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Top Forums Shell Programming and Scripting Identifying same character and ouput the sum Post 302108306 by anbu23 on Monday 26th of February 2007 04:32:51 AM
Old 02-26-2007
Code:
nawk '{a[$1]+=$3}END{for(x in a){ NUM=a[x]; print NUM}}' Input > Output

 

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LOGFILE(1)							       mrtg								LOGFILE(1)

NAME
logfile - description of the mrtg-2 logfile format SYNOPSIS
This document provides a description of the contents of the mrtg-2 logfile. OVERVIEW
The logfile consists of two main sections. A very short one at the beginning: The first Line It stores the traffic counters from the most recent run of mrtg The rest of the File Stores past traffic rate averates and maxima at increassing intervals The first number on each line is a unix time stamp. It represents the number of seconds since 1970. DETAILS
The first Line The first line has 3 numbers which are: A (1st column) A timestamp of when MRTG last ran for this interface. The timestamp is the number of non-skip seconds passed since the standard UNIX "epoch" of midnight on 1st of January 1970 GMT. B (2nd column) The "incoming bytes counter" value. C (3rd column) The "outgoing bytes counter" value. The rest of the File The second and remaining lines of the file 5 numbers which are: A (1st column) The Unix timestamp for the point in time the data on this line is relevant. Note that the interval between timestamps increases as you prograss through the file. At first it is 5 minutes and at the end it is one day between two lines. This timestamp may be converted in EXCEL by using the following formula: =(x+y)/86400+DATE(1970,1,1) you can also ask perl to help by typing perl -e 'print scalar localtime(x)," "' x is the unix timestamp and y is the offset in seconds from UTC. (Perl knows y). B (2nd column) The average incoming transfer rate in bytes per second. This is valid for the time between the A value of the current line and the A value of the previous line. C (3rd column) The average outgoing transfer rate in bytes per second since the previous measurement. D (4th column) The maximum incoming transfer rate in bytes per second for the current interval. This is calculated from all the updates which have occured in the current interval. If the current interval is 1 hour, and updates have occured every 5 minutes, it will be the biggest 5 minute transferrate seen during the hour. E (5th column) The maximum outgoing transfer rate in bytes per second for the current interval. AUTHOR
Butch Kemper <kemper@bihs.net> and Tobias Oetiker <oetiker@ee.ethz.ch> 3rd Berkeley Distribution 2.9.17 LOGFILE(1)
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