$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'.

在这里:

$file = 'http://www.example.com/somefile.jpg';
$file_headers = @get_headers($file);
if(!$file_headers || $file_headers[0] == 'HTTP/1.1 404 Not Found') {
    $exists = false;
}
else {
    $exists = true;
}

从这里和上面帖子的正下方，有一个卷曲的解决方案:

function url_exists($url) {
    return curl_init($url) !== false;
}

get_headers()返回一个数组，其中包含服务器响应HTTP请求时发送的报头。

$image_path = 'https://your-domain.com/assets/img/image.jpg';

$file_headers = @get_headers($image_path);
//Prints the response out in an array
//print_r($file_headers); 

if($file_headers[0] == 'HTTP/1.1 404 Not Found'){
   echo 'Failed because path does not exist.</br>';
}else{
   echo 'It works. Your good to go!</br>';
}

$headers = @get_headers($this->_value);
if(strpos($headers[0],'200')===false)return false;

所以任何时候你接触一个网站，得到200个以上的东西，它会工作

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;
        }
    }

检查url是在线还是离线——

function get_http_response_code($theURL) {
    $headers = @get_headers($theURL);
    return substr($headers[0], 9, 3);
}