在Bash中使用getopts检索单个选项的多个参数

Question

我需要getopts的帮助。

我创建了一个Bash脚本，在运行时看起来像这样：

$ foo.sh -i env -d directory -s子目录-f file

从每个标志处理一个参数时，它可以正常工作。但是当我从每个标志调用多个参数时，我不确定如何从getopts中的变量中提取多个变量信息。

while getopts ":i:d:s:f:" opt
   do
     case $opt in
        i ) initial=$OPTARG;;
        d ) dir=$OPTARG;;
        s ) sub=$OPTARG;;
        f ) files=$OPTARG;;

     esac
done

在获取选项后，我想从变量

构建目录结构

foo.sh -i test -d directory -s subdirectory -s subdirectory2 -f file1 file2 file3

然后目录结构将是

/test/directory/subdirectory/file1
/test/directory/subdirectory/file2
/test/directory/subdirectory/file3
/test/directory/subdirectory2/file1
/test/directory/subdirectory2/file2
/test/directory/subdirectory2/file3

有什么想法吗？

Answer 1

您可以多次使用相同的选项，并将所有值添加到数组。

对于这里非常具体的原始问题，Ryan的mkdir -p解决方案显然是最好的。

但是，对于使用getopts 从同一选项中获取多个值的更一般性问题，这里是：

#!/bin/bash

while getopts "m:" opt; do
    case $opt in
        m) multi+=("$OPTARG");;
        #...
    esac
done
shift $((OPTIND -1))

echo "The first value of the array 'multi' is '$multi'"
echo "The whole list of values is '${multi[@]}'"

echo "Or:"

for val in "${multi[@]}"; do
    echo " - $val"
done

输出结果为：

$ /tmp/t
The first value of the array 'multi' is ''
The whole list of values is ''
Or:

$ /tmp/t -m "one arg with spaces"
The first value of the array 'multi' is 'one arg with spaces'
The whole list of values is 'one arg with spaces'
Or:
 - one arg with spaces

$ /tmp/t -m one -m "second argument" -m three
The first value of the array 'multi' is 'one'
The whole list of values is 'one second argument three'
Or:
 - one
 - second argument
 - three

Answer 2

我知道这个问题已经过时了，但我想在这里提出这个答案，以防有人来寻找答案。

像BASH一样支持像这样递归制作目录，因此不需要脚本。例如，原始海报需要以下内容：

$ foo.sh -i test -d directory -s subdirectory -s subdirectory2 -f file1 file2 file3
/test/directory/subdirectory/file1
/test/directory/subdirectory/file2
/test/directory/subdirectory/file3
/test/directory/subdirectory2/file1
/test/directory/subdirectory2/file2
/test/directory/subdirectory2/file3

使用此命令行可以轻松完成此操作：

pong:~/tmp
[10] rmclean$ mkdir -pv test/directory/{subdirectory,subdirectory2}/{file1,file2,file3}
mkdir: created directory ‘test’
mkdir: created directory ‘test/directory’
mkdir: created directory ‘test/directory/subdirectory’
mkdir: created directory ‘test/directory/subdirectory/file1’
mkdir: created directory ‘test/directory/subdirectory/file2’
mkdir: created directory ‘test/directory/subdirectory/file3’
mkdir: created directory ‘test/directory/subdirectory2’
mkdir: created directory ‘test/directory/subdirectory2/file1’
mkdir: created directory ‘test/directory/subdirectory2/file2’
mkdir: created directory ‘test/directory/subdirectory2/file3’

甚至更短一些：

pong:~/tmp
[12] rmclean$ mkdir -pv test/directory/{subdirectory,subdirectory2}/file{1,2,3}
mkdir: created directory ‘test’
mkdir: created directory ‘test/directory’
mkdir: created directory ‘test/directory/subdirectory’
mkdir: created directory ‘test/directory/subdirectory/file1’
mkdir: created directory ‘test/directory/subdirectory/file2’
mkdir: created directory ‘test/directory/subdirectory/file3’
mkdir: created directory ‘test/directory/subdirectory2’
mkdir: created directory ‘test/directory/subdirectory2/file1’
mkdir: created directory ‘test/directory/subdirectory2/file2’
mkdir: created directory ‘test/directory/subdirectory2/file3’

