我想同时使用getopts
和位置参数,但是如果我将位置参数传递给程序getopts
迷路。
directory=$1
while getopts l: flag; do
case "$flag" in
l) level=$OPTARG;;
esac
done
if [ -n "$level" ]; then
echo "Level exist!"
else
echo "Level doesn't exist!"
fi
所以当我像这样运行程序时:
sh myprogram.sh ~/documents -l 2
我预计:
Level exist!
相反,它返回:
Level doesn't exist!
问题是,如果我在没有位置参数(~/documents)的情况下运行程序,如下所示:
sh myprogram.sh -l 2
我得到正确的输出:
Level exist!
这是为什么?我如何同时使用位置参数和getopts
在bash中?
Thanks!
大多数工具都是以以下形式编写的:tool [options] arg ...
所以你会这样做:
# first, parse the options:
while getopts l: flag; do
case "$flag" in
l) level=$OPTARG;;
\?) exit 42;;
esac
done
# and shift them away
shift $((OPTIND - 1))
# validation
if [ -n "$level" ]; then
echo "Level exist!"
else
echo "Level doesn't exist!"
fi
# THEN, access the positional params
echo "there are $# positional params remaining"
for ((i=1; i<=$#; i++)); do
printf "%d\t%s\n" $i "${!i}"
done
Use the \?
如果用户提供未知选项或未能提供所需参数,则中止脚本
并像这样调用它:
$ bash test.sh
Level doesn't exist!
there are 0 positional params remaining
$ bash test.sh -l 2
Level exist!
there are 0 positional params remaining
$ bash test.sh -l 2 foo bar
Level exist!
there are 2 positional params remaining
1 foo
2 bar
$ bash test.sh -x
test.sh: illegal option -- x
$ bash test.sh -l
test.sh: option requires an argument -- l
But you cannot将选项放在参数后面:找到第一个非选项参数时 getopts 停止
$ bash test.sh foo bar -l 2
Level doesn't exist!
there are 4 positional params remaining
1 foo
2 bar
3 -l
4 2
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