[英]cURL Commandline to php version
當我在命令行上使用以下內容時,我得到了完美的輸出。
curl -X GET -H "Authorization: sso-key API_KEY:API_SECRET" "https://api.godaddy.com/v1/domains/mydomain.com"
但是當我嘗試使用以下代碼從 PHP 獲取它時
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$URL);
curl_setopt($ch, CURLOPT_TIMEOUT, 30); //timeout after 30 seconds
curl_setopt($ch, CURLOPT_RETURNTRANSFER,true);
curl_setopt($ch, CURLOPT_HTTPHEADER, [ "Authorization: sso-key API_KEY:API_SECRET"]);
curl_setopt($ch, CURLOPT_HTTPAUTH, CURLAUTH_ANY);
curl_setopt($ch, CURLOPT_USERPWD, "API_KEY:API_SECRET");
$result=curl_exec ($ch);
curl_close ($ch);
var_dump($result);
我什么都沒得到。
我在這里錯過了什么?
$URL = "https://api.godaddy.com/v1/domains/mydomain.com";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$URL);
curl_setopt($ch, CURLOPT_TIMEOUT, 30);
curl_setopt($ch, CURLOPT_RETURNTRANSFER,true);
curl_setopt($ch, CURLOPT_HTTPHEADER, [ "Authorization: sso-key API_KEY:API_SECRET"]);
curl_setopt($ch, CURLOPT_HTTPAUTH, CURLAUTH_ANY);
$result=curl_exec ($ch);
$httpCode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
var_dump($result);
var_dump($httpCode);
curl_close ($ch);
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