比如,我有一个脚本,用这行代码调用:

./myscript -vfd ./foo/bar/someFile -o /fizz/someOtherFile

或者这个:

./myscript -v -f -d -o /fizz/someOtherFile ./foo/bar/someFile 

什么是公认的解析方式,以便在每种情况下(或两者的某种组合)$v、$f和$d都将设置为true,而$outFile将等于/fiz/someOtherFile?


当前回答

我想提交我的项目:https://github.com/flyingangel/argparser

source argparser.sh
parse_args "$@"

就这么简单。环境将填充与参数同名的变量

其他回答

来自digitalpeer.com,稍作修改:

用法myscript.sh-p=my_prefix-s=dirname-l=libname

#!/bin/bash
for i in "$@"
do
case $i in
    -p=*|--prefix=*)
    PREFIX="${i#*=}"

    ;;
    -s=*|--searchpath=*)
    SEARCHPATH="${i#*=}"
    ;;
    -l=*|--lib=*)
    DIR="${i#*=}"
    ;;
    --default)
    DEFAULT=YES
    ;;
    *)
            # unknown option
    ;;
esac
done
echo PREFIX = ${PREFIX}
echo SEARCH PATH = ${SEARCHPATH}
echo DIRS = ${DIR}
echo DEFAULT = ${DEFAULT}

要更好地理解${i#*=},请在本指南中搜索“Substring Removal”。它在功能上等同于调用一个不需要的子流程的‘sed’s/[^=]*=//'<<<“$i”`或调用两个不需要子流程的“echo”$i”|sed’s/[^=]*=//'`。

使用bash模块中的模块“参数”

例子:

#!/bin/bash
. import.sh log arguments

NAME="world"

parse_arguments "-n|--name)NAME;S" -- "$@" || {
  error "Cannot parse command line."
  exit 1
}

info "Hello, $NAME!"

根据这里的其他答案,这是我的版本:

#!/bin/bash
set -e

function parse() {
    for arg in "$@"; do # transform long options to short ones
        shift
        case "$arg" in
            "--name") set -- "$@" "-n" ;;
            "--verbose") set -- "$@" "-v" ;;
            *) set -- "$@" "$arg"
        esac
    done

    while getopts "n:v" optname  # left to ":" are flags that expect a value, right to the ":" are flags that expect nothing
    do
        case "$optname" in
            "n") name=${OPTARG} ;;
            "v") verbose=true ;;
        esac
    done
    shift "$((OPTIND-1))" # shift out all the already processed options
}


parse "$@"
echo "hello $name"
if [ ! -z $verbose ]; then echo 'nice to meet you!'; fi

用法:

$ ./parse.sh
hello
$ ./parse.sh -n YOUR_NAME
hello YOUR_NAME
$ ./parse.sh -n YOUR_NAME -v
hello YOUR_NAME
nice to meet you!
$ ./parse.sh -v -n YOUR_NAME
hello YOUR_NAME
nice to meet you!
$ ./parse.sh -v
hello 
nice to meet you!

我认为这个很简单:

#!/bin/bash
#

readopt='getopts $opts opt;rc=$?;[ "$rc$opt" = "0?" ]&&exit 1;[ $rc = 0 ]||{ shift $[OPTIND-1];false; }'

opts=vfdo:

# Enumerating options
while eval "$readopt"
do
    echo OPT:$opt ${OPTARG+OPTARG:$OPTARG}
done

# Enumerating arguments
for arg
do
    echo ARG:$arg
done

调用示例:

./myscript -v -do /fizz/someOtherFile -f ./foo/bar/someFile
OPT:v 
OPT:d 
OPT:o OPTARG:/fizz/someOtherFile
OPT:f 
ARG:./foo/bar/someFile

另一个Shell参数分析器(ASAP)

符合POSIX,无getopt

我受到@bronson相对简单的回答的启发,并试图改进它(不增加太多复杂性)。结果如下:

使用-n[arg]、-abn[arg],--name[arg]和--name=arg样式中的任意一种选项;参数可以按任何顺序出现,循环后$@中只留下位置参数;使用--强制将剩余的参数视为位置参数;检测无效选项和缺少的参数;不依赖于getopt或外部工具(一个功能使用简单的sed命令);便携式,紧凑,可读性强,具有独立功能。

# Convenience functions.
usage_error () { echo >&2 "$(basename $0):  $1"; exit 2; }
assert_argument () { test "$1" != "$EOL" || usage_error "$2 requires an argument"; }

# One loop, nothing more.
if [ "$#" != 0 ]; then
  EOL=$(printf '\1\3\3\7')
  set -- "$@" "$EOL"
  while [ "$1" != "$EOL" ]; do
    opt="$1"; shift
    case "$opt" in

      # Your options go here.
      -f|--flag) flag='true';;
      -n|--name) assert_argument "$1" "$opt"; name="$1"; shift;;

      # Arguments processing. You may remove any unneeded line after the 1st.
      -|''|[!-]*) set -- "$@" "$opt";;                                          # positional argument, rotate to the end
      --*=*)      set -- "${opt%%=*}" "${opt#*=}" "$@";;                        # convert '--name=arg' to '--name' 'arg'
      -[!-]?*)    set -- $(echo "${opt#-}" | sed 's/\(.\)/ -\1/g') "$@";;       # convert '-abc' to '-a' '-b' '-c'
      --)         while [ "$1" != "$EOL" ]; do set -- "$@" "$1"; shift; done;;  # process remaining arguments as positional
      -*)         usage_error "unknown option: '$opt'";;                        # catch misspelled options
      *)          usage_error "this should NEVER happen ($opt)";;               # sanity test for previous patterns

    esac
  done
  shift  # $EOL
fi

# Do something cool with "$@"... \o/

注:我知道。。。二进制模式为0x01030307的参数可能会破坏逻辑。但是,如果有人在命令行中传递这样的参数,他们应该得到它。