Bash – Script Launched from Another Script Not Visible in ps Command

bash

I have one script called test.sh. This calls another script foo.sh, like so:

#!/bin/bash
echo "starting other script"
. ./foo.sh

The foo.sh script looks like this:

#!/bin/bash
sleep 10
echo "running"
sleep 10

Now, when I do the following from the terminal:

ps -def | grep foo.sh | grep -v grep

I never get a result. I can see the output from the scripts in the terminal, no problem, but how can I check if foo.sh is running at all?

Best Answer

You launch foo.sh as . ./foo.sh.

A shell script is not a process by itself. It is interpreted by an instance of the shell, which is the process.

. is an alias of the source command that executes commands in the current shell environment. This means ./foo.sh is executed by the same shell that executes the script that launches it. It does not create a new process and this is why you cannot find it using ps.

You can launch foo.sh as ./foo.sh. In this case it will be executed by a new instance of the shell, i.e. into a new process.

But, depending on what it does, the execution of the original shell after ./foo.sh completes might be different in the two cases. When foo.sh is executed in the same shell, the environment variables it changes belong to the shell that also executes the original script. Changing them could affect the original script.

When you run foo.sh in a separate script the changes it operates on the environment variable do not affect the original script. They are different processes, the do not share anything.

but how can I check if foo.sh is running at all?

This is easy from the script that launched foo.sh because it waits until foo.sh completes before resuming its execution.

It is probably impossible (or almost there) from outside.

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