在 bash 中使用 getopts 處理可選參數

我有一個 bash 腳本，它處理帶有可選參數的輸入文件。腳本看起來像這樣

#!/bin/bash
while getopts a:b:i: option
do
case "${option}"
in
a) arg1=${OPTARG};;
b) arg2=${OPTARG};;
i) file=${OPTARG};;
esac
done

[ -z "$file" ] && { echo "No input file specified" ; exit; }

carry out some stuff

腳本執行良好，但我需要像這樣指定輸入文件

sh script.sh -a arg1 -b arg2 -i filename

我希望能夠在沒有-i選項的情況下呼叫腳本，就像這樣

sh script.sh -a arg1 -b arg2 filename

當沒有指定輸入文件時仍然有錯誤消息。有沒有辦法做到這一點？

#!/bin/sh -

# Beware variables can be inherited from the environment. So
# it's important to start with a clean slate if you're going to
# dereference variables while not being guaranteed that they'll
# be assigned to:
unset -v file arg1 arg2

# no need to initialise OPTIND here as it's the first and only
# use of getopts in this script and sh should already guarantee it's
# initialised.
while getopts a:b:i: option
do
 case "${option}" in
   (a) arg1=${OPTARG};;
   (b) arg2=${OPTARG};;
   (i) file=${OPTARG};;
   (*) exit 1;;
 esac
done

shift "$((OPTIND - 1))"
# now "$@" contains the rest of the arguments

if [ -z "${file+set}" ]; then
 if [ "$#" -eq 0 ]; then
   echo >&2 "No input file specified"
   exit 1
 else
   file=$1 # first non-option argument
   shift
 fi
fi

if [ "$#" -gt 0 ]; then
 echo There are more arguments:
 printf ' - "%s"\n' "$@"
fi

我將其更改為bash，因為該程式碼中sh沒有任何特定內容。bash

引用自：https://unix.stackexchange.com/questions/617037