java - 无法使用 java 显示来自 URL 的 json 数据

Question

我遇到了问题，但不知道如何解决。

我有以下代码:

public static void main(String[] args) 
{
    BufferedReader br = null;
    StringBuilder sb = null;
    String line = null;
    Scanner scanner = new Scanner(System.in);


    //print the menu
    System.out.println("Choose from the menu:");
    System.out.println("1 -> Town weather ");
    System.out.println("2 -> About");
    System.out.println("3 -> Exit");

    try 
    {
        //read from keyboard the value
        //int choice = scanner.nextInt();
        int choice = 1;

        switch (choice)
        {
            case 1:
                System.out.println("Give desired town:");
                String town = str.nextLine();
                URL json = new URL("http://api.openweathermap.org/data/2.5/weather?q=" + town + "&APPID=");
                HttpURLConnection url = (HttpURLConnection) json.openConnection();

                url.setRequestMethod("GET");
                url.connect();

                //read the data from url
                br = new BufferedReader(new InputStreamReader(url.getInputStream()));
                sb = new StringBuilder();

                while ((line = br.readLine()) != null)
                {
                    sb.append(line + '\n'); 
                }                   

                String txt = sb.toString();

                break;

            case 2:
                break;

            case 3:
                System.exit(0);         
        }
    }
    catch (InputMismatchException i)
    {
        System.out.println("Wrong choice!");
    }
    catch (MalformedURLException m)
    {
        System.out.println("Wrong URL!");
    }
    catch (IOException io)
    {
        System.out.println("Wrong town! The url shows 404 not found");
    }
    catch (NullPointerException np)
    {
        System.out.println("Null exception!");
        np.printStackTrace();
    }
    catch (Exception e) //catch all exception where not previous caught.
    {
        e.printStackTrace();
    }
}

所以我将 json 数据放入 txt 变量中。问题是，我的项目需要将所有数据(或其中的一部分)显示为列表。我必须以人们可以阅读的方式展示它们。

我尝试了 pretty-print ，但我不想显示符号 { } , :

理想情况下，我想将这些数据存储到数据库中，然后在维护它们的同时显示其中的一些数据。第一步是将它们拆分为一个数组，仅包含字符串，不包含任何特殊字符。

有人可以帮我解决这个问题吗？我搜索了 stackoverflow，发现了很多回复，但没有一个对我有用。

编辑:我更新了代码，包含完整的主代码，下面您可以看到 json 响应:

{
  "coord": {
    "lon": -86.97,
    "lat": 34.8
  },
  "weather": [
    {
      "id": 800,
      "main": "Clear",
      "description": "clear sky",
      "icon": "01d"
    }
  ],
  "base": "stations",
  "main": {
    "temp": 270.48,
    "pressure": 1033,
    "humidity": 30,
    "temp_min": 270.15,
    "temp_max": 271.15
  },
  "visibility": 16093,
  "wind": {
    "speed": 2.6,
    "deg": 340
  },
  "clouds": {
    "all": 1
  },
  "dt": 1514921700,
  "sys": {
    "type": 1,
    "id": 226,
    "message": 0.0021,
    "country": "US",
    "sunrise": 1514897741,
    "sunset": 1514933354
  },
  "id": 4830668,
  "name": "Athens",
  "cod": 200
}

预先感谢您的帮助。

Answer 1

尝试下面的代码将您的响应转换为 json 对象，您可以将其转换为所需的格式或稍后保存到数据库

package test;

import javax.net.ssl.HttpsURLConnection;
import org.apache.http.HttpResponse;
import org.apache.http.HttpStatus;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.HttpParams;
import com.google.gson.JsonObject;
import com.google.gson.JsonParser;
public class Test {

    String url="your URL"
    DefaultHttpClient httpClient = new DefaultHttpClient();
    httpget = new HttpGet(url);
    HttpResponse httpResponse  = httpClient.execute(httpget);


    if (httpResponse.getStatusLine().getStatusCode() == HttpStatus.SC_OK) {
        //read response from body
        ResponseHandler<String> handler = new BasicResponseHandler();
        String respBody = handler.handleResponse(httpResponse);
        if (respBody != null && !"".equals(respBody)) {
            JsonObject responseJson =  new 
        JsonParser().parse(respBody).getAsJsonObject();

        }
    }

}