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我在 Andrew Birkett 的博客Applicative arrows for XML &&& return to pure上读到，我们可以混合使用箭头和应用函子。

我自己尝试过，但我没有我所期望的。我想要这个结果：

[Scenario {scenario = "11111", origin = "333", alarm = "Sonde1"},
 Scenario {scenario = "22222", origin = "444", alarm = "Sonde2"}]

但我得到了这个：

[Scenario {scenario = "11111", origin = "333", alarm = "Sonde1"},
 Scenario {scenario = "11111", origin = "333", alarm = "Sonde2"},
 Scenario {scenario = "11111", origin = "444", alarm = "Sonde1"},
 Scenario {scenario = "11111", origin = "444", alarm = "Sonde2"},
 Scenario {scenario = "22222", origin = "333", alarm = "Sonde1"},
 Scenario {scenario = "22222", origin = "333", alarm = "Sonde2"},
 Scenario {scenario = "22222", origin = "444", alarm = "Sonde1"},
 Scenario {scenario = "22222", origin = "444", alarm = "Sonde2"}]

我认为我的代码中有一个转折，但我不知道在哪里搜索。

如果有人可以提出一些帮助，下面是我的代码。

{-# LANGUAGE Arrows, NoMonomorphismRestriction #-}

import Text.XML.HXT.Core
import Control.Applicative
import Text.XML.HXT.Arrow.ReadDocument
import Data.Maybe
import Text.XML.HXT.XPath.Arrows
import Text.Printf


data Scenario = Scenario
  { scenario, origin, alarm    :: String
  }
  deriving (Show, Eq)


xml= "<DATAS LANG='en'>\
    \ <SCENARIO ID='11111'>\
    \   <ORIGIN ID='333'>\
    \       <SCENARIO_S ERR='0'></SCENARIO_S>\
    \       <SCENARIO_S ERR='2'></SCENARIO_S>\
    \       <ALARM_M NAME='Sonde1'></ALARM_M>\
    \   </ORIGIN>\
    \ </SCENARIO>\
    \ <SCENARIO ID='22222'>\
    \   <ORIGIN ID='444'>\
    \       <SCENARIO_S ERR='10'></SCENARIO_S>\
    \       <SCENARIO_S ERR='12'></SCENARIO_S>\
    \       <ALARM_M NAME='Sonde2'></ALARM_M>\
    \   </ORIGIN>\
    \ </SCENARIO>\
    \</DATAS>"

parseXML string = readString [ withValidate no
                         , withRemoveWS yes  -- throw away formating WS
                         ] string


parseVal tag name = WrapArrow $ getXPathTrees (printf "/DATAS/%s" tag) >>>  getAttrValue name

parseDatas = unwrapArrow $ Scenario <$> parseVal "SCENARIO"      "ID"
                                 <*> parseVal "SCENARIO/ORIGIN"        "ID"
                                 <*> parseVal "SCENARIO/ORIGIN/ALARM_M"        "NAME"

testarr1= runX (parseXML xml >>> parseDatas)
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1 回答 1

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正如 rampion 所指出的，问题在于 list monad 如何与 applicative 一起工作。看看这个：

λ *Main > (+) <$> [1,2,3] <*> [1,2,3]
[2,3,4,3,4,5,4,5,6]

结果是应用于 [1,2,3] 和 [1,2,3] 的 (+) 的笛卡尔积：结果列表有 9 个元素。

在您的代码中，parseVal "SCENARIO" "ID"将返回一个包含 2 个元素的列表，parseVal "SCENARIO/ORIGIN" "ID"并且parseVal "SCENARIO/ORIGIN/ALARM_M" "NAME". 因此，结果将有 8 个元素。

相反，这就是我将如何更改您的代码：

--- parse a generic tag
parseVal tag name = WrapArrow $ getXPathTrees (printf "%s" tag) >>>  getAttrValue name

--- parse a "SCENARIO" xml element
parseScenario = unwrapArrow $ Scenario
        <$> (WrapArrow $ getAttrValue "ID")
        <*> (parseVal "SCENARIO/ORIGIN" "ID")
        <*> (parseVal "SCENARIO/ORIGIN/ALARM_M" "NAME")

--- parse the XML, extract a list of SCENARIOS and, for each, apply parseScenario
testarr1= runX (parseXML xml >>> getXPathTrees (printf "/DATAS/SCENARIO" ) >>> parseScenario)

结果如所愿：

λ *Main > testarr1 
[Scenario {scenario = "11111", origin = "333", alarm = "Sonde1"},Scenario {scenario = "22222", origin = "444", alarm = "Sonde2"}]
于 2011-12-24T20:53:27.777 回答