Question

我一直收到此错误消息，我不明白为什么：

src\Parsing.hs:21:18: parse error on input `='

它来自这条线：

tokens  = map strip $ splitWhen (== delimiter) line

我的代码：

tokenize :: String -> HashMap String String
tokenize line = 
   let (delimiter, fieldOrder) = delimiterAndFieldOrderFor line
           tokens  = map strip $ splitWhen (== delimiter) line
   in Map.fromList $  zip fieldOrder tokens

delimiterAndFieldOrderFor :: String -> (Char, [String])
delimiterAndFieldOrderFor line    
       | isInfixOf "," line = (',', ["LastName", "FirstName", "Gender", "FavoriteColor", "BirthDate"])    
       | isInfixOf "|" line = ('|', ["LastName", "FirstName", "Ignored", "Gender", "FavoriteColor", "BirthDate"])    
       | otherwise          = (' ', ["LastName", "FirstName", "Ignored", "Gender", "BirthDate", "FavoriteColor"])

感谢您的帮助！

Answer 1

压痕。

在第4行，tokens应缩进到与开始(delim...的parens相同的深度。 E.g。

tokenize line = 
   let (delimiter, fieldOrder) = delimiterAndFieldOrderFor line
       tokens  = map strip $ splitWhen (== delimiter) line
   in Map.fromList $  zip fieldOrder tokens

您也可以考虑使用where子句：

tokenize line = Map.fromList $ zip f tokens
   where
       (sep, f) = delimiterAndFieldOrderFor line
       tokens   = map strip $ splitWhen (== sep) line

作为一种风格，那些非常长的变量名称有点混淆，IMO。

Answer 2

此代码中有额外的缩进：

let (delimiter, fieldOrder) = delimiterAndFieldOrderFor line
       tokens  = map strip $ splitWhen (== delimiter) line
in Map.fromList $  zip fieldOrder tokens

这会导致编译器将该行视为其上方行上的表达式的延续。它应该是

let (delimiter, fieldOrder) = delimiterAndFieldOrderFor line
    tokens  = map strip $ splitWhen (== delimiter) line
in Map.fromList $  zip fieldOrder tokens

这个Haskell有什么问题？ “输入上的解析错误'='”

2 个答案: