[mrtg] Correct way to do Target math?

Koelstra, J. (Jan) JKoelstra at MINSZW.NL
Wed Jul 25 11:32:29 CEST 2007


In case you need a zero in the second half of the & statement you can use PseudoZero.
So your targetline woul look like
Target[name]: <oid>&<oid>:public@<ip> - <oid>&PseudoZero:public@<ip>
PseudoOne is also recognized by MRTG (usefull when you want to multiply values...)

One note about using named oid's in mathemetical expressions: In early versions of MRTG the only way to get the right result was by using the full numerical oid's. I don't know if this has changed after MRTG 2.9.x



-----Original Message-----
From: mrtg-bounces at lists.oetiker.ch [mailto:mrtg-bounces at lists.oetiker.ch] On Behalf Of Ray Van Dolson
Sent: Wednesday, July 25, 2007 2:43 AM
To: mrtg at lists.oetiker.ch
Subject: [mrtg] Correct way to do Target math?

I'm trying to generate a graph using SNMP of actual memory used on a Linux system running Net-SNMP.  In theory:

  memTotalReal.0 - memBuffer.0 - memCached.0 - memAvailReal.0

Is the real memory available.

I would like to graph this and also memTotalReal.0.  I am having a hard time figuring out the right way to do this however... I am currently

  Target[memory_used]: memTotalReal.0&memTotalReal.0:public at localhost -
    memBuffer.0&memIndex.0:public at localhost -
    memCached.0&memIndex.0:public at localhost -
    memAvailReal.0&memIndex.0:public at localhost

(Pardon the word wrapping).  memIndex.0 evaluates to 0 so that my second variable doesn't continue to get subtracted.

Is there a better way to do this?  I tried doing something like:

  memTotalReal.0&memTotalReal.0:<stuff> - memBuffer.0&0:<stuff> ....

But this doesn't evaluate.  Other than the first group, I just want the second half of the & statement to be 0.


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