cURL命令行到PHP版本

Question

当我在命令行上使用以下命令时，我得到的输出很完美。

curl -X GET -H "Authorization: sso-key API_KEY:API_SECRET" "https://api.godaddy.com/v1/domains/mydomain.com" 

但是当我尝试从PHP得到这个使用下面的代码

$ch = curl_init(); 
curl_setopt($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); 

我没有得到任何东西。

我在这里错过了什么？

Answer 1

$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);