regular expression for find and replace

Question

I've got strings like:

('Michael Herold','Michael Herold'),

but I need to remove the last parts so I end up with:

('Michael Herold'),

I'm still new to Regular Expressions so they confuse me. I'm using Notepad++.

Answer 1

find: \('([^']*)','\1'\) 
Replace: ('\1')

Answer 2

So the actual function you use will depend on the language. Notepad++ is a text editor, not a language.

The regular expression that you will want will be ",'Michael Herold'" and you'll replace any matches with "", the empty string.

So in PHP for example, you'll have

    $source = "('Michael Herold','Michael Herold')";
    $pattern = "/(,'Michael Herold')+/";
    $newString = $preg_replace($pattern, $source, ""); 

Do the equivalent in whatever language you use.

Answer 3

I'm not sure what flavor of regular expressions Notepad++ uses, but try replacing this expression:

\('([^']*)','\1'\)

with this one:

('$1')

The \1 matches whatever was found in the first set of single quotes ( Michael Herold in your example), and $1 is replaced with that same string. (Try \1 if $1 doesn't work in Notepad++.)

See it in action here .