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f# - 匹配多个 ParserResult 并提取值

问题描述

这是一个关于使用 FParsec 的 ParserResult 的问题。

是否有更简洁的实现match_result来拉出XKEYVALUE或XATTRIBUTES包含在ParserResult没有两个嵌套匹配项的内部？

以下代码用作 F# 控制台应用程序...

// Learn more about F# at http://fsharp.org

open System
open FParsec

// Types
type XKEYVALUE = string * string
type XATTRIBUTES = Map< string, string>

type PARSER_RESULT_XML = 
    | PR_XKEYVALUE of ParserResult<XKEYVALUE, unit>
    | PR_XATTRIBUTES of ParserResult<XATTRIBUTES, unit>

// Parser Trace
let (<!>) (p: Parser<_,_>) label : Parser<_,_> =
    fun stream ->
        printfn "%A: Entering %s" stream.Position label
        let reply = p stream
        do
            match (reply.Status) with
                | Ok  -> printfn "%A: Leaving %s (%A) - %A" stream.Position label reply.Status reply.Result
                | Error -> printfn "%A: Leaving %s (%A) - %A" stream.Position label reply.Status reply.Error
                | FatalError -> printfn "%A: Leaving %s with FatalError (%A)" stream.Position label reply.Status
                | _ -> printfn "%A: Leaving %s with unknown status" stream.Position label
        reply

// Parsers
let ws = spaces
let str = pstring
let xKey : Parser<string,unit> = (ws >>. regex "[a-zA-Z][a-zA-Z0-9:]*") <!> "xKey"
let xStringLiteral = regex "[^\"]+" <!> "xStringLiteral"
let xKeyValue : Parser<XKEYVALUE, unit> = 
    ws >>. xKey .>>. (ws >>. str "=" >>. ws >>. str "\"" >>. xStringLiteral .>> str "\"" .>> ws) |>> XKEYVALUE <!> "xKeyValue"
let xAttributes : Parser<XATTRIBUTES, unit> = sepEndBy xKeyValue ws |>> XATTRIBUTES <!> "xAttributes"

// Test Data
let xKeyValue_text = """    key =  "value"    aa"""
let xAttributes_text = "key1=\"value1\" key2=\"value2\""

let match_result ( result : PARSER_RESULT_XML) : string =
    match result with 
        | PR_XKEYVALUE(a) ->    match a with  
                                    | Success( ((key : string), (value : string)), x2, x3) -> sprintf "%s=\"%s\"" key value
                                    | Failure( _, _, _) -> sprintf "FAILURE"
        | PR_XATTRIBUTES( a) -> match a with
                                    | Success( (x1 : Map<string, string>), x2, x3) -> sprintf "x1=%A, x2=%A, x3=%A" x1 x2 x3
                                    | Failure( _, _, _) -> sprintf "FAILURE"

[<EntryPoint>]
let main argv =
    run xKeyValue xKeyValue_text |> PR_XKEYVALUE |> match_result |> printfn "%A"
    run xAttributes xAttributes_text |> PR_XATTRIBUTES |> match_result |> printfn "%A"
    0 // return an integer exit code

但match_result它的嵌套匹配似乎很笨拙。

一个失败的实验是使用AND 模式匹配将匹配项PARSER_RESULT_XML & Success( …)放在同一个匹配表达式中，但我无法让两个匹配表达式的类型一致。

您将如何修改match_result以更好或更清洁地做到这一点？

标签： f#fparsec

解决方案

这对我来说是主观的，但是如果你去掉不需要的类型注释并更改 match_result 的缩进，我认为它会提高可读性。

let match_result result =
    match result with 
    | PR_XKEYVALUE a ->
        match a with  
        | Success ((key, value), _, _) -> sprintf "%s=\"%s\"" key value
        | Failure _ -> sprintf "FAILURE"
    | PR_XATTRIBUTES a -> 
        match a with
        | Success (x1, x2, x3) -> sprintf "x1=%A, x2=%A, x3=%A" x1 x2 x3
        | Failure _ -> sprintf "FAILURE"

如果您仍然对此不满意，您可能会发现在此处使用活动模式很有帮助：

let (|KeyValue|Attributes|Failed|) result =
    match result with
    | PR_XKEYVALUE a ->
        match a with  
        | Success ((key, value), _, _) -> KeyValue (key, value)
        | Failure _ -> Failed
    | PR_XATTRIBUTES a -> 
        match a with
        | Success (x1, x2, x3) -> Attributes (x1, x2, x3)
        | Failure _ -> Failed

let match_result result =
    match result with 
    | KeyValue (key, value) -> sprintf "%s=\"%s\"" key value
    | Attributes (x1, x2, x3) -> sprintf "x1=%A, x2=%A, x3=%A" x1 x2 x3
    | Failed -> sprintf "FAILURE"

后者更好，因为您将用域术语解释结果以及如何处理它（在这种情况下为打印消息）。

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