在PHP中测试404的URL的简便方法？

Question

我正在教自己一些基本的刮擦,我发现有时我输入到我的代码中的URL返回404,这会使我的所有其余代码变得粗糙.

所以我需要在代码顶部进行测试,以检查URL是否返回404.

这似乎是一项相当直接的任务,但谷歌没有给我任何答案.我担心我在寻找错误的东西.

一篇博客推荐我用这个:

$valid = @fsockopen($url, 80, $errno, $errstr, 30);

然后测试以查看$ valid是否为空.

但我认为给我问题的URL有一个重定向,所以$ valid对所有值都是空的.或许我做错了什么.

我也查看了"头部请求",但我还没有找到任何可以使用或尝试的实际代码示例.

建议？这是关于卷曲的？

Answer 1

如果您使用的是PHP的curl绑定,则可以使用以下方法检查错误代码curl_getinfo:

$handle = curl_init($url);
curl_setopt($handle,  CURLOPT_RETURNTRANSFER, TRUE);

/* Get the HTML or whatever is linked in $url. */
$response = curl_exec($handle);

/* Check for 404 (file not found). */
$httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
if($httpCode == 404) {
    /* Handle 404 here. */
}

curl_close($handle);

/* Handle $response here. */

Answer 2

如果您运行的php5可以使用:

$url = 'http://www.example.com';
print_r(get_headers($url, 1));

或者使用php4,用户提供了以下内容:

/**
This is a modified version of code from "stuart at sixletterwords dot com", at 14-Sep-2005 04:52. This version tries to emulate get_headers() function at PHP4. I think it works fairly well, and is simple. It is not the best emulation available, but it works.

Features:
- supports (and requires) full URLs.
- supports changing of default port in URL.
- stops downloading from socket as soon as end-of-headers is detected.

Limitations:
- only gets the root URL (see line with "GET / HTTP/1.1").
- don't support HTTPS (nor the default HTTPS port).
*/

if(!function_exists('get_headers'))
{
    function get_headers($url,$format=0)
    {
        $url=parse_url($url);
        $end = "\r\n\r\n";
        $fp = fsockopen($url['host'], (empty($url['port'])?80:$url['port']), $errno, $errstr, 30);
        if ($fp)
        {
            $out  = "GET / HTTP/1.1\r\n";
            $out .= "Host: ".$url['host']."\r\n";
            $out .= "Connection: Close\r\n\r\n";
            $var  = '';
            fwrite($fp, $out);
            while (!feof($fp))
            {
                $var.=fgets($fp, 1280);
                if(strpos($var,$end))
                    break;
            }
            fclose($fp);

            $var=preg_replace("/\r\n\r\n.*\$/",'',$var);
            $var=explode("\r\n",$var);
            if($format)
            {
                foreach($var as $i)
                {
                    if(preg_match('/^([a-zA-Z -]+): +(.*)$/',$i,$parts))
                        $v[$parts[1]]=$parts[2];
                }
                return $v;
            }
            else
                return $var;
        }
    }
}

两者的结果都类似于:

Array
(
    [0] => HTTP/1.1 200 OK
    [Date] => Sat, 29 May 2004 12:28:14 GMT
    [Server] => Apache/1.3.27 (Unix)  (Red-Hat/Linux)
    [Last-Modified] => Wed, 08 Jan 2003 23:11:55 GMT
    [ETag] => "3f80f-1b6-3e1cb03b"
    [Accept-Ranges] => bytes
    [Content-Length] => 438
    [Connection] => close
    [Content-Type] => text/html
)

因此,您可以检查标题响应是否正常,例如:

$headers = get_headers($url, 1);
if ($headers[0] == 'HTTP/1.1 200 OK') {
//valid 
}

if ($headers[0] == 'HTTP/1.1 301 Moved Permanently') {
//moved or redirect page
}

W3C代码和定义

Answer 3

使用strager的代码,您还可以检查CURLINFO_HTTP_CODE以获取其他代码.有些网站不报告404,而是简单地重定向到自定义404页面并返回302(重定向)或类似的东西.我用它来检查服务器上是否存在实际文件(例如robots.txt).很明显,这种文件如果存在则不会导致重定向,但如果不存在,则会重定向到404页面,正如我之前所说,可能没有404代码.

function is_404($url) {
    $handle = curl_init($url);
    curl_setopt($handle,  CURLOPT_RETURNTRANSFER, TRUE);

    /* Get the HTML or whatever is linked in $url. */
    $response = curl_exec($handle);

    /* Check for 404 (file not found). */
    $httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
    curl_close($handle);

    /* If the document has loaded successfully without any redirection or error */
    if ($httpCode >= 200 && $httpCode < 300) {
        return false;
    } else {
        return true;
    }
}

Answer 4

正如strager所暗示的那样,请考虑使用cURL.您可能还有兴趣使用curl_setopt设置CURLOPT_NOBODY 以跳过下载整个页面(您只需要标题).

