4

运行以下代码

var_dump(get_headers("http://www.domainnnnnnnnnnnnnnnnnnnnnnnnnnnn.com/CraxyFile.jpg"));

对于任何不存在的域或 URL，返回 HTTP 200 而不是 404

Array
(
    [0] => HTTP/1.1 200 OK
    [1] => Server: nginx/1.1.15
    [2] => Date: Mon, 08 Oct 2012 12:29:13 GMT
    [3] => Content-Type: text/html; charset=utf-8
    [4] => Connection: close
    [5] => Set-Cookie: PHPSESSID=3iucojet7bt2peub72rgo0iu21; path=/; HttpOnly
    [6] => Expires: Thu, 19 Nov 1981 08:52:00 GMT
    [7] => Cache-Control: no-store, no-cache, must-revalidate, post-check=0, pre-check=0
    [8] => Pragma: no-cache
    [9] => Set-Cookie: bypassStaticCache=deleted; expires=Thu, 01-Jan-1970 00:00:01 GMT; path=/; httponly
    [10] => Set-Cookie: bypassStaticCache=deleted; expires=Thu, 01-Jan-1970 00:00:01 GMT; path=/; httponly
    [11] => Vary: Accept
)

如果你跑

var_dump(get_headers("http://www.domain.com/CraxyFile.jpg"));

你得到

Array
(
    [0] => HTTP/1.1 404 Not Found
    [1] => Date: Mon, 08 Oct 2012 12:32:18 GMT
    [2] => Content-Type: text/html
    [3] => Content-Length: 8727
    [4] => Connection: close
    [5] => Server: Apache
    [6] => Vary: Accept-Encoding
)

它们有很多实例get_headers已被证明是验证现有 URL 的解决方案

这是错误还是 get_headers 不是验证 URL 的可靠方法

观看现场演示

更新 1

必须发现 CURL 也有同样的问题

$curl = curl_init();
curl_setopt_array($curl, array(CURLOPT_RETURNTRANSFER => true,CURLOPT_URL => 'idontexist.tld'));
curl_exec($curl);
$info = curl_getinfo($curl);
curl_close($curl);
var_dump($info);

也返回相同的结果

4

1 回答 1

11

问题与域名的长度无关，只是域名是否存在。

您正在使用将不存在的域解析为服务器的 DNS 服务，该服务器为您提供“友好”错误页面，并返回 200 响应代码。这意味着它也不是get_headers()具体的问题，它是任何潜在地依赖于合理的 DNS 查找的程序。

一种无需为您工作的每个环境硬编码变通的方法即可处理此问题的方法可能如下所示：

// A domain that definitely does not exist. The easiest way to guarantee that
// this continues to work is to use an illegal top-level domain (TLD) suffix
$testDomain = 'idontexist.tld';

// If this resolves to an IP, we know that we are behind a service such as this
// We can simply compare the actual domain we test with the result of this
$badIP = gethostbyname($testDomain);

// Then when you want to get_headers()
$url = 'http://www.domainnnnnnnnnnnnnnnnnnnnnnnnnnnn.com/CraxyFile.jpg';

$host = parse_url($url, PHP_URL_HOST);
if (gethostbyname($host) === $badIP) {
  // The domain does not exist - probably handle this as if it were a 404
} else {
  // do the actual get_headers() stuff here
}

您可能希望以某种方式缓存第一次调用的返回值gethostbyname()，因为您知道您正在查找一个不存在的名称，这通常需要几秒钟。

于 2012-10-08T13:02:33.313 回答