AWK/SED print if 2nd character in a column is greater than 0

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# 1  
Old 08-18-2011
AWK/SED print if 2nd character in a column is greater than 0

We have an access log where column 8 displays the time in seconds like below:
Tj8nQAoNgwsAABov9cIAAAFL - - - [07/Aug/2011:17:01:04 -0700] (0) -  "GET /aaaaa/bbbb/bbbb

where column 8 is printed (0). We are trying to find how many entries are there that has column 8 greater than 0.

Remember $8 is (0) and not 0. It has parenthesis around the number.
I tried various commands like below which did not work. Some help might be greatly appreciated.
awk '{print "time=",($8 != "(0)")}' access.log
awk '{print "time="$8 !~/^(0)$/}' access.log
awk '{print $8 !~ /(0)/}' access.log

Last edited by Franklin52; 08-19-2011 at 03:04 PM.. Reason: Please use code tags for data and code samples, thank you
# 2  
Old 08-18-2011
awk '$8!~"(0)"' access.log

This User Gave Thanks to bartus11 For This Post:
# 3  
Old 08-18-2011

Last edited by bvsa; 08-18-2011 at 06:50 PM..
# 4  
Old 08-18-2011
Don't hijack other people's threads. Post your problem as a new thread.
# 5  
Old 08-18-2011
That helped me to some extent. Thank you. Is there a way we can print rows whose $8 is greater than 10 secs. $8 is time in seconds, format = (n), parenthesis included.

I did a awk '$8 >"(10)"' access.log which also printed rows < (10) value. Any ideas?
# 6  
Old 08-18-2011
awk '{x=$8;gsub("[()]","",x);if (x>10) print}' access.log

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