英文:
how to read\print args in bash script with a login
问题
给定一个名为 foo.sh 的 Bash 脚本，用于执行登录操作。
虽然 foo.sh 被调用时带有参数，但这些参数并没有被打印出来。
如何打印传递给 foo.sh 的参数？例如：
$ ./foo.sh arg1 arg2
我期望得到的输出是：
arg1 arg2
英文:
given a bash script named foo.sh that performs a login
#!/usr/bin/env bash -l
bar() {
echo "$@"
}
bar "$@"
although foo.sh is invoked with arguments, the arguments are not being printed.
how to print the arguments that are passed to foo.sh? e.g.
$ ./foo.sh arg1 arg2
i would expect to get the output
arg1 arg2
答案1
得分: 2
If you have a recent /usr/bin/env which accepts the -S (--split-string) option :
#!/usr/bin/env -S bash -l
bar() {
echo "$@"
}
bar "$@"
英文:
If you have a recent /usr/bin/env which accept -S (--split-string) option :
#!/usr/bin/env -S bash -l
bar() {
echo "$@"
}
bar "$@"
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