Re: find/replace in place

[Almut Behrens <almut_behrens@gmx.net>]

On Mon, Dec 19, 2005 at 03:38:23PM -0500, Tony Heal wrote:
> I have a database name I want to replace inside of an xml file. I can do
> this in several step using sed, but I would like to do it in a single step
> using perl. This is what I have in sed and what does not work in perl.
>  
> SED
> #!/bin/bash
> echo -n "Please enter the name of the new database: "
> read syelledb
> dbchange=`cat /tmp/data-sources.xml|grep database|cut -d ">" -f2|cut -d "<"
> -f1`
> sed s/$dbchange/$syelledb/ /tmp/data-sources.xml > /tmp/data-sources.xml.tmp
> mv /tmp/data-sources.xml /tmp/data-sources.xml.orig
> mv /tmp/data-sources.xml.tmp /tmp/data-sources.xml
> 
> PERL (single line in bash script)
> #!/bin/bash
> echo -n "Please enter the name of the new database: "
> read syelledb
> 
> dbchange=`cat /tmp/data-sources.xml|grep database|cut -d ">" -f2|cut -d "<"
> -f1`
> /usr/bin/perl -pi -w -e 's/$dbchange/$syelledb/'

use double quotes:

/usr/bin/perl -pi -w -e "s/$dbchange/$syelledb/"

so the shell will interpolate the contents of the variables into the
s/// expression (with single quotes you'd replace the literal string
'$dbchange' by '$syelledb'...)

Cheers,
Almut