Convert Curl Anfrage Adresse Um httpUrlConnection Anfrage und Antwort erhalten

Question

Wie kann ich Client-ID und Client geheim in 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"); 

Erste httpstatus: 400

Answer 1

Content-Typeapplication/x-www-form-urlencoded sein sollte. Details finden Sie unter 4.4.2. Access Token Request von RFC 6749 (OAuth 2.0 Authorization Framework).