Question

当我在命令行上使用以下内容时，我将完美地获得输出。

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

但是当我尝试使用以下代码从PHP获取此内容时

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

cURL命令行到php版本

1 个答案: