Question

I’m trying to do some manipulation with WordPress and I’m trying to write a script for it…

# cat /usr/local/uftwf/_wr.sh 
#!/bin/sh
# $Id$
#

table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`
echo $table_prefix
# 

Yet I’m getting following output

# /usr/local/uftwf/_wr.sh 
 ABSPATH ABSPATH wp-settings.php_KEY LOGGED_IN_KEY NONCE_KEY AUTH_SALT SECURE_AUTH_SALT LOGGED_IN_SALT NONCE_SALT wp_0zw2h5_ de_DE WPLANG WP_DEBUG s all, stop editing! Happy blogging. */
# 

Running from command line, I get the correct output that I’m looking for:

# grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'
wp_0zw2h5_
# 

What is going wrong in the script?

Answer 1

The problem is the grep command:

table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`

It either needs three backslashes – not one – or you need to use single quotes (which is much simpler):

table_prefix=$(grep '^$table_prefix' wp-config.php | awk -F"'" '{print $2}')

It’s also worth using the $( ... ) notation in general.

The trouble is that the backquotes removes the backslash, so the shell variable is evaluated, and what’s passed to grep is, most likely, just ^, and each line starts with a beginning of line.