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Perl or sed command ?

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# 1  
Old 12-20-2013
Perl or sed command ?
Hi Guys

Am working on a bash script but got stuck, in this line:

Code:
32 $configValues['CONFIG_DB_PASS'] = '';

What would be the best command to enter the password between the " Perl or sed ?

Been trying with Perl using this command:

Code:
perl -pi -e 's/''/Seattle#1669!/g' /var/www/html/daloradius/library/daloradius.conf.php

Which of course is not working.

Really appreciate any help.
# 2  
Old 12-20-2013
Like so:
Code:
 sed "s/''/'password'/" file
32 $configValues['CONFIG_DB_PASS'] = 'password';

This User Gave Thanks to RudiC For This Post:
Tox
# 3  
Old 12-20-2013
Hi RudiC thank you for the help, when running command this is what I get:

Code:
[root@p1 db]# sed "s/''/'Seattle#1669!'/" /var/www/html/daloradius/library/daloradius.conf.php
-bash: !'/": event not found

Any ideas ?
# 4  
Old 12-20-2013
Hi, you must protect ! , so as \!:
Code:
sed "s/''/'Seattle#1669\!'/" /var/www/html/daloradius/library/daloradius.conf.php

Regards.
This User Gave Thanks to disedorgue For This Post:
Tox
# 5  
Old 12-20-2013
Hi disedorgue

That did the trick, thank you for the help.

@ RudiC thank you for the help.
# 6  
Old 12-21-2013
For modifying the source file add the -i option.
Also, ensuring the line has got PASS seems safer.
Code:
sed -i "/PASS/ s/''/'Seattle#1669\!'/" /var/www/html/daloradius/library/daloradius.conf.php

Code:
sed -i -pe "s/''/'Seattle#1669\!'/ if /PASS/;" /var/www/html/daloradius/library/daloradius.conf.php

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