Re: quick script question

M E Fieu wrote:
Hi.. Need to ask you a simple script question, I used the date command to get the yesterday web
log file using the following command echo ex`date -d yesterday +%y%m%d.log` in Linux. I want to
create a simple script to cat /home/IT/ex060530.log > kr.log using the following script but it is
not working. May I know how do I put the output of echo ex`date -d yesterday +%y%m%d.log into cat
/home/IT/xxxx > kr.log ?

[root@w2 krweblog3]# echo ex`date -d yesterday +%y%m%d.log`
ex060530.log

[root@w2 krweblog3]# vi krweblog.sh

#!/bin/sh
#
KRWEBLOG3= echo ex`date -d yesterday +%y%m%d.log`
cat /home/IT/$KRWEBLOG3 > kr.log

For thing like this, I like to use something like:

KRWEBLOG=$(date -d yesterday +ex%y%m%d.log)
cat /home/IT/$KRWEBLOG3 > kr.log

KRWEBLOG will have the value of ex060530.log if I ran it now on my
system. Enclosing a command inside $() returns the output of the
command. You could also use:
KRWEBLOG=`date -d yesterday +ex%y%m%d.log`
but I find the $() format easyer to read.

Mikkel
--

Do not meddle in the affairs of dragons,
for thou art crunchy and taste good with Ketchup!

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