shell script to recursively search and find/replace text in target files

Hello All,

This is probaably a 1 minute solution for many script maseters, but for me is quite muddy.

Task:
1. Recursively search  all child directories for file WP-CONFIG.PHP
2. Find all "Lines" that begin with define('DB_PASSWORD', 'xxxxxx');   -- note 'xxxxxx' can be any length text.
3. Replace the 'xxxxxxx' text with new text that I would supply as paramter argument upon running the script.
4. Produce a count of the number of times the script found and editied a WP-CONFIG.PHP file.

Thats it.
Louis CapeceAsked:
Who is Participating?
 
florjanCommented:
find . -iname "WP-CONFIG.PHP" -print | xargs sed -i "s/('DB_PASSWORD', '.*')/('DB_PASSWORD', 'newpassword')/g"
grep -lr "newpas" .

Open in new window

0
 
de2ZotjesCommented:
#!/usr/bin/env bash

REPLACE=$1
n=0
for entry in $(find . -name WP-CONFIG.PHP -a -type f); do
    n=$(($n+1))
    sed -e "s/^define\('DB_PASSWORD','[^']*'/define('DB_PASSWORD','$REPLACE'/" -i $entry
done

echo $n

Open in new window


Line 3 takes the first command line argument and puts it in the REPLACE var
Line 5 uses the find utility to locate every file named WP-CONFIG.PHP under the current dir
Line 7 replaces every occurence of ....DB_PASSWORD with the text in REPLACE var
 

0
 
florjanCommented:
find . -iname "WP-CONFIG.PHP" -print | xargs sed -i 's/oldpass/newpass/g'
grep -lr "newpas" .

Open in new window

Make sure to escape(backslash\ any special characters) and to tell me if there is any other line same as password so I can adapt the script/
0
Ultimate Tool Kit for Technology Solution Provider

Broken down into practical pointers and step-by-step instructions, the IT Service Excellence Tool Kit delivers expert advice for technology solution providers. Get your free copy now.

 
farzanjCommented:
1.  ls -R WP-CONFIG.PHP
or  find . -type f -name WP-CONFIG.PHP
2. grep -R '^define(\'DB_PASSWORD\''

3.  sed "s/xxxxx/yyyyy/g" $1
4.  #Before modifying the script
     Precount=$(grep -c "XXXX" WP-CONFIG.PHP)
     #Edit using above
     Postcount=$(grep -c "XXXX" WP-CONFIG.PHP)
     Edited=$(( Precount - Postcount ))
0
 
Louis CapeceAuthor Commented:
Could someone please confirm this, I suspect there may be confusion...
the existing password is NOT known, it could be anything that is beween the single quotes.

In a nutshell, what I an trying to do is alter all the WP_CONFIG.PHP "Define Password" statement to a single value....the existing passwords are all different.
0
 
tel2Commented:
Hi louiscapece,

de2Zotjes's solution seems to cater for any existing password.
But I don't think it caters for the space you have between the comma and the 'xxxxxx' (old password).  To make it flexible in this regard (i.e. to handle 0 or more spaces there), maybe change line 7 to something like this:
    sed -e "s/^define\('DB_PASSWORD', *'[^']*'/define('DB_PASSWORD', '$REPLACE'/" -i $entry
0
 
florjanCommented:
I have tested script from my last post on gentoo on my test wp install in bash and it worked.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.