Using preg_replace to replace all occurrences in p

Question

Regex is absolutely my weak point and this one has me completely stumped. I am building a fairly basic search functionality and I need to be able to alter my user input based on the following pattern:

Subject:

%22first set%22 %22second set%22-drupal -wordpress

Desired output:

+"first set" +"second set" -drupal -wordpress

I wish I could be more help as I normally like to at least post the solution I have so far, but on this one I'm at a loss.

Any help is appreciated. Thank you.

Answer 1

Seems your data is URL encoded. If you apply urldecode, you will get

"first set" "second set" -drupal -wordpress

(I assume you have a space before -drupal).

Now you have to add +. Again, I assume you have to add those before all words that don't have a - and that are not inside quotes:

$str = '"first set" "second set" -drupal -wordpress foo';
echo preg_replace('#( |^)(?!(?:\w+"|-| ))#','\1+', $str));
// prints +"first set" +"second set" -drupal -wordpress +foo

Update: If you cannot use urldecode, you could just use str_replace to replace %22 with ".

Answer 2

preg_replace('/%22((?:[^%]|%[^2]|%2[^2])*)%22/', '+"$1"', $str);

Explanation: The $1 is a backreference, which references the first ()-section in the regular expression, in this case, ((?:[^%]|%[^2]|%2[^2])*). And the [^%] and the alternations (...|...|...) after it prevents %22 in between from being matched due to greediness. See http://en.wikipedia.org/wiki/Regular_expression#Lazy_quantification.

I found that technique in a JavaCC example of matching block comments (/* */), and I can't find any other webpages explaining it, so here is a cleaner example: To match a block of text between 12345 12345........12345 with no 12345 in between: /12345([^1]|1[^2]|12[^3]|123[^4]|1234[^5])*12345/

Answer 3

Is this what you're looking for?

<?php
  $input = "%22first set%22 %22second set%22-drupal -wordpress";
  $res = preg_replace( "/\%22(.+?)\%22/","+\"(\\1)\" ", $input);
  print $res;
?>