PHP simplexml_load_file捕获403 [英] PHP simplexml_load_file catch 403

问题描述

我正在使用以下PHP:

I am using the following PHP:

$xml = simplexml_load_file($request_url) or die("url not loading");

我使用:

$status = $xml->Response->Status->code;

检查响应状态. 200一切正常，继续.

To check the status of the response. 200 bening everything is ok, carry on.

但是，如果出现403访问被拒绝的错误，如何在PHP中捕获此错误，以便返回用户友好的警告?

However if I get a 403 access denied error, how do I catch this in PHP so I can return a user friendly warning?

推荐答案

要从对simplexml_load_file()的调用中检索HTTP响应代码，我知道的唯一方法是使用鲜为人知的PHP $http_response_header.每次通过HTTP包装器发出HTTP请求时，都会自动将该变量自动创建为包含每个响应标头的数组.换句话说，每次您使用simplexml_load_file()或file_get_contents()以及以"http://"开头的URL

To retrieve the HTTP response code from a call to simplexml_load_file(), the only way I know is to use PHP's little known $http_response_header. This variable is automagically created as an array containing each response header separately, everytime you make a HTTP request through the HTTP wrapper. In other words, everytime you use simplexml_load_file() or file_get_contents() with a URL that starts with "http://"

您可以使用print_r()检查其内容，例如

You can inspect its content with a print_r() such as

$xml = @simplexml_load_file($request_url);
print_r($http_response_header);

但是，在您的情况下，您可能想使用 file_get_contents() 然后，测试您是否收到了4xx响应，如果没有，则将正文传递给simplexml_load_string().例如:

In your case, though, you might want to retrieve the XML separately with file_get_contents() then, test whether you got a 4xx response, then if not, pass the body to simplexml_load_string(). For instance:

$response = @file_get_contents($request_url);
if (preg_match('#^HTTP/... 4..#', $http_response_header[0]))
{
    // received a 4xx response
}

$xml = simplexml_load_string($response);

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