Answer 5

如果您正在寻找一个最简单的解决方案,那么您可以尝试使用php5

file_get_contents('www.yoursite.com');
//and check by echoing
echo $http_response_header[0];

Answer 6

此函数返回 PHP 中 URL 的状态代码，而无需使用 HEAD 请求下载正文（html 代码）：

\n

function isHttpStatusCode200(string $url): bool\n{\n    return getHttpResponseCode($url) === 200;\n}\n\nfunction getHttpResponseCode(string $url): int\n{\n   $context = stream_context_create(\n\xc2\xa0\xc2\xa0\xc2\xa0   \xc2\xa0array(\n\xc2\xa0\xc2\xa0   \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0'http'\xc2\xa0=>\xc2\xa0array(\n\xc2\xa0\xc2\xa0\xc2\xa0   \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0'method'\xc2\xa0=>\xc2\xa0'HEAD'\n\xc2\xa0\xc2\xa0  \xc2\xa0\xc2\xa0 \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0)\n\xc2\xa0\xc2\xa0   \xc2\xa0\xc2\xa0)\n   );\n    $headers = get_headers($url, false, $context);\n    return substr($headers[0], 9, 3);\n}\n

\n

例子：

\n

echo isHttpStatusCode200('https://www.google.com');\n//displays: true\n

\n

Answer 7

如果 url 不返回 200 OK，这将为您提供 true

function check_404($url) {
   $headers=get_headers($url, 1);
   if ($headers[0]!='HTTP/1.1 200 OK') return true; else return false;
}

Answer 8

我在这里找到了这个答案:

if(($twitter_XML_raw=file_get_contents($timeline))==false){
    // Retrieve HTTP status code
    list($version,$status_code,$msg) = explode(' ',$http_response_header[0], 3);

    // Check the HTTP Status code
    switch($status_code) {
        case 200:
                $error_status="200: Success";
                break;
        case 401:
                $error_status="401: Login failure.  Try logging out and back in.  Password are ONLY used when posting.";
                break;
        case 400:
                $error_status="400: Invalid request.  You may have exceeded your rate limit.";
                break;
        case 404:
                $error_status="404: Not found.  This shouldn't happen.  Please let me know what happened using the feedback link above.";
                break;
        case 500:
                $error_status="500: Twitter servers replied with an error. Hopefully they'll be OK soon!";
                break;
        case 502:
                $error_status="502: Twitter servers may be down or being upgraded. Hopefully they'll be OK soon!";
                break;
        case 503:
                $error_status="503: Twitter service unavailable. Hopefully they'll be OK soon!";
                break;
        default:
                $error_status="Undocumented error: " . $status_code;
                break;
    }

实质上,您使用"文件获取内容"方法来检索URL,该URL会自动使用状态代码填充http响应标头变量.

Answer 9

附录；考虑到性能，测试了这 3 种方法。

结果，至少在我的测试环境中：

卷曲获胜

这个测试是在只需要头文件（noBody）的情况下完成的。测试自己：

$url = "http://de.wikipedia.org/wiki/Pinocchio";

$start_time = microtime(TRUE);
$headers = get_headers($url);
echo $headers[0]."<br>";
$end_time = microtime(TRUE);
echo $end_time - $start_time."<br>";


$start_time = microtime(TRUE);
$response = file_get_contents($url);
echo $http_response_header[0]."<br>";
$end_time = microtime(TRUE);
echo $end_time - $start_time."<br>";

$start_time = microtime(TRUE);
$handle = curl_init($url);
curl_setopt($handle,  CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($handle, CURLOPT_NOBODY, 1); // and *only* get the header 
/* Get the HTML or whatever is linked in $url. */
$response = curl_exec($handle);
/* Check for 404 (file not found). */
$httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
// if($httpCode == 404) {
    // /* Handle 404 here. */
// }
echo $httpCode."<br>";
curl_close($handle);
$end_time = microtime(TRUE);
echo $end_time - $start_time."<br>";