Re: quick script question

From: Todd Zullinger <tmz@xxxxxxxxx>
Date: Wed, 31 May 2006 01:22:43 -0400

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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

Get rid of the echo, it's not needed.

KRWEBLOG3="ex`date -d yesterday +%y%m%d.log`

If you ever do have to run a few commands to create a variable, you
can use bash's command substitution to nest commands. Instead of
using backquotes (`command`), you use $(command).

KRWEBLOG3=$(echo ex$(date -d yesterday +%y%m%d.log))

See the section on Expansion in the bash man page for more details.

- --
Todd OpenPGP -> KeyID: 0xD654075A | URL: www.pobox.com/~tmz/pgp
======================================================================
A free society is one where it is safe to be unpopular.
-- Adlai Stevenson

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References:
- quick script question
  - From: M E Fieu

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