[rrd-users] Re: simple graphing

Mike Hunter mhunter at ack.Berkeley.EDU
Tue Jan 13 20:29:18 MET 2004


On Jan 13, "Dave Stern - Former Rocket Scientist" wrote:

> I'm exceedingly new to RRDtool and am trying to find the simplest way to
> graph data from an existing file of the format
> MMM DD HH:MIN  datapoint
>    eg
> Jan  1 12:00   123
> Jan  1 12:15    80
> Jan  1 12:30   257
> 
> I initialized the database using
> 
> rrdtool create test.rrd --start N DS:in:GAUGE:900:U:U RRA:AVERAGE:0.5:1:432
> 
> To insert the data,  I would like to ideally do something like
>     rrdtool update test.rrd `cat /tmp/data.2`
> but that cleayly wont work even if I massage the data to look like
> 1200:123
> 1215:80
> 1230:257
> 
> For that matter,making a script that says
> rrdtool update test.rrd U:123
> rrdtool update test.rrd U:80
> rrdtool update test.rrd U:257
>  wont do it
> 
> Clearly the problem is my time format. Is there some way to tell rrdtool to
> just take the timestamp as is without calculating seconds from an epoch?

rrdtool is extremely fond of epoch seconds.  Here's a fragment of perl
code that uses `date` to convert a filename such as this:

ft-v05.2004-01-13.110001-0800

to epoch seconds.

The -0800 is my GMT offset (have a look at date -u if you need to generate 
that).

        $time =~ s/^.*ft-v05\.(.*)/$1/;
        my ($y, $m, $d, $mess, $g) = split /[-.]/, $time;
        my $H = substr ($mess, 0, 2);
        my $M = substr ($mess, 2, 2);
        my $S = substr ($mess, 4, 2);
        $g =~ s/^0//;
        $g =~ s/0*$//;
        my $epoch_time_cmd = "date -v-$g"."H -v$y"."y -v$m"."m -v$d"."d -v$H".
                             "H -v$M"."M -v$S"."S \"+%s\"";
        #print                "date -v-$g"."H -v$y"."y -v$m"."m -v$d"."d -v$H".
                             #"H -v$M"."M -v$S"."S \"+%s\"";
        my $epoch_time = `$epoch_time_cmd`;
        chomp $epoch_time;

Mike

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