我正在开发一个程序,它带有另一个带有参数的命令:
$ myprog record -f 1.txt ls
$ myprog record -f 1.txt ls -l
像sudo这样的东西:
$ sudo ls
$ sudo ls -l
这是我使用cmdargs的代码:
#!/usr/bin/env stack
-- stack --resolver lts-9.20 script --package cmdargs
{-# LANGUAGE DeriveDataTypeable #-}
module Main where
import System.Console.CmdArgs as Args
data Tape = Record {file :: Maybe String, command :: String, commandArgs :: [String]}
| Replay
deriving (Data, Typeable, Show, Eq)
main :: IO ()
main = do
m <- cmdArgs $ modes
[ Record { file = def &= typ "FILE"
, command = def &= typ "COMMAND" &= argPos 0
, commandArgs = def &= typ "ARGS" &= args }
, Replay ]
print m
它工作得很好,但仅适用于没有自己标志的命令:
$ ./cmdargs.hs record -f 1.txt ls
Record {file = Just "1.txt", command = "ls", commandArgs = []}
$ ./cmdargs.hs record -f 1.txt ls 1 2 3
Record {file = Just "1.txt", command = "ls", commandArgs = ["1","2","3"]}
$ ./cmdargs.hs record -f 1.txt ls -l
Unknown flag: -l
问:如何指示cmdargs按原样保留“commandArgs”,而不查找其中的标志?
传统方法不在解析器中,而是使用--
参数,指示标志的结束:
$ ./args record -- something -l
Record {file = Nothing, command = "something", commandArgs = ["-l"]}
不幸的是,我不知道如何告诉getArgs在任意地方这样做,但我想它会与模式有关。