Haskell错误"不在范围内:数据构造函数"

Jef*_*ner 7 haskell

我已经回顾了关于这个错误的其他帖子,我不认为我犯了这些错误.

不在范围内:数据构造函数'Extraction'.

Configuration.hs:

module Configuration
(Config
 , columns
 , headers
 , types
 , totals
 , extractions,
 Extraction
 , xState
 , xDivisions
 , xOffice
 ...) where

...

data Extraction = Extraction { xState     :: String
                             , xDivisions :: Maybe [String]
                             , xOffice    :: Maybe String } deriving Show


data Config = Config { columns     ::  String
                     , headers     :: [String]
                     , types       :: [String]
                     , totals      :: [String]
                     , extractions :: [Extraction] } deriving Show

...
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PIF.hs:

module PIF (...) where

import Configuration

...

data Report = Report { division  :: String
                     , state     :: String
                     , office    :: String
                     , inSection :: Bool
                     , content   :: [String] } deriving Show

...

extract :: Config -> [Report] -> [Report]
extract c = filter f
  where f Report { division=d, state=s, office=o, inSection=_, content=_ } =
          map or $ map isMatch $ extractions c
          where isMatch
                  | Extraction { xState=xS, xDivisions=Just xD, xOffice=Nothing } = s==xS && (map or $ map (==d) xD)
                  | Extraction { xState=xS, xDivisions=Nothing, xOffice=Just xO } = s==xS && o==xO
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如果您需要更多信息,请与我们联系.谢谢.

这是我的更正extract:

extract c = filter f
  where f Report { division=d, state=s, office=o, inSection=_, content=_ } =
          or $ map isMatch $ extractions c
          where isMatch x =
                  case ((xDivisions x), (xOffice x)) of (Nothing, Just y) -> s==(xState x) && o==y
                                                        (Just y, Nothing) -> s==(xState x) && (or $ map (==d) y)
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scl*_*clv 14

将导出行更改ExtractionExtraction(..).

没有它,您将导出类型而不是数据构造函数.由于您的类型和构造函数共享相同的名称,因此在这种情况下不太明显.