如何使用参数传递多个Shell脚本

Question

I have scripts a.sh and b.sh in which i have pass the ip as argument.

I tried running it as

sh -x a.sh 172.19.57.21 & b.sh 172.19.57.21 &

But I see only first script runs.

Answer 1

When you run sh -x a.sh 172.19.57.21 & b.sh 172.19.57.21 & :

• sh -x a.sh 172.19.57.21 is one command, & sends it to background immediately

• b.sh 172.19.57.21 is another command, again & puts it in background

The problem seems to me is that the script b.sh is not executable and as you are not running it as an argument to shell (unlike a.sh ), it is failed in the PATH search.

You can run b.sh as shell's argument as well eg:

sh a.sh 172.19.57.21 & sh b.sh 172.19.57.21 &

Or if both scripts are executables and have proper shebang:

./a.sh 172.19.57.21 & ./b.sh 172.19.57.21 &

---

I would recommend a wrapper to get the argument IP address once, and call required scripts from the wrapper, something like a tiny function would do:

wrapper() {
    /path/to/a.sh "$@" &
    /path/to/b.sh "$@" &
}

Now, you can just do eg:

wrapper 172.19.57.21

Answer 2

use ; between commands like

sh -x a.sh 172.19.57.21 &; sh -x b.sh 172.19.57.21 &