Date: Mon, 12 Sep 2005 20:37:22 +0200 From: Frank Mueller - emendis GmbH <Frank.Mueller@emendis.de> To: Paul Schmehl <pauls@utdallas.edu> Cc: freebsd-questions@freebsd.org Subject: Re: Shell scripting question Message-ID: <4325CAE2.3020906@emendis.de> In-Reply-To: <01A14A33D6971135F96609A2@utd59514.utdallas.edu> References: <01A14A33D6971135F96609A2@utd59514.utdallas.edu>
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To get the date in the right format you could simply use
date +%H
Greetz,
Ice
Paul Schmehl schrieb:
> I've written a script to check apache to make sure it's running *and*
> logging. One of the variables I create is named DATEHOUR, and it's
> created by parsing the output of date in such a way that all I get is
> the hour (using awk and cut.) I'm comparing DATEHOUR to LOGHOUR, which
> represents the the most recent hour that the log was written to
>
> I've run in to a small problem I'm not sure how to solve. When the hour
> is less than 10, the script generates an arithmetic expression error.
>
> Here's part of the script so you can visualize what I'm trying to do:
>
> PROG=/usr/local/sbin/apachectl
> LOG=/var/log/httpd-access.log
> PID=`/bin/ps -auxw | grep http | grep root | grep -v grep | awk '{print
> $2}'`
> DATE=`date | awk '{print $4}' | cut -d':' -f1,2`
> LOGDATE=`ls -lsa ${LOG} | awk '{print $9}'`
> DATEHOUR=`echo ${DATE} | cut -d':' -f1`
> LOGHOUR=`echo ${LOGDATE} | cut -d':' -f1`
> DATEMIN=`echo ${DATE} | cut -d':' -f2`
> LOGMIN=`echo ${LOGDATE} | cut -d':' -f2`
> LOGGING=1
>
> if [ $((DATEMIN)) -gt $((LOGMIN+15)) ]; then
> LOGGING=0
> elif [ $((DATEHOUR)) -ne $((LOGHOUR)) ] && [ $((DATEMIN+60)) -gt
> $((LOGMIN+15)) ]; then
> LOGGING=0
> fi
>
> When DATEHOUR is less than 10 (01-09), the script generate an arithmetic
> expression, variable substition error. I'm pretty certain it's because
> of the leading zero, so I'm trying to figure out how to strip that out.
> I thought that parameter expansion would do it, but I get some odd (to
> me) results.
>
> Assume DATE is 09.
>
> echo ${DATE:0,0}
> 09
> echo ${DATE:0,1}
> 9
> echo ${DATE:1,1}
> 9
>
> I would have thought that 0,0 would return only the first character and
> 1,1 would return only the second, but that is obviously not the case.
>
> How can I strip the leading character from the string so that I can test
> to see if it's zero?
>
> Paul Schmehl (pauls@utdallas.edu)
> Adjunct Information Security Officer
> University of Texas at Dallas
> AVIEN Founding Member
> http://www.utdallas.edu/ir/security/
> _______________________________________________
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--
Frank Mueller
eMail: Frank.Mueller@emendis.de
Mobil: +49.177.6858655
Fax: +49.951.3039342
emendis GmbH
Hofmannstr. 89, 91052 Erlangen, Germany
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