或更短，更符合：

pong:~/tmp
[14] rmclean$ mkdir -pv test/directory/subdirectory{1,2}/file{1,2,3}
mkdir: created directory ‘test’
mkdir: created directory ‘test/directory’
mkdir: created directory ‘test/directory/subdirectory1’
mkdir: created directory ‘test/directory/subdirectory1/file1’
mkdir: created directory ‘test/directory/subdirectory1/file2’
mkdir: created directory ‘test/directory/subdirectory1/file3’
mkdir: created directory ‘test/directory/subdirectory2’
mkdir: created directory ‘test/directory/subdirectory2/file1’
mkdir: created directory ‘test/directory/subdirectory2/file2’
mkdir: created directory ‘test/directory/subdirectory2/file3’

或者最后，使用序列：

pong:~/tmp
[16] rmclean$ mkdir -pv test/directory/subdirectory{1..2}/file{1..3}
mkdir: created directory ‘test’
mkdir: created directory ‘test/directory’
mkdir: created directory ‘test/directory/subdirectory1’
mkdir: created directory ‘test/directory/subdirectory1/file1’
mkdir: created directory ‘test/directory/subdirectory1/file2’
mkdir: created directory ‘test/directory/subdirectory1/file3’
mkdir: created directory ‘test/directory/subdirectory2’
mkdir: created directory ‘test/directory/subdirectory2/file1’
mkdir: created directory ‘test/directory/subdirectory2/file2’
mkdir: created directory ‘test/directory/subdirectory2/file3’

Answer 3

getopts选项只能采用零个或一个参数。您可能希望更改接口以删除-f选项，并迭代剩余的非选项参数

usage: foo.sh -i end -d dir -s subdir file [...]

所以，

while getopts ":i:d:s:" opt; do
  case "$opt" in
    i) initial=$OPTARG ;;
    d) dir=$OPTARG ;;
    s) sub=$OPTARG ;;
  esac
done
shift $(( OPTIND - 1 ))

path="/$initial/$dir/$sub"
mkdir -p "$path"

for file in "$@"; do
  touch "$path/$file"
done

Answer 4

我解决了你这样的问题：

而不是：

foo.sh -i test -d directory -s subdirectory -s subdirectory2 -f file1 file2 file3

这样做：

foo.sh -i test -d directory -s "subdirectory subdirectory2" -f "file1 file2 file3"

使用空间分隔符，您可以使用基本循环运行它。 这是代码：

while getopts ":i:d:s:f:" opt
   do
     case $opt in
        i ) initial=$OPTARG;;
        d ) dir=$OPTARG;;
        s ) sub=$OPTARG;;
        f ) files=$OPTARG;;

     esac
done

for subdir in $sub;do
   for file in $files;do
      echo $subdir/$file
   done   
done

以下是示例输出：

$ ./getopts.sh -s "testdir1 testdir2" -f "file1 file2 file3"
testdir1/file1
testdir1/file2
testdir1/file3
testdir2/file1
testdir2/file2
testdir2/file3

Answer 5

实际上有一种方法可以使用getopts检索多个参数，但它需要使用getopts'OPTIND变量进行一些手动黑客攻击。

请参阅以下脚本（转载如下）：https://gist.github.com/achalddave/290f7fcad89a0d7c3719。这可能是一种更简单的方法，但这是我能找到的最快捷方式。

#!/bin/sh

usage() {
cat << EOF
$0 -a <a1> <a2> <a3> [-b] <b1> [-c]
    -a      First flag; takes in 3 arguments
    -b      Second flag; takes in 1 argument
    -c      Third flag; takes in no arguments
EOF
}

is_flag() {
    # Check if $1 is a flag; e.g. "-b"
    [[ "$1" =~ -.* ]] && return 0 || return 1
}

# Note:
# For a, we fool getopts into thinking a doesn't take in an argument
# For b, we can just use getopts normal behavior to take in an argument
while getopts "ab:c" opt ; do
    case "${opt}" in
        a)
            # This is the tricky part.

