realcoder Posted September 2, 2010 Share Posted September 2, 2010 $html = @file_get_contents($urlw) or problem('Can\'t open Your Remote URL!'); $html = strtolower($html); $site_url = strtolower($set['site_url']); if (preg_match_all('/<a\s[^>]*href=([\"\']??)([^" >]*?)\\1([^>]*)>/siU', $html, $matches, PREG_SET_ORDER)) { foreach($matches as $match) { if ($match[2] == $set['site_url'] || $match[2] == $set['site_url'].'/') { $found = 1; if (strstr($match[3],'nofollow')) { $nofollow = 1; } break; } } } if ($found == 0) { echo "<center><h2>Our URL Not FOund on your page please submit or check again</h2></center>"; } else next proceggerr.... in $html we getting the content of that page which that user give where he put our link what is this actually this all code is doing infact i was making a link exchange site where on a page people can submit their link for exchange and give their site link and the that page link where they have put our link this code verify at the sPot did they have put our code or not but m facing some problem.. beacuse it's not working good Quote Link to comment Share on other sites More sharing options...
realcoder Posted September 2, 2010 Author Share Posted September 2, 2010 or tell me any function or method through that i can check my $site_url in $html in which m getting all the content of page $site_url is the url of my site Quote Link to comment Share on other sites More sharing options...
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