Shell Programming and Scripting

Hello,

I have a log file for the year, which contains lines starting with the data in the format of YYYY-MM-DD. I need to get all the lines that contain the DD being 04, how would I do this? I tried using grep "*-*04" but it didn't work.

Any quick one liners I should know about?

Thank you.

Try:

Code:

awk -F- '$3~"04"' file

grep takes regular expressions, not globs, and in regular expressions, * means "zero or more of the previous character". So the first * did nothing, and the second * means "match any number of - characters". If you want to match zero or more of any character you can use the special "." character, like:

Code:

grep ".*-.*-04"

though I'd specify it a bit further so those *'s don't get out of control and try to match the entire line, like

Code:

grep "[0-9][0-9][0-9][0-9]-[0-9][0-9]-04"