Shell Programming and Scripting

Hi Friends,

I have the following logfile. Currently time in india is 07/31/2014 12:33:34 and i have the following content in logfile. I want to display only those entries which contain string 'Exception' within last 3 hours. In this case, it would be the last line only

Code:

[7/30/14 4:59:30 Exception Found
[7/30/14 5:18:55 Result is OK
[7/30/14 8:45:22 Exception found
[7/30/14 9:22:26 Exception found
[7/31/14 11:55:66 Exception found

I can get the timestamp in logfile as

Code:

awk '/Exception/ {print $1,$2}' trial.txt | sed 's/^.//' which gives below results

7/30/14 4:59:30
7/30/14 8:45:22
7/30/14 9:22:26
7/31/14 11:55:66

And i can get current timestamp as

Code:

date +"%m/%d/%Y %T"
07/31/2014 12:33:34

So, how can i compare current timestamp with timestamp in log file and display 'Exception' entries until last 3 hours.

---------- Post updated at 02:40 AM ---------- Previous update was at 02:07 AM ----------

I am trying this command. But it is throwing exception saying syntax error. What am i doing wrong?

Code:

awk '/Exception/ if ($1,$2 > $(date +"%m/%d/%Y %T" -d  "3 hour ago")) print $1,$2 ' trial.txt  | sed 's/^.//'