Android：如何从Zomato API获取JSON对象？

2 回答

• 2

  您可以使用Rest Client通过标头和URL调用请求 . 我建议使用Volley或Retrofit来做 . 以下是使用Volley的示例：

  RequestQueue queue = Volley.newRequestQueue(this);
          String url = "https://developers.zomato.com/api/v2.1/search?&lat=27&lon=153";
          JsonObjectRequest postRequest = new JsonObjectRequest(Request.Method.GET, url, null,
                  new Response.Listener<JSONObject>() {
                      @Override
                      public void onResponse(JSONObject response) {
                          // response
                          Log.d("Response", response.toString());
                      }
                  },
                  new Response.ErrorListener() {
                      @Override
                      public void onErrorResponse(VolleyError error) {
                          // TODO Auto-generated method stub
                          Log.d("ERROR", "error => " + error.toString());
                      }
                  }
          ) {
              @Override
              public Map<String, String> getHeaders() throws AuthFailureError {
                  Map<String, String> params = new HashMap<String, String>();
                  params.put("user-key", "55a1d18014dd0c0dac534c02598a3368");
                  params.put("Accept", "application/json");
  
                  return params;
              }
          };
          queue.add(postRequest);
  

  我得到的回应是：

  {"restaurants":[],"results_found":0,"results_shown":0,"results_start":0}
  

  回复于 2024-04-27T10:55:19+08:00
• 1

  我注意到Curl包含 --header 所以我对URL中的 Headers 进行了一些研究，并在 url.openConnection(); 之后添加了这两行

  URLConnection urlConnection = url.openConnection();
  urlConnection.setRequestProperty("Accept", " application/json");
  urlConnection.setRequestProperty("user-key", " "+API_KEY);
  

  我得到了我需要的东西 .

  回复于 2024-04-27T10:55:19+08:00