Regex: find last occurance of word in string

Question

I need to find last occurance of a word in a string (and replace it). So in following sentence I would be looking for the second "chocolate".

I love milk chocolate but I hate white chocolate.

How can that be achieved with regular expression? Could you please give me some explanation? Thanks.

Answer 1

If you want to match the second occurrence of any distinct word, you may be able to use a backreference, depending on the language and regex implementation you're in.

For example, in sed, you might do the following:

sed 's/\(.*\([[:<:]][[:alpha:]]*[[:>:]]\).*\)\(\2\)\(.*\)/\1russians\4/'

Breaking this down for easier reading, it looks like this:

• s/ - substitute in sed
• \(.*\([[:<:]][[:alpha:]]*[[:>:]]\).*\)\(\2\)\(.*\) - the search RE. Not really so complex....
  • [[:<:]] and [[:>:]] are portable word boundaries,
  • [[:alpha:]] is the class of alphabetical characters (words)
  • \( and \) surround atoms for use in backreferences, in BRE (this is sed, remember)
• \1russians\4 - replacement string consists of the first (outer) parenthesized backreference from the RE, followed by the replacement word, followed by the trailing characters.

For example:

$ t="I love milk chocolate but I hate white chocolate."
$ sed 's/\(.*\([[:<:]][[:alpha:]]*[[:>:]]\).*\)\(\2\)\(.*\)/\1russians\4/' <<<"$t"
I love milk chocolate but I hate white russians.
$ t="In a few years, your twenty may be worth twenty bucks."
$ sed 's/\(.*\([[:<:]][[:alpha:]]*[[:>:]]\).*\)\(\2\)\(.*\)/\1fifty\4/' <<<"$t"
In a few years, your twenty may be worth fifty bucks.
$

Answer 2

If you want to use a regex you could use something like this:

(.*)chocolate

And the replacement string would be:

$1banana
  ^-- whatever you want

working demo

Update: as Lucas pointed out in his comment, you can improve the regex by using:

(.*)\bchocolate\b

This allows you to avoid false positives like chocolateeejojo

Answer 3

PCRE would look like this:

/^(.*)chocolate/$1replace/sm