Regex Find and Replace: Deleting everything except words starting with "#"?

Question

I'm a novice at regex and can't find a way to do this easily. I'd like to delete every word that isn't starting with # and put a comma between them, so for example if I have:

Cookie Recipe for n00bs
#cookie #recipe #chocolate To do this you have to etc...
Bla bla bla mumbo jumbo

I'd like to get as a result:

cookie, recipe, chocolate

If you could help me it'd be great, thanks and have a good day!

Answer 1

You missed to tell which programming language you are using. Here comes an example in PHP, which uses perl compatible regular expressions:

$text = <<<EOF
Cookie Recipe for n00bs
#cookie #recipe #chocolate To do this you have to etc...
Bla bla bla mumbo jumbo
EOF;

$pattern = '/((?<=#)\w+)/';
preg_match_all($pattern, $text, $matches);

echo implode(', ', $matches[0]);

I'm using a so called positive lookbehind assertion (?<=#) which ensures that only words are matched which are preceded by a # but, and this is important, it does not include the # itself into the match. After the lookbehind expression, I'm matching as many word characters \w as possible.

After that implode() is used to concatenate the resulting matches with a ,. A regex can't be used for that part of the job.

You can see how this regex works at Regex101.com

Answer 2

Try this:

$re = "/#\\w+/";
$str = "#cookie #recipe #chocolate To do this you have to etc...";
$str .= "#cookie #recipe #chocolate To do this you have to etc...";

preg_match_all($re, $str, $matches);
$r=@implode(", ",@$matches[0]);  // for adding comma(,) and space( ).
var_dump( $matches,$r);

output:

array (size=1)
  0 => 
    array (size=6)
      0 => string '#cookie' (length=7)
      1 => string '#recipe' (length=7)
      2 => string '#chocolate' (length=10)
      3 => string '#cookie' (length=7)
      4 => string '#recipe' (length=7)
      5 => string '#chocolate' (length=10)

string '#cookie, #recipe, #chocolate, #cookie, #recipe, #chocolate' (length=58)

live demo