PHP: Parse_url w/ preg_replace to remove portion of referrer URL string

Question

I'm trying to extract portions of a URL in PHP using parse_url and remove a portion of the [path] if it exists.

For example, if my URL is "http://localhost/website/section/type/name/" and I want to remove the /website/ portion from the url [path]:

$url = wp_get_referer(); // using WordPress to get referring URL
$parsed = parse_url($url);
$parsed = preg_replace('/website/', '', $parsed[path]);

If I echo $parsed I get //section/type/name/ as the result, whereas I'd like to return simply section/type/name as the result when "/website/" is present in the URL path string. How might I achieve this result?

Phil · Accepted Answer · 2017-11-09T23:30:58.373

3

preg_replace is treating the / characters as the expression delimiter.

Since your search term is not a regular expression, you can simply use str_replace instead. Add in rtrim to remove the trailing slash.

$parsed = rtrim(str_replace('/website/', '', $parsed['path']), '/');

Note that I've also quoted the path parameter so as not to trigger an "unknown constant" notice

edited Nov 09 '17 at 23:30

answered Nov 09 '17 at 23:25

Phil

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Ah! Ok thank you! Would `preg_replace` be required to remove a trailing slash if present? – nickpish Nov 09 '17 at 23:28
@nickpish you can simply use `rtrim`. I'll add it to my answer – Phil Nov 09 '17 at 23:30

PHP: Parse_url w/ preg_replace to remove portion of referrer URL string

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