Find and kill a process in one line using bash and regex
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How can I extract the process id automatically and kill it in the same line?
In bash, you should be able to do:
Details on its workings are as follows:
and you can see it terminating all the sleepers. Explaining the grep '[p]ython csp_build.py' bit in a bit more detail:When you do sleep 3600 & followed by ps -ef | grep sleep, you tend to get two processes with sleep in it, the sleep 3600 and the grep sleep (because they both have sleep in them, that's not rocket science).However, ps -ef | grep '[s]leep' won't create a process with sleep in it, it instead creates grep '[s]leep' and here's the tricky bit: the grep doesn't find it because it's looking for the regular expression "any character from the character class [s] (which is s) followed by leep.In other words, it's looking for sleep but the grep process is grep '[s]leep' which doesn't have sleep in it.When I was shown this (by someone here on SO), I immediately started using it because
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