            # $OPTIND has the index of the _next_ parameter; so "\${$((OPTIND))}"
            # will give us, e.g., ${2}. Use eval to get the value in ${2}.
            # The {} are needed in general for the possible case of multiple digits.

            eval "a1=\${$((OPTIND))}"
            eval "a2=\${$((OPTIND+1))}"
            eval "a3=\${$((OPTIND+2))}"

            # Note: We need to check that we're still in bounds, and that
            # a1,a2,a3 aren't flags. e.g.
            #   ./getopts-multiple.sh -a 1 2 -b
            # should error, and not set a3 to be -b.
            if [ $((OPTIND+2)) -gt $# ] || is_flag "$a1" || is_flag "$a2" || is_flag "$a3"
            then
                usage
                echo
                echo "-a requires 3 arguments!"
                exit
            fi

            echo "-a has arguments $a1, $a2, $a3"

            # "shift" getopts' index
            OPTIND=$((OPTIND+3))
            ;;
        b)
            # Can get the argument from getopts directly
            echo "-b has argument $OPTARG"
            ;;
        c)
            # No arguments, life goes on
            echo "-c"
            ;;
    esac
done

Answer 6

如果要为选项指定任意数量的值，可以使用简单循环查找它们并将它们填充到数组中。例如，让我们修改OP的示例以允许任意数量的-s参数：

unset -v sub
while getopts ":i:d:s:f:" opt
   do
     case $opt in
        i ) initial=$OPTARG;;
        d ) dir=$OPTARG;;
        s ) sub=("$OPTARG")
            until [[ $(eval "echo \${$OPTIND}") =~ ^-.* ]] || [ -z $(eval "echo \${$OPTIND}") ]; do
                sub+=($(eval "echo \${$OPTIND}"))
                OPTIND=$((OPTIND + 1))
            done
            ;;
        f ) files=$OPTARG;;
     esac
done

这将获取第一个参数（$ OPTARG）并将其放入数组$ sub中。然后它将继续搜索剩余的参数，直到它遇到另一个虚线参数或者没有更多的参数要评估。如果它找到了更多不是虚线参数的参数，它会将它添加到$ sub数组并突破$ OPTIND变量。

所以在OP的例子中，可以运行以下命令：

foo.sh -i test -d directory -s subdirectory1 subdirectory2 -f file1

如果我们将这些行添加到脚本中以演示：

echo ${sub[@]}
echo ${sub[1]}
echo $files

输出结果为：

subdirectory1 subdirectory2
subdirectory2
file1

Answer 7

最初的问题涉及到getopts，但是还有另一种解决方案可以在没有getopts的情况下提供更灵活的功能（这可能有点冗长，但提供了更灵活的命令行界面）。这是一个例子：

while [[ $# > 0 ]]
do
    key="$1"
    case $key in
        -f|--foo)
            nextArg="$2"
            while ! [[ "$nextArg" =~ -.* ]] && [[ $# > 1 ]]; do
                case $nextArg in
                    bar)
                        echo "--foo bar found!"
                    ;;
                    baz)
                        echo "--foo baz found!"
                    ;;
                    *)
                        echo "$key $nextArg found!"
                    ;;
                esac
                if ! [[ "$2" =~ -.* ]]; then
                    shift
                    nextArg="$2"
                else
                    shift
                    break
                fi
            done
        ;;
        -b|--bar)
            nextArg="$2"
            while ! [[ "$nextArg" =~ -.* ]] && [[ $# > 1 ]]; do
                case $nextArg in
                    foo)
                        echo "--bar foo found!"
                    ;;
                    baz)
                        echo "--bar baz found!"
                    ;;
                    *)
                        echo "$key $nextArg found!"
                    ;;
                esac
                if ! [[ "$2" =~ -.* ]]; then
                    shift
                    nextArg="$2"
                else
                    shift
                    break
                fi
            done
        ;;
        -z|--baz)
            nextArg="$2"
            while ! [[ "$nextArg" =~ -.* ]] && [[ $# > 1 ]]; do

                echo "Doing some random task with $key $nextArg"

                if ! [[ "$2" =~ -.* ]]; then
                    shift
                    nextArg="$2"
                else
                    shift
                    break
                fi
            done
        ;;
        *)
            echo "Unknown flag $key"
        ;;
    esac
    shift
done

