function varify_url($url){ $check = @fopen($url,"r"); if($check){$status = true; }else{$status = false; }return $status; }接着在需要的地方直接调用即可$url = "http://www.baidu.com"; if(varify_url($url)){echo "<div>Congratulation ! Your URL <a href=$url>$url</a> : is <b>valid </b></div>"; }else{echo "<div>Error ! Your URL : <a href=$url>$url</a> is <b>invalid </b></div>"; }希望本文所述对大家的php程序设计有所帮助。