在shell脚本中引用命令行参数

The following shell script takes a list of arguments, turns Unix paths into WINE/Windows paths and invokes the given executable under WINE.

以下shell脚本获取参数列表,将Unix路径转换为WINE / Windows路径并在WINE下调用给定的可执行文件。

#! /bin/sh

if [ "${1+set}" != "set" ]
then 
  echo "Usage; winewrap EXEC [ARGS...]"
  exit 1
fi

EXEC="$1"
shift

ARGS=""

for p in "$@";
do
  if [ -e "$p" ]
  then
    p=$(winepath -w $p)
  fi
  ARGS="$ARGS '$p'"
done

CMD="wine '$EXEC' $ARGS"
echo $CMD
$CMD

However, there's something wrong with the quotation of command-line arguments.

但是,命令行参数的引用有问题。

$ winewrap '/home/chris/.wine/drive_c/Program Files/Microsoft Research/Z3-1.3.6/bin/z3.exe' -smt /tmp/smtlib3cee8b.smt
Executing: wine '/home/chris/.wine/drive_c/Program Files/Microsoft Research/Z3-1.3.6/bin/z3.exe' '-smt' 'Z: mp\smtlib3cee8b.smt'
wine: cannot find ''/home/chris/.wine/drive_c/Program'

Note that:

The path to the executable is being chopped off at the first space, even though it is single-quoted.

可执行文件的路径在第一个空格处被切断,即使它是单引号。

The literal "\t" in the last path is being transformed into a tab character.

最后一个路径中的文字“\ t”正在转换为制表符。

Obviously, the quotations aren't being parsed the way I intended by the shell. How can I avoid these errors?

显然,引用并没有按照shell的意图进行解析。我该如何避免这些错误?

EDIT: The "\t" is being expanded through two levels of indirection: first, "$p" (and/or "$ARGS") is being expanded into Z:\tmp\smtlib3cee8b.smt; then, \t is being expanded into the tab character. This is (seemingly) equivalent to

编辑:“\ t”正在通过两个间接级别进行扩展:首先,“$ p”(和/或“$ ARGS”)正在扩展为Z:\ tmp \ smtlib3cee8b.smt;然后,\ t正在扩展到制表符。这(看似)相当于

Y='y\ty'
Z="z${Y}z"
echo $Z

which yields

zy\tyz

and not

zy  yz

UPDATE: eval "$CMD" does the trick. The "\t" problem seems to be echo's fault: "If the first operand is -n, or if any of the operands contain a backslash ( '\' ) character, the results are implementation-defined." (POSIX specification of echo)

更新:eval“$ CMD”可以解决问题。 “\ t”问题似乎是echo的错误:“如果第一个操作数是-n,或者如果任何操作数包含反斜杠('\')字符,则结果是实现定义的。” (回声的POSIX规范)

4 个解决方案

#1

I you do want to have the assignment to CMD you should use

我确实想要你应该使用CMD的作业

eval $CMD

instead of just $CMD in the last line of your script. This should solve your problem with spaces in the paths, I don't know what to do about the "\t" problem.

而不是在脚本的最后一行只有$ CMD。这应该可以解决路径中空格的问题,我不知道如何处理“\ t”问题。

#2

bash’s arrays are unportable but the only sane way to handle argument lists in shell

bash的数组是不可移植的,但是在shell中处理参数列表的唯一理智方式

The number of arguments is in ${#}

参数的数量是$ {#}

Bad stuff will happen with your script if there are filenames starting with a dash in the current directory

如果在当前目录中有以破折号开头的文件名,则脚本会发生错误

If the last line of your script just runs a program, and there are no traps on exit, you should exec it

如果脚本的最后一行只是运行一个程序,并且退出时没有陷阱,那么你应该执行它

With that in mind

考虑到这一点

#! /bin/bash

# push ARRAY arg1 arg2 ...
# adds arg1, arg2, ... to the end of ARRAY
function push() {
    local ARRAY_NAME="${1}"
    shift
    for ARG in "${@}"; do
        eval "${ARRAY_NAME}[\${#${ARRAY_NAME}[@]}]=\${ARG}"
    done
}

PROG="$(basename -- "${0}")"

if (( ${#} < 1 )); then
  # Error messages should state the program name and go to stderr
  echo "${PROG}: Usage: winewrap EXEC [ARGS...]" 1>&2
  exit 1
fi

EXEC=("${1}")
shift

for p in "${@}"; do
  if [ -e "${p}" ]; then
    p="$(winepath -w -- "${p}")"
  fi
  push EXEC "${p}"
done

exec "${EXEC[@]}"

#3

replace the last line from $CMD to just

将$ CMD中的最后一行替换为just

wine '$EXEC' $ARGS

葡萄酒'$ EXEC'$ ARGS

You'll note that the error is ''/home/chris/.wine/drive_c/Program' and not '/home/chris/.wine/drive_c/Program'

您会注意到错误是'/home/chris/.wine/drive_c/Program'而不是'/home/chris/.wine/drive_c/Program'

The single quotes are not being interpolated properly, and the string is being split by spaces.

单引号没有正确插值,字符串被空格分割。

#4

You can try preceeding the spaces with \ like so:

您可以尝试在空格前面加上\ like,如下所示:

/home/chris/.wine/drive_c/Program Files/Microsoft\ Research/Z3-1.3.6/bin/z3.exe

You can also do the same with your \t problem - replace it with \\t.

你也可以对你的问题做同样的事情 - 用\\ t替换它。

#1