以下是提供 JSON 響應的鏈接：https://www.instagram.com/p/CDq1KJPJcTN/ ?__a=1簡單地將 URL 復制并粘貼到瀏覽器中即可得到 JSON 響應。但是當我嘗試使用 file_get_contents 或 cURL 在 PHP 中訪問相同內容時，它顯示 Instagram 的 html 頁面代碼ini_set("allow_url_fopen", 1);ini_set("allow_url_include", 1);$context = [ 'http' => [ 'method' => 'GET', 'header' => 'User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.47 Safari/537.36', ], ];$context = stream_context_create($context);$data = file_get_contents('https://www.instagram.com/p/CDq1KJPJcTN/?__a=1', false, $context);卷曲：$curl_handle=curl_init();curl_setopt($curl_handle, CURLOPT_URL,'https://www.instagram.com/p/CDq1KJPJcTN/?__a=1');curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, 1);curl_setopt($curl_handle, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.47 Safari/537.36');$query = curl_exec($curl_handle);curl_close($curl_handle);echo $query;當我使用rest api測試工具（https://resttesttest.com/）時。它僅給出 JSON 響應。我認為缺少一些重要的標題，我無法弄清楚。給出了 Content-Type:application/json 和 Access-Control-Allow-Origin: * 。
1 回答
揚帆大魚
TA貢獻1799條經驗 獲得超9個贊
嘗試這個
$curl_handle=curl_init();
curl_setopt($curl_handle, CURLOPT_URL,'https://www.instagram.com/p/CDq1KJPJcTN/?__a=1');
curl_setopt($curl_handle, CURLOPT_HTTPHEADER, array('Content-Type: application/json; charset=utf-8'));
curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($curl_handle, CURLOPT_HEADER, FALSE);
curl_setopt($curl_handle, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.47 Safari/537.36');
curl_setopt($curl_handle, CURLOPT_SSL_VERIFYPEER, FALSE);
$query = curl_exec($curl_handle);
curl_close($curl_handle);
echo $query;
使用 file_get_contents 在本地對我有用
ini_set("allow_url_fopen", TRUE);
ini_set("allow_url_include", TRUE);
$context = [
'http' => [
'method' => 'GET',
'header' => 'User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.47 Safari/537.36',
],
];
$context = stream_context_create($context);
$data = file_get_contents('https://www.instagram.com/p/CDq1KJPJcTN/?__a=1', false, $context);
echo $data;
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