带有 GSON 的 Java JSON [英] Java JSON with GSON

问题描述

问题来了，我使用 Wunderground 的天气 API，但在使用 GSON 获取天气时遇到问题.

Here's the problem, I'm using a weather APi from Wunderground and am having trouble using GSON to fetch the weather.

import java.net.*;
import java.io.*;
import com.google.gson.*;

public class URLReader {

    public static URL link;

    public static void main(String[] args) {
        try{
            open();
            read();
        }catch(IOException e){}
    }

    public static void open(){
        try{
            link = new URL("http://api.wunderground.com/api/54f05b23fd8fd4b0/geolookup/conditions/forecast/q/US/CO/Denver.json");
        }catch(MalformedURLException e){}
    }

    public static void read() throws IOException{
        Gson gson = new Gson();

        // Code to get variables like humidity, chance of rain, ect...
    }
} 

这就是我得到的，所以有人可以帮助我让这个东西工作.我需要能够解析特定变量，而不仅仅是整个 JSON 文件.

That is as far as I got, so can someone please help me to get this thing working. I need to be able to parse specific variables not just the entire JSON file.

我使用了缓冲读取器，但它读取了整个 JSON 文件.

I used a buffered reader but it read the entire JSON file.

提前致谢，我是初学者，所以请放轻松并非常简单地解释事情:)

Thanks in advance, I'm a beginner so go easy and explain things very straightforward :)

推荐答案

你可以解析整个流，然后拉出你需要的.下面是一个解析它的例子:

You can parse the entire stream, and then pull out what you need. Here's an example of parsing it:

    String url=getUrl();
    JSONObject jsonObject = new JSONObject();
    StringBuilder stringBuilder=new StringBuilder();
    try 
    {
        HttpGet httpGet = new HttpGet(url);
        HttpClient client = new DefaultHttpClient();
        HttpResponse response = client.execute(httpGet);
        HttpEntity entity = response.getEntity();
        InputStream stream = entity.getContent();
        int b;
        while ((b = stream.read()) != -1) {
            stringBuilder.append((char) b);
        }

        jsonObject = new JSONObject(stringBuilder.toString());

    } catch (JSONException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
    } catch (IOException e) {    }

阅读它，您必须在表格中导航，如下所示:

Reading it, you have to navigate your way through the table, something like this:

jsonObject.getJSONObject("Results");

以下是我使用此库的一个程序的示例:

Here's an example from one of my programs using this library:

int statusCode=jobj.getJSONObject("info").getInt("statuscode");

我大量使用以下内容来使其正确:

I make heavy use out of the following to get it right:

jsonObject.names();

这将为您提供所有键的名称.从那里，您必须弄清楚它是数组、对象还是原始类型.我需要一点时间才能把它做好，但是一旦你做了一次，它就会永远完成，希望如此.在他们的 JSON 库中查看 Android 文档.

That will give you the name of all of the keys. From there, you have to figure out if it's an array, an object, or a primitive type. It takes me a bit to get it right, but once you've done it once, it's done forever, hopefully. Take a look at the Android documents on their JSON library.

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