Shell Programming and Scripting

I need to get the lines in a log file matching a particular string, and print out the output to the screen from the point of the match, to the end of the file.

I know this code does it: awk '/regex/,0'

but the string I need to search for inside the log file looks something like this: 09/Oct/2009:13:51

how do I use awk to search for the above string without encountering errors due to the "/" that is in the string itself?

thanks in advance for your help guys.

You have to escape special characters

Code:

awk '/09\/Oct\/2009:13:51/,0' file

I tried that, but i got an error:

Code:

bash-3.00# 
bash-3.00# awk '/09\/Oct\/2009:13:51/,0' /opt/logs/access
awk: syntax error near line 1
awk: bailing out near line 1
bash-3.00# 
bash-3.00#

Please advise. thanks

Hi.

On Solaris you will.

If you are using Solaris, use /usr/xpg4/bin/awk.

thank you so much guys. this works.