Question

如何让evancz / url-parser匹配路径的开头并将其余路径作为String返回？例如：

s "search" </> string |> format Search

如果路径类似＆＃34; search / text / with / slashes＆＃34;

，则

不匹配

我希望用＆＃34; text / with / slashes＆＃34;字符串作为有效负载。

Answer 1

evancz / url-parser似乎只处理由'/'分隔的段。我想你必须编写自己的解析器。（这里，我使用url-parser集成了yourparser，这是你现有的解析器）

urlParser : Navigation.Parser (Result String String)
urlParser =
  let
    parser : Navigation.Location -> (Result String String)
    parser location =
      Maybe.oneOf
        [ Result.toMaybe (yourparser location)
        , Result.toMaybe (extractSearch location.pathname)
        ]
      |> Result.fromMaybe "parse error"
    extractSearch pathname = 
      if String.startsWith "/search/" pathname
      then Ok (String.dropLeft 8 pathname)
      else Err "failed to extract search string"
  in Navigation.makeParser yourparser

如何让evancz / url-parser以String的形式返回路径的其余部分？

1 个答案: