How to grep search for string but omit string from result

Question

I'm trying to find the expiration date of an SSL Certificate using the following command:

openssl s_client -connect ${CERT}:443 </dev/null 2>&1 | openssl x509 -text | grep "Not After"

This outputs a result to the terminal in the format:

Not After : Apr 14 10:06:09 2018 GMT

How do I remove the "Not After : " part of the result and just be shown:

Apr 14 10:06:09 2018 GMT

Any help appreciated!

score 2 · Answer 1 · answered Apr 04 '18 at 09:22

2

Just use a lookbehind to get rid of it:

$ grep -Po '(?<=Not After : ).*' <<< "Not After : Apr 14 10:06:09 2018 GMT"
Apr 14 10:06:09 2018 GMT

With (?<=Not After : ).* we are telling grep to print everything that comes after the fixed string "Not After : ".

Note this requires a grep with the -P option to load the Perl regex.

answered Apr 04 '18 at 09:22

fedorqui

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thanks for the help, but the -P flag doesn't work for me. Fixed now – tnoel999888 Apr 04 '18 at 10:03

score 1 · Accepted Answer · answered Apr 04 '18 at 09:29

1

A simple approach:-

   openssl s_client -connect ${CERT}:443 </dev/null 2>&1 | openssl x509 -text | grep "Not After" | sed 's/Not After : //g'

Here I have added | sed 's/Not After : //g'. Now here sed command will search for string Not After : and replace it from the original string redirected to it. And final output would be:-

Apr 14 10:06:09 2018 GMT

Hope this will help you.

answered Apr 04 '18 at 09:29

Abhijit Pritam Dutta

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why `/g`? If you just want to replace once, there is no need to use the global flag. – fedorqui Apr 04 '18 at 09:55
@Abhijit Pritam thanks, this worked! – tnoel999888 Apr 04 '18 at 10:03
why grep, sed is sufficient? – kyodev Apr 04 '18 at 10:09

How to grep search for string but omit string from result

2 Answers2