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I want to write in a bash script a piece of code that checks if a program is already running. I have the following in order to search whether bar is running
foo=`ps -ef | grep bar | grep -v grep` The
grep -v grep part is to ensure that the "grep bar" is not taken into account in ps results
When bar isn't running, foo is correctly empty. But my problem lies in the fact tha the script has
set -e which is a flag to terminate the script if some command returns an error. It turns out that when bar isn't running, "grep -v grep" doesn't match with anything and grep returns an error. I tried using -q or -s but to no avail.
Is there any solution to that? Thx
314
If there's no match, that should generally be considered a failure, so a return of 0 would not be appropriate. Indeed, grep returns 0 if it matches, and non-zero if it does not.
The quiet option ( -q ), causes grep to run silently and not generate any output. Instead, it runs the command and returns an exit status based on success or failure. The return status is 0 for success and nonzero for failure.
-v means "invert the match" in grep, in other words, return all non matching lines.
The grep manual at the exit status section report: EXIT STATUS The exit status is 0 if selected lines are found, and 1 if not found. If an error occurred the exit status is 2.
Sure:
ps -ef | grep bar | { grep -v grep || true; } Or even:
ps -ef | grep bar | grep -v grep | cat If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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