Vim Regular Expression demonstrating use of look-ahead zero-width assertion to replace all the commas on a line that aren't a part of a quoted string

vim_re.txt

This uses positive look-ahead to check if there's a string ahead to substitute commas present outside quoted strings.

    s/\v,(([^"]*"[^"]*")*[^"]*$)@=/|/g

Explanation:

* "[^"]*"

Match a double-quoted string

* [^"]*"[^"]*"

Match some characters along with the previous string

* ([^"]*"[^"]*")*

Check for the previous match 0 or more times

* ([^"]*"[^"]*")*[^"]*$

From a certain position ( a comma in this case ) check that we get some characters which aren't part of a string, a string, followed by some more non-string characters till the end of the line.

* (([^"]*"[^"]*")*[^"]*$)@=

Converts the previous match to a [look-ahead zero-width assertion](http://perldoc.perl.org/perlretut.html#Looking-ahead-and-looking-behind) and its (Vim counterpart](http://vimdoc.sourceforge.net/htmldoc/pattern.html#/zero-width)


Note:

* Deciding if a comma is inside or outside a string is done by checking the number of quotes ahead. If there are even number of quotes then we say that we're not inside a string. 
* This does not handle cases like "abcd\"efghi" where the quote character is also a part of the string