当我从命令行进行操作时,它有效:
$ curl -X "POST" "https://ABCD/login/oauth2/access_token" -H "Authorization: Basic XXXX=" -H "Content-Type:application/x-www-form-urlencoded" --data-urlencode "realm=XXX" --data-urlencode "XXX=XXX"
但是,当我从php做到这一点时,它不起作用:
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, 'https://ABCD/login/oauth2/access_token');
curl_setopt($ch, CURLOPT_POST, 1);
$ss = array(
'realm' => 'XXX',
'XXX'=>'XXX'
);
curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($ss));
curl_setopt($ch, CURLOPT_TIMEOUT, 30);
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch, CURLOPT_HTTPHEADER,
array(
'Authorization: Basic XXXX=',
'Content-Type: application/x-www-form-urlencoded',
));
$result=curl_exec ($ch);
$http_status = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close ($ch);
知道我做错了什么?我获得了http状态,"所需的参数或身体丢失或不正确。"
您的命令行说--data-urlencode
- URL编码
您的php说'Content-Type: application/x-www-form-urlencoded',
- 所以您说您正在编码数据
它也说curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($ss));
- 所以您是 JSON编码 it而不是编码URL。
发送您声称要发送的URL编码数据。