[SOLVED] How to remove quotes with grep and cut in one command

Issue

This is what I do currently:

pass1=$(grep DB_PASSWORD /home/user/public_html/wp-config.php | cut -d ' ' -f3)
# it echoes 'password' with start and end single quotes
pass=${pass1:1:-1}
# it remove single quotes from start and end

But I want to combine these two commands in a single go.

I tried some wrong commands which showed me the help of grep.

Solution

You could use awk:

pass=$(awk -F"'" '/DB_PASSWORD/{print $4}' /home/user/public_html/wp-config.php)

I simply defined ' to be the field separator, which makes the password the 4th field.

Answered By - tink

Answer Checked By - Timothy Miller (WPSolving Admin)