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How to print the entire line if the mentioned match is found?

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Top Forums Shell Programming and Scripting How to print the entire line if the mentioned match is found?
# 8  
Old 03-02-2015
You said that you wanted to print lines where the elapsed time is greater than 24 hours. The code:
Code:
awk '{split($4,a,":"); if (a[1]>=24) print}' file

will also print lines where the time is exactly 24:00:00 (which is NOT greater than 24 hours). If you really want to only print lines where the elapsed time is greater than 24 hours you could try either of the following:
Code:
awk '{split($4,a,":"); if (a[1]>=24 || (a[1]==24 && (a[2] + a[3]))) print}' file

or:
Code:
awk '(h=($4+0))>24 || (h==24 && substr($4,length($4)-4)!="00:00")' file

If you want to try any of these on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk.

If file contains:
Code:
xyz abc status 23:00:00 idle
abc def status 24:00:00 idle
def  gji status  27:00:02 idle
fgh  gty status 00:00:00 idle
dwc u&lf status 24:00:10 idle
dwc u&lf status 100:01:23 idle

the last two scripts above produce the output:
Code:
def  gji status  27:00:02 idle
dwc u&lf status 24:00:10 idle
dwc u&lf status 100:01:23 idle

while the 1st script above produces the output:
Code:
abc def status 24:00:00 idle
def  gji status  27:00:02 idle
dwc u&lf status 24:00:10 idle
dwc u&lf status 100:01:23 idle

This User Gave Thanks to Don Cragun For This Post:
rahul2662
# 9  
Old 03-04-2015
Thanks a lot Sir.
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