如何在java中打印完整的json http请求结果

Question

无法打印完整的JSON结果。我的计划如下：

这个程序的新手。无法找到错误的原因。

按如下方式查询程序：

curl -X POST -H 
"Content-Type:application/json" --header 
'X-Auth-Token:IEkmVGHsa4R3cGPw56MkfQ' -d '{
"sensor_key":"e4aa3e35t675fc57ce81f3dd6e2dcdef492at4f7", 
"date_ranges":[{
"from":"2015/04/02 17:05:00", 
"to":"2015/04/02 17:10:00"
}], 
"time_zone":"Mumbai" , 
"time_format":"str", 
"per":"50"
}' 
'https://api.datonis.io/api/v2/datonis_query/sensor_event_raw_data


    package IOT;

    import java.util.HashMap;
    import java.util.Map;
    import org.apache.http.HttpResponse;
    import org.apache.http.client.methods.HttpPost;
    import org.apache.http.entity.StringEntity;
    import org.apache.http.impl.client.DefaultHttpClient;
    import org.apache.http.message.BasicHeader;
    import org.apache.http.protocol.HTTP;
    import org.json.JSONObject;
    import org.json.JSONArray;

    public class HttpPostWithBody {

    public static void main(String args[]) {
    String Message = "6f2159f998";

    try {
        new HttpPostWithBody().sendJSONData(Message);
    } catch (Exception E) {
    System.out.println("Exception Occured. " + E.getMessage());
    }
    }

    public String sendJSONData(String message) throws Exception {

    //creating map object to create JSON object from it
    Map< String, Object >jsonValues = new HashMap< String, Object >();
    jsonValues.put("sensor_key",message);
    jsonValues.put("from", "2016/08/29 16:55:00");
    jsonValues.put("to", "2016/08/29 17:05:00");
    jsonValues.put("time_zone", "Mumbai");
    jsonValues.put("per", "50");
    jsonValues.put("metrics", "1st data");

    JSONObject json = new JSONObject(jsonValues);


    String url = "https://api.datonis.io/api/v2/datonis_query/sensor_event_raw_data";

    DefaultHttpClient client = new DefaultHttpClient();
    HttpPost post = new HttpPost(url);
    post.setHeader("Content-Type", "application/json");
    post.setHeader("Accept", "application/json"); 
    post.setHeader("X-Auth-Token", "9v8IjBku0a9y-D7SpLq6ZA");

    //setting json object to post request.
    StringEntity entity = new StringEntity(json.toString(), "UTF8");
    entity.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));

    post.setEntity(entity);
    //this is your response:

    HttpResponse response = client.execute(post);

    //JSONObject myObject1 = new JSONObject(response);

    JSONArray ja = new JSONArray(response);

    //JSONObject jo = ja.getJSONObject();
    System.out.println("Response: " + ja.getJSONObject(0));

    System.out.println("Response: " + response.getStatusLine());
    return response.getStatusLine().toString();
    }
    }

Answer 1

试试这个：

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonElement;
import com.google.gson.JsonParser;
...
Gson gson = new GsonBuilder().setPrettyPrinting().create();
JsonParser jp = new JsonParser();
JsonElement je = jp.parse(json.toString());
String prettyJsonString = gson.toJson(je);
System.out.println(prettyJsonString);

输出：

{
  "sensor_key": "6f2159f998",
  "from": "2016/08/29 16:55:00",
  "to": "2016/08/29 17:05:00",
  "metrics": "1st data",
  "time_zone": "Mumbai",
  "per": "50"
}

查看整个代码 here

有关GSON库的详细信息，请查看 this

要下载GSON，请检查 this

Answer 2

可以，你应该使用相同的解决方案： Pretty-Print JSON in Java

此外，GSON总体上非常好