我如何检查一个URL是否存在(不是404)在PHP?


当前回答

cURL可以返回HTTP代码,我不认为所有额外的代码是必要的?

function urlExists($url=NULL)
    {
        if($url == NULL) return false;
        $ch = curl_init($url);
        curl_setopt($ch, CURLOPT_TIMEOUT, 5);
        curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
        $data = curl_exec($ch);
        $httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
        curl_close($ch); 
        if($httpcode>=200 && $httpcode<300){
            return true;
        } else {
            return false;
        }
    }

其他回答

简单的方法是卷曲(和更快)

<?php
$mylinks="http://site.com/page.html";
$handlerr = curl_init($mylinks);
curl_setopt($handlerr,  CURLOPT_RETURNTRANSFER, TRUE);
$resp = curl_exec($handlerr);
$ht = curl_getinfo($handlerr, CURLINFO_HTTP_CODE);


if ($ht == '404')
     { echo 'OK';}
else { echo 'NO';}

?>

在某些服务器中不能使用curl 你可以用这个代码

<?php
$url = 'http://www.example.com';
$array = get_headers($url);
$string = $array[0];
if(strpos($string,"200"))
  {
    echo 'url exists';
  }
  else
  {
    echo 'url does not exist';
  }
?>
function url_exists($url) {
    $headers = @get_headers($url);
    return (strpos($headers[0],'200')===false)? false:true;
}
$url = 'http://google.com';
$not_url = 'stp://google.com';

if (@file_get_contents($url)): echo "Found '$url'!";
else: echo "Can't find '$url'.";
endif;
if (@file_get_contents($not_url)): echo "Found '$not_url!";
else: echo "Can't find '$not_url'.";
endif;

// Found 'http://google.com'!Can't find 'stp://google.com'.
function URLIsValid($URL)
{
    $exists = true;
    $file_headers = @get_headers($URL);
    $InvalidHeaders = array('404', '403', '500');
    foreach($InvalidHeaders as $HeaderVal)
    {
            if(strstr($file_headers[0], $HeaderVal))
            {
                    $exists = false;
                    break;
            }
    }
    return $exists;
}