PHP：如何获取请求页面并获取正文和http错误代码

问题描述：

I want know if is possible to get the HTTP error codes and response in case of error instead of false (file get contents error) and error throwed. I'm using file_get_contents on PHP 7.2

I already tried doing this: $r = file_get_contents("https://somewebsite");

Output: PHP Warning: file_get_contents(https://somewebsite): failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request I can get the response code with $http_response_header but $r is false and i want get the error response page.

我想知道是否可以获取HTTP错误代码和响应，以防错误而不是false（文件获取）内容错误）和错误抛出。我在PHP 7.2上使用file_get_contents p>

我已经尝试过这样做： $ r = file_get_contents（“https：// somewebsite”）; code> p>

输出：代码> PHP警告：file_get_contents（https：// somewebsite）：无法打开流：HTTP请求失败！ HTTP / 1.0 400错误请求 code> 我可以使用$ http_response_header获取响应代码，但$ r为false，我想获取错误响应页面。 p> div>

答

You should try with curl

$ch = curl_init('https://httpstat.us/404');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
// if you want to follow redirections
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
// you may want to disable certificate verification
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, false);

$response = curl_exec($ch);
$httpCode = curl_getinfo($ch, CURLINFO_HTTP_CODE);

if ($httpCode >= 400) {
    // error
    // but $response still contain the response
} else {
    // everything is fine
}

curl_close($ch);

答

Using file get contents

$context = stream_context_create(array(
    'http' => array('ignore_errors' => true),
));

$result = file_get_contents('http://your/url', false, $context);

PHP：如何获取请求页面并获取正文和http错误代码

相关推荐