Replace unknown string using sed in bash script

Question

Let's say I have a wp-config.php like so:

define('DB_NAME', 'db_name'); define('DB_USER', 'user'); define('DB_PASSWORD', 'pass'); 

In my bash script I have a construct similar to:

#!/usr/bin/bash # Database config echo -e "\ndatabase name: \c" read DB_NAME sed -i -e "s/define('DB_NAME', );/define('DB_NAME', $DB_NAME);/g" "C:/Apache24/htdocs/test.txt" 

It doesn't work however. How could I edit (using sed) the variables db_name, user & pass in test.txt given that these variables would typically be unknown in advance?

Answer 1

eval "`echo 'NL=qsq' | tr 'qs' '\047\012'`"; # newline echo DB_NAME; read DB_NAME; db_name_esc=$DB_NAME db_name_esc=${db_name_esc//\\/\\\\\\\\} # escape backslash db_name_esc=${db_name_esc//\'/\\\\\'} # escape single quote db_name_esc=${db_name_esc//\"/\\\"} # escape double quote db_name_esc=${db_name_esc//&/\\&} # escape & special to sed db_name_esc=${db_name_esc//\//\\\/} # escape / special to s/// db_name_esc=${db_name_esc//${NL}/\\${NL}} # esc literal newline for sed # and then plug in your variables properly escaped for both PHP+sed sed -e " s/\\(define('DB_NAME',\\).*/\\1 '$db_name_esc');/ " yourfile