在这个例子中，我们循环遍历所有命令行选项，寻找与我们接受的命令行标志匹配的参数（例如-f或-foo）。一旦找到标志，我们就会遍历每个参数，直到我们用完参数或遇到另一个标志。这会让我们回到我们的外循环中，它只处理标志。

使用此设置，以下命令是等效的：

script -f foo bar baz
script -f foo -f bar -f baz
script --foo foo -f bar baz
script --foo foo bar -f baz

您还可以解析令人难以置信的无组织参数集，例如：

script -f baz derp --baz herp -z derp -b foo --foo bar -q llama --bar fight

获得输出：

--foo baz found!
-f derp found!
Doing some random task with --baz herp
Doing some random task with -z derp
--bar foo found!
--foo bar found!
Unknown flag -q
Unknown flag llama
--bar fight found!

Answer 8

由于您没有展示您希望如何构建列表

/test/directory/subdirectory/file1
. . .
test/directory/subdirectory2/file3

有点不清楚如何继续，但基本上你需要继续将任何新值附加到适当的变量，即

 case $opt in
    d ) dirList="${dirList} $OPTARG" ;;
 esac

请注意，在第一次传递时，dir将为空，并且最终会在${dirList}的最终值的前面开出一个空格。 （如果你真的需要不包含任何额外空格的代码，无论是正面还是背面，都有一个我可以告诉你的命令，但它很难理解，而且你似乎不需要它，但是让我知道）

然后，您可以将列表变量包装在for循环中以发出所有值，即

for dir in ${dirList} do
   for f in ${fileList} ; do
      echo $dir/$f
   done
done

最后，在您的案例陈述中“捕获”任何未知输入被认为是一种良好做法，即

 case $opt in
    i ) initial=$OPTARG;;
    d ) dir=$OPTARG;;
    s ) sub=$OPTARG;;
    f ) files=$OPTARG;;
    * ) 
       printf "unknown flag supplied "${OPTARG}\nUsageMessageGoesHere\n"
       exit 1
    ;;

 esac 

我希望这会有所帮助。

Answer 9

#!/bin/bash
myname=$(basename "$0")

# help function
help () { cat <<EOP
   $myname: -c cluster [...] -a action [...] -i instance [...]
EOP
}

# parse sub options
get_opts () {
  rs='' && rc=0 # return string and return code
  while [[ $# -gt 0 ]]; do
    shift
    [[ "$1" =~ -.* ]] && break ||  rs="$rs $1" && rc=$((rc + 1))
  done
  echo "$rs"
}

#parse entire command-line
while [[ $# -gt 0 ]]; do
    case $1 in
        "-a") ACTS="$(get_opts $@)"
           ;;
        "-i") INSTS=$(get_opts $@)
           ;;
        "-c") CLUSTERS=$(get_opts $@)
           ;;
        "-h") help
           ;;
        ?) echo "sorry, I dont do $1"
           exit
           ;;
    esac
    shift
done

Answer 10

以下链接应该是此要求的通用解决方案。

易于注入且清晰易懂，还能最大限度地减少对原始代码的影响。

Multiple option arguments using getopts (bash)

function getopts-extra () {
    declare i=1
    # if the next argument is not an option, then append it to array OPTARG
    while [[ ${OPTIND} -le $# && ${!OPTIND:0:1} != '-' ]]; do
        OPTARG[i]=${!OPTIND}
        let i++ OPTIND++
    done
}

# Use it within the context of `getopts`:
while getopts s: opt; do
    case $opt in
       s) getopts-extra "$@"
          args=( "${OPTARG[@]}" )
    esac
done