Check_Log - SOLUTION

[James Turnbull]

Mike Koponick wrote:
> All,
>
> Thanks for all the help/information regarding this subject.
>
> I have found a solution that works in this case.
>
> The problem is that the "%" is a directive for printf. Since the "%" was
> in the string, printf thinks it should be a directive.
>
> Since I had no need for the "%" in the output string, I removed it.
>
> I changed the check_log script with the following:
>
> OLD:
> $ECHO "($count) $lastentry" 
>
> NEW:
> $ECHO "($count) $lastentry" | /bin/sed 's/%//'
>
> I'm sure there is probably a more efficient way of doing this, but it
> was a quick fix for me.
>   
I guess the easier way might be to escape the % in printf.

Regards

James Turnbull

-- 
James Turnbull <james at lovedthanlost.net>
---
Author of Pro Nagios 2.0
(http://www.amazon.com/gp/product/1590596099/)

Hardening Linux
(http://www.amazon.com/gp/product/1590594444/)
---
PGP Key (http://pgp.mit.edu:11371/pks/lookup?op=get&search=0x0C42DF40)



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