How do I filter this line "XX<i>\r</i><i>\n</i>" to XX?

Question

Question is in the title. I tried strip_tags, but that shows up blank and I tried using preg_replace, but I don't quite understand the syntax apparently..

Tried this

    $test = "Home<i>\r</i><i>\n</i>";       
$patterns = array();
$patterns[0] = '<i>';
$test = preg_replace($patterns, '', $test);

but echo $test is still the same!

Answer 1

strip_tags would return XX\r\n, so preg_replace would probably be what you're looking for, assuming it can actually do what you want to do. I don't know myself whether or not it will strip the tags like you want, but if you just don't understand the syntax, read up here.

Hope this helps.

Answer 2

if you want to just get the first part of the string up to the first tag, how about something like:

$test = "Home<i>\r</i><i>\n</i>";
preg_match('/^[^<]+/', $test, $match);
echo $match[0];

if you want to remove all tags, strip_tags should do it, or you can use preg_replace like:

$test = "Home<i>\r</i><i>\n</i>";
$test = preg_replace('/<[^>]+?>/', '', $test);
echo $test;

Answer 3

The problem with your syntax on preg_replace() is that you did not define the pattern as an accurate regular expression. Your method would work if you used str_replace(), but for preg_replace(), the correct pattern for identifying <i> is '/<i>/'. Basically, you need to enclose it in / chars to define it as a regex.

Jonathan's expression is much better than a pattern by pattern implementation, if you are looking to remove all tags. If not, you can handle specific tags like this:

$test = "Home<i>\r</i><i>\n</i>";       
$patterns = array();
$patterns[0] = '/<\/?i>/'; //removes <i> and </i>
$patterns[1] = '/\r|\n/'; //removes \r and \n
$test = preg_replace($patterns, '', $test);