Grep date with regex

Question

I have files with lines in the following formats:

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12/16/09 17:56:30.211 rest of line...
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12/17/09 05:34:10.809 rest of line...
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How do I grep the lines out of this file that are between two dates, i.e. all the values for one day? I need to create a daily file with the contents from all the different server logs for that day.

See also: http://serverfault.com/questions/40681/fast-script-to-cut-section-of-logfile-based-on-timestamp — MikeyB, Dec 17 '09 at 21:36

score 4 · Accepted Answer · answered Dec 17 '09 at 19:46

4

Brackets can be used to give you a range of numbers. The carrot can match the start of a line. So..

grep "^12\/1[6-8]\/09" will give you everything that starts with 12/16/09, 12/17/09, and 12/18/09

answered Dec 17 '09 at 19:46

Christopher Karel

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Thanks, I was so hung up on the timestamp that I didn't think of just using the day. – Lance Roberts Dec 17 '09 at 19:52
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You could add a space after the year also, like `grep "^12\/1[6-8]\/09 "` to be sure. – raphink Dec 17 '09 at 19:55

Grep date with regex

1 Answers1