how can I do preg_replace?

Question

I'm trying to get data from a site but there are some problems. I want to change the expression begining with <div class="phraserec" and ending with </div></div> how can I do it?

$source = "<div class="gwblock" id="1936036"><div class="phraserec"bla bla...</div></div>"
$output = preg_replace('/'. preg_quote('<div class="phraserec" '.'(.*?)'.'</div></div>','/') .'/', '</div>' , $ll);

I wanted output;

<div class="gwblock" id="1936036"><div class="phraserec"bla bla...</div>

[Parsing HTML with regex is a hard job](https://stackoverflow.com/a/4234491/372239) HTML and regex are not good friends. Use a parser, it is simpler, faster and much more maintainable. See: http://php.net/manual/en/class.domdocument.php — Toto, Oct 18 '19 at 18:47
I'm not sure I understand why you're trying to do. Just get rid of one of the two `` at the end? — Barmar, Oct 18 '19 at 18:56
@Toto It seems like he's trying to either parse or create invalid HTML, so DOMDocument won't work. — Barmar, Oct 18 '19 at 18:57
@Barmar yes if "phraserec" into "gwblock" i wanted only on "/div" sorry for my bad english. — Semih Gürler, Oct 18 '19 at 19:07

score 0 · Accepted Answer · answered Oct 18 '19 at 19:14

There's no reason to call preg_quote(), and it's preventing (.*?) from being treated as a pattern. And instead of putting the capture group just around .*?, you can put it around everything before the second </div>.

$output = preg_replace('#(<div class="phraserec".*?</div>)</div>#', '$1', $source);

preg_quote() is needed when you want to turn dynamic input into a literal value in a regular expression.

how can I do preg_replace?

1 Answers1