PHP：如何獲取請求頁面並獲取正文和http錯誤代碼

Question

我想知道是否有可能獲得HTTP錯誤代碼並在錯誤而不是false（文件獲取內容錯誤）和錯誤引發的情況下做出響應。 我在PHP 7.2上使用file_get_contents

我已經嘗試過這樣做： $r = file_get_contents("https://somewebsite");

輸出： PHP Warning: file_get_contents(https://somewebsite): failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request PHP Warning: file_get_contents(https://somewebsite): failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request我可以使用$ http_response_header獲取響應代碼，但是$ r為false，我想獲取錯誤響應頁面。

Answer 1

你應該嘗試卷曲

$ch = curl_init('https://httpstat.us/404');
curl_setopt($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);

Answer 2

使用文件獲取內容

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

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