在 bash 中检查 getopts 状态的最佳方法是什么？

发布于 2024-10-04 13:42:24 字数 851 浏览 1 评论 0原文

我正在使用以下脚本：

#!/bin/bash
#script to print quality of man
#unnikrishnan 24/Nov/2010
shopt -s -o nounset
declare -rx SCRIPT=${0##*/}
declare -r OPTSTRING="hm:q:"
declare SWITCH
declare MAN
declare QUALITY
if [ $# -eq 0 ];then
printf "%s -h for more information\n" "$SCRIPT"
exit 192
fi
while getopts "$OPTSTRING" SWITCH;do
case "$SWITCH" in
h) printf "%s\n" "Usage $SCRIPT -h -m MAN-NAME -q MAN-QUALITY"
   exit 0
   ;;
m) MAN="$OPTARG"
   ;;
q) QUALITY="$OPTARG"
   ;;
\?) printf "%s\n" "Invalid option"
    printf "%s\n" "$SWITCH"
    exit 192
    ;;
*) printf "%s\n" "Invalid argument"
   exit 192
    ;;
esac
done
printf "%s is a %s boy\n" "$MAN" "$QUALITY"
exit 0

在这个脚本中，如果我给出垃圾选项：

./getopts.sh adsas
./getopts.sh: line 32: MAN: unbound variable

您可以看到它失败。看来虽然不工作。解决它的最好方法是什么。

原文

I am using following script :

#!/bin/bash
#script to print quality of man
#unnikrishnan 24/Nov/2010
shopt -s -o nounset
declare -rx SCRIPT=${0##*/}
declare -r OPTSTRING="hm:q:"
declare SWITCH
declare MAN
declare QUALITY
if [ $# -eq 0 ];then
printf "%s -h for more information\n" "$SCRIPT"
exit 192
fi
while getopts "$OPTSTRING" SWITCH;do
case "$SWITCH" in
h) printf "%s\n" "Usage $SCRIPT -h -m MAN-NAME -q MAN-QUALITY"
   exit 0
   ;;
m) MAN="$OPTARG"
   ;;
q) QUALITY="$OPTARG"
   ;;
\?) printf "%s\n" "Invalid option"
    printf "%s\n" "$SWITCH"
    exit 192
    ;;
*) printf "%s\n" "Invalid argument"
   exit 192
    ;;
esac
done
printf "%s is a %s boy\n" "$MAN" "$QUALITY"
exit 0

In this if I am giving the junk option :

./getopts.sh adsas
./getopts.sh: line 32: MAN: unbound variable

you can see it fails. it seems while is not working. What is the best way to solve it.

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惯饮孤独 2024-10-11 13:42:24

当没有选项参数时，getopts 内置函数返回 1（“false”）。

因此，除非您有以 - 开头的选项参数，否则您的 while 永远不会执行。

请注意 bash(1) 的 getopts 部分的最后一段：

          getopts  returns true if an option, specified or unspecified, is
          found.  It returns false if the end of options is encountered or
          an error occurs.

The getopts builtin returns 1 ("false") when there are no option arguments.

So, your while never executes unless you have option arguments beginning with a -.

Note the last paragraph in the getopts section of bash(1):

          getopts  returns true if an option, specified or unspecified, is
          found.  It returns false if the end of options is encountered or
          an error occurs.

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冷夜 2024-10-11 13:42:24

如果您绝对需要 MAN，那么我建议您不要将其设为选项参数，而是将其设为位置参数。选项应该是可选的。

但是，如果您想将其作为一个选项来执行，那么请执行以下操作：

# initialise MAN to the empty string
MAN=
# loop as rewritten by DigitalRoss
while getopts "$OPTSTRING" SWITCH "$@"; do
  case "$SWITCH" in
    m) MAN="$OPTARG" ;;
  esac
done
# check that you have a value for MAN
[[ -n "$MAN" ]] || { echo "You must supply a MAN's name with -m"; exit 1; }

更好的是，在退出之前打印使用消息 - 将其拉出到函数中，以便您可以与 -h 选项的情况共享它。

If you absolutely require MAN, then i suggest you don't make it an option parameter, but a positional parameter. Options are supposed to be optional.

However, if you want to do it as an option, then do:

# initialise MAN to the empty string
MAN=
# loop as rewritten by DigitalRoss
while getopts "$OPTSTRING" SWITCH "$@"; do
  case "$SWITCH" in
    m) MAN="$OPTARG" ;;
  esac
done
# check that you have a value for MAN
[[ -n "$MAN" ]] || { echo "You must supply a MAN's name with -m"; exit 1; }

Even better, print the usage message before exiting - pull it out into a function so you can share it with the -h option's case.

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