Question

我正在尝试创建一个带有可选选项标志的脚本。使用getopts，可以在标志之后指定一个强制参数（使用冒号），但我想保持它是可选的。

会是这样的：

./install.sh -a 3

或

./install.sh -a3

其中'a'是标志，'3'是跟随。

的可选参数

提前致谢。

Answer 1

getopt外部程序允许选项通过在选项名称中添加双冒号来拥有单个可选参数。

# Based on a longer example in getopt-parse.bash, included with
# getopt
TEMP=$(getopt -o a:: -- "$@")
eval set -- "$TEMP"
while true ; do
   case "$1" in
     -a)
        case "$2" in 
          "") echo "Option a, no argument"; shift 2 ;;
          *) echo "Option a, argument $2"; shift 2;;
        esac ;;
     --) shift; break ;;
     *) echo "Internal error!"; exit 1 ;;
   esac
done

Answer 2

以下是没有getopt的，它带有一个带-a标志的可选参数：

for WORD; do
        case $WORD in
            -a?)  echo "single arg Option"
                SEP=${WORD:2:1}
                echo $SEP
                shift ;;
            -a) echo "split arg Option"
                if [[ ${2:0:1} != "-" && ${2:0:1} != ""]] ; then
                 SEP=$2
                 shift 2
                 echo "arg present"
                 echo $SEP
                else
                 echo "optional arg omitted"
                fi ;;
            -a*) echo "arg Option"
                SEP=${WORD:2}
                echo $SEP
                shift ;;
            -*) echo "Unrecognized Short Option"
                echo "Unrecognized argument"
            ;;
        esac
done

其他选项/标志也可以轻松添加。

Answer 3

使用getopt功能。在大多数系统上，man getopt将为其生成文档，甚至是在脚本中使用它的示例。从我系统的手册页：

以下代码片段显示了如何处理参数对于可以使用选项-a和-b以及选项-o的命令，这需要一个论点。

       args=`getopt abo: $*`
       # you should not use `getopt abo: "$@"` since that would parse
       # the arguments differently from what the set command below does.
       if [ $? != 0 ]
       then
               echo 'Usage: ...'
               exit 2
       fi
       set -- $args
       # You cannot use the set command with a backquoted getopt directly,
       # since the exit code from getopt would be shadowed by those of set,
       # which is zero by definition.
       for i
       do
               case "$i"
               in
                       -a|-b)
                               echo flag $i set; sflags="${i#-}$sflags";
                               shift;;
                       -o)
                               echo oarg is "'"$2"'"; oarg="$2"; shift;
                               shift;;
                       --)
                               shift; break;;
               esac
       done
       echo single-char flags: "'"$sflags"'"
       echo oarg is "'"$oarg"'"

此代码将接受以下任何内容：

       cmd -aoarg file file
       cmd -a -o arg file file
       cmd -oarg -a file file
       cmd -a -oarg -- file file

Answer 4

在bash中有一些隐式变量：

    $#: contains number of arguments for a called script/function

    $0: contains names of script/function
    $1: contains first argument
    $2: contains second argument
    ...
    $n: contains n-th argument

例如：

    #!/bin/ksh

    if [ $# -ne 2 ]
    then
        echo "Wrong number of argument - expected 2 : $#"
    else
        echo "Argument list:"
        echo "\t$0"
        echo "\t$1"
        echo "\t$2"
    fi

Answer 5

我的解决方案：

#!/bin/bash
count=0
skip=0
flag="no flag"
list=($@) #put args in array
for arg in $@ ; do #iterate over array
    count=$(($count+1)) #update counter
    if [ $skip -eq 1 ]; then #check if we have to skip this args
        skip=0
        continue
    fi
    opt=${arg:0:2} #get only first 2 chars as option
    if [ $opt == "-a" ]; then #check if option equals "-a"
       if [ $opt == $arg ] ; then #check if this is only the option or has a flag
            if [ ${list[$count]:0:1} != "-" ]; then #check if next arg is an option
                skip=1 #skip next arg
                flag=${list[$count]} #use next arg as flag
            fi
       else
            flag=${arg:2} #use chars after "-a" as flag
       fi 
    fi
done

echo $flag

如何使用标志的可选参数创建bash脚本

5 个答案: