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using awk to get specific section of lines in logs


 
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# 1  
Old 07-17-2012
using awk to get specific section of lines in logs

i have a log file that has the date and time that looks like this:

Code:
Wed Jun 28 15:46:21 2012 test failed tailed passed passed not error panic


what we want to focus on is the first 5 columns because they contain the date and time.

the date and time can be anywhere on the line. in this particular case, the date and time is at the beginning of the line. my question is, what happens if it is somewhere else on that line? how can i make it so that when i run a one liner awk command, it finds the date and time wherever it is on the line?

OS: linux and sunos
shell: bash
# 2  
Old 07-17-2012
What do you want to do with it? Output just the datim, or check for lines that contain a specified date?
# 3  
Old 07-17-2012
Quote:
Originally Posted by CarloM
What do you want to do with it? Output just the datim, or check for lines that contain a specified date?

the lines that contain a specified date. thank you
# 4  
Old 07-17-2012
Like this?

Code:
awk '/([A-Z][a-z]{2} ){2}[0-9]{2} ([0-9]{2}:){2}[0-9]{2} [0-9]{4}/' infile

# 5  
Old 07-17-2012
Code:
awk '/Wed Jun 28 15:46:21 2012/' file

or just
Code:
grep 'Wed Jun 28 15:46:21 2012' file

Neither care where the matching string is on the line.
# 6  
Old 07-17-2012
Quote:
Originally Posted by elixir_sinari
Like this?

Code:
awk '/([A-Z][a-z]{2} ){2}[0-9]{2} ([0-9]{2}:){2}[0-9]{2} [0-9]{4}/' infile


this might be the answer. but when i run it i get nothing back
# 7  
Old 07-17-2012
What's the data in the input file?

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