将 Curl 请求地址转换为 httpUrlConnection 请求并获得响应

Question

如何将客户端 ID 和客户端密码放入 httprequest？

           /*
            curl -X POST -H "Content-Type: application/json" \
           --user "<client_id>:<client_secret>" \
        -d '{"grant_type": "client_credentials", "scope": "public"}' \
        'https://api.lyft.com/oauth/token'
      */

  URL url = new URL("https://api.lyft.com/oauth/token");
           HttpURLConnection conn = (HttpURLConnection) url.openConnection();
           conn.setRequestProperty("Content-Type", "application/json");
           conn.setRequestProperty("grant_type", "client_credentials");
           conn.setRequestProperty("scope", "public");
           //conn.setRequestProperty("Authorization", "Token token=" + accessToken);
           conn.setRequestMethod("POST");

获取 httpStatus：400

Answer 1

Content-Type 应该是application/x-www-form-urlencoded。有关详细信息，请参阅 RFC 6749 的 4.4.2. Access Token Request（OAuth 2.0 授权框架）。