将命令行选项传递给 bash 中调用的脚本
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Passing command line options to invoked script in bash
提问by Michael
Suppose I have a script a.shto be invoked with options
假设我有一个a.sh要使用选项调用的脚本
a.sh -a1 a1option -a2 a2option
Suppose also I have a script b.sh, which invokes a.shand uses its own options. So user executes the scripts as follows:
假设我还有一个 script b.sh，它调用a.sh并使用自己的选项。所以用户执行脚本如下：
b.sh -b1 b1option -b2 b2option -a1 a1option -a2 a2option
Now I wonder how to parse the command line options in b.sh.
现在我想知道如何解析b.sh.
I do not need to parse the entirecommand line. I do not want b.shto be aware of options a1and a2. I would like to get only options b1and b2and pass the restto a.sh.
我不需要解析整个命令行。我不想b.sh知道选项a1和a2. 我只想获得选项b1，b2并将其余部分传递给a.sh.
How would you do it ?
你会怎么做？
回答by dangenet
As requested, this method avoids parsing the entire command line. Only the arguments up to --are collected for b.sh. Then the arguments for b are stripped and only the remaining arguments are passed to a.sh.
根据要求，此方法避免解析整个命令行。只--收集 到的参数b.sh。然后去除 b 的参数，只将剩余的参数传递给a.sh。
b.shis invoked with b.sh -b b1option -B b2option -- -a1 a1option -a2 a2option. In this line, the double dash --indicates the end of options for b.sh. The following parses the options before the --for use by b.sh, then removes the b arguments from the $@so you can pass it to a.shwithout worrying about what errors a.shmight give you.
b.sh用 调用b.sh -b b1option -B b2option -- -a1 a1option -a2 a2option。在这一行中，双破折号--表示 的选项结束b.sh。下面解析--for use by之前的选项b.sh，然后从 the 中删除 b 参数，$@以便您可以将其传递给，a.sh而不必担心a.sh可能会给您带来什么错误。
while getopts ":b:B:" opt; do
case $opt in
b) B1=${OPTARG}
;;
B) B2=${OPTARG}
;;
esac
done
## strips off the b options (which must be placed before the --)
shift $(({OPTIND}-1))
a.sh "$@"
A note: This method utilizes the bash builtin getopts. Getopts (as opposed to getopt, no s) takes only single-character options; hence, I have used band Binstead of b1and b2.
注意：此方法使用 bash 内置的 getopts。Getopts（与 getopt 相对，没有 s）只接受单字符选项；因此，我使用了bandB而不是b1and b2。
My favorite getoptsreference.
我最喜欢的getopts参考。
回答by konsolebox
You can do something like this:
你可以这样做：
#!/bin/bash
while [[ $# -gt 0 ]]; do
case "" in
-b1)
B1=true
B1OPT=
shift
;;
-b2)
B2=true
B2OPT=
shift
;;
--)
shift
break
;;
*)
echo "Invalid option: "
exit 1 ## Could be optional.
;;
esac
shift
done
bash a2.sh "$@"
Note that you should place your variable $@inside doublequotes to prevent word splitting when expanded.
请注意，您应该将变量$@放在双引号内，以防止展开时分词。
回答by danadam
If a.sh can ignore options it doesn't know you can just call it with all the options b.sh was called:
如果 a.sh 可以忽略选项，它不知道您可以使用 b.sh 调用的所有选项调用它：
a.sh "${@}"
