
            case val of
               []   -> ""
               [""] -> ""
               _    -> head val

         -- some maybe wrapping
         getMayVal name = do
            val <- getV name
            case val of
               []   -> Nothing
               [""] -> Nothing
               _    -> Just (head val)

      name     <- getVal "name"
      lastname <- getVal "lastname"
      email    <- getVal "email"
      phone    <- getMayVal "phone"
      home     <- getMayVal "home"

      return $ Recipient name lastname email phone home

关于如何映射树的任何线索？

Answer 1

发现没有必要遍历树。 HXT已经做到了。从xml构造数据的方法比我编写的天真的方法更简单我用以下方法替换了整个readMyXml函数：

readMyXml path = do
   fi <- readFile path
   let
      doc = readString [withValidate no] fi

   return =<< runX $ getRecipients doc

wrapStr a = if null a
               then Nothing
               else Just a

getD a = deep (hasName a)
   >>> removeAllWhiteSpace
   >>> deep getText

getMD a = getD a
   >>^ wrapStr

getRecipients doc = doc
   >>> deep (hasName "list")
   >>> multi (hasName "recipient")
   >>> proc y -> do
       nime <- getD "name"     -< y
       lstn <- getD "lastname" -< y
       mail <- getD "email"    -< y
       phon <- getMD "phone"   -< y
       homi <- getMD "home"    -< y
       returnA -< Recipient nime lstn mail phon homi

现在，应用于问题中定义的文档的getRecipients的返回值是[收件人]
干杯

你如何映射HXT树？

1 个答案: