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Printing a Line from a file

 
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# 1  
Old 08-30-2007
Printing a Line from a file
I have a log file with several lines as follows:

Aug 30 06:35:08 trnwvltfit1 /usr/lib/snmp/snmpdx: [ID 702911 daemon.error] Agent snmpd appeared dead but responded to ping

I am using the date of the line to determine which lines to print. However, I am only trying to print the parts of the line that are NOT highlighted in red. I have no knowledge of how long the line may be. How can this be done? So far I have:

Code:
$cat messages.log | grep "Aug 30" |

# 2  
Old 08-30-2007
Code:
sed -n -e "s/Aug 30 \(..:..:..\) trnwvltfit1 \(.*\)/\1 \2/p" log.file

# 3  
Old 08-30-2007
It seems to be printing out the right output. Say I wanted to replace the 30 with a variable $previousday, would this still hold up?
Last edited by Nysif Steve; 08-30-2007 at 01:05 PM..
# 4  
Old 08-31-2007
Quote:
Originally Posted by Nysif Steve
It seems to be printing out the right output. Say I wanted to replace the 30 with a variable $previousday, would this still hold up?
Yes it will hold up. Make sure the expression is within double quotes and not single quotes.
# 5  
Old 08-31-2007
Thank you for your help!
 
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