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Re: quick script question

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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.

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