Shell Script to get exception from logs for last one hour

问题

I am developing script which will grep logs of last one hour and check any exception and send email for solaris platform.

I did following steps

grep -n -h date +'%Y-%m-%d %H:%M' test.logs

above command gives me line number and then i do following

tail +6183313 test.log | grep 'exception'

sample logs

2014-02-17 10:15:02,625 | WARN  | m://mEndpoint | oSccMod | 262 - com.sm.sp-client - 0.0.0.R2D03-SNAPSHOT | 1201 or 101 is returned as exception code from SP, but it is ignored
2014-02-17 10:15:02,625 | WARN  | m://mEndpoint | oSccMod | 262 - com.sm.sp-client - 0.0.0.R2D03-SNAPSHOT | SP error ignored and mock success returned
2014-02-17 10:15:02,626 | INFO  | 354466740-102951 | ServiceFulfill | 183 - org.apache.cxf | Outbound Message

Please suggest any better alternative to perform above task.

回答1:

With GNU date, one can use:

 grep "^$(date -d -1hour +'%Y-%m-%d %H')" test.logs | grep 'exception'| mail -s "exceptions in last hour of test.logs" ImranRazaKhan

The first step above is to select all log entries from the last hour. This is done with grep by looking for all lines beginning with the year-month-day and hour that matches one hour ago:

grep "^$(date -d -1hour +'%Y-%m-%d %H')" test.logs

The next step in the pipeline is to select from those lines the ones that have exceptions:

grep 'exception'

The last step in the pipeline is to send out the mail:

mail -s "exceptions in last hour of test.logs" ImranRazaKhan

The above sends mail to ImranRazaKhan (or whatever email address you chose) with the subject line of "exceptions in last hour of test.logs".

The convenience of having the -d option to date should not be underestimated. It might seem simple to subtract 1 from the current hour but, if the current hour is 12am, then we need to adjust both the day and the hour. If the hour was 12am on the first of the month, we would also have to change the month. And likewise for year. And, of course, February requires special consideration during leap years.

Adapting the above to Solaris:

Consider three cases:

1. Under Solaris 11 or better, the GNU date utility is available at /usr/gnu/bin/date. Thus, we need simply to specify a path for date:

    grep "^$(/usr/gnu/bin/date -d -1hour +'%Y-%m-%d %H')" test.logs | grep 'exception'| mail -s "exceptions in last hour of test.logs" ImranRazaKhan
   

2. Under Solaris 10 or earlier, one can download & install GNU date

3. If GNU date is still not available, we need to find another way to find the date and time for one hour ago. The simplest workaround is likely to select a timezone that is one hour behind your timezone. If that timezone was, say, Hong Kong, then use:

    grep "^$(TZ=HongKong date +'%Y-%m-%d %H')" test.logs | grep 'exception'| mail -s "exceptions in last hour of test.logs" ImranRazaKhan
   

回答2:

You can do like this:

dt="$(date -d '1 hour ago' "+%m/%d/%Y %H:%M:%S")"

awk -v dt="$dt" '$0 ~ dt && /exceltion/' test.logs

回答3:

Scanning through millions lines of log sounds terribly inefficient. I would suggest changing log4j (what it looks like) configuration of your application to cut a new log file every hour. This way, tailing the most recent file becomes a breeze.

来源：https://stackoverflow.com/questions/21821968/shell-script-to-get-exception-from-logs-for-last-one-hour