-8

可能重复：
如何从 Java HTTPResponse 解析 JSON？
如何在Android中解析json字符串？

我如何解析这个 json 字符串

{
"apiVersion": "2.1",
"data": {
    "id": "pHuoDqcIyqk",
    "uploaded": "2012-10-29T16:08:15.000Z",
    "updated": "2012-11-02T08:48:28.000Z",
    "uploader": "googlenexus",
    "category": "Tech",
    "title": "Nexus: Ask Me Anything",
    "description": "The Best of Google, now in 3 sizes. Introducing Nexus 4, Nexus 7 and Nexus 10. The new smartphone and tablets from Google. Shop now at play.google.com/nexus",
    "thumbnail": {
        "sqDefault": "http://i.ytimg.com/vi/pHuoDqcIyqk/default.jpg",
        "hqDefault": "http://i.ytimg.com/vi/pHuoDqcIyqk/hqdefault.jpg"
    },
    "player": {
        "default": "http://www.youtube.com/watch?v=pHuoDqcIyqk&feature=youtube_gdata_player",
        "mobile": "http://m.youtube.com/details?v=pHuoDqcIyqk"
    },
    "content": {
        "5": "http://www.youtube.com/v/pHuoDqcIyqk?version=3&f=videos&app=youtube_gdata",
        "1": "rtsp://v8.cache5.c.youtube.com/CiILENy73wIaGQmpyginDqh7pBMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp",
        "6": "rtsp://v7.cache4.c.youtube.com/CiILENy73wIaGQmpyginDqh7pBMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp"
    },
    "duration": 61,
    "aspectRatio": "widescreen",
    "rating": 4.8985643,
    "likeCount": "5227",
    "ratingCount": 5363,
    "viewCount": 1038854,
    "favoriteCount": 0,
    "commentCount": 1442,
    "accessControl": {
        "comment": "allowed",
        "commentVote": "allowed",
        "videoRespond": "moderated",
        "rate": "allowed",
        "embed": "allowed",
        "list": "allowed",
        "autoPlay": "allowed",
        "syndicate": "allowed"
    }
}
}

谁能告诉我如何调用上面的 URL 并获取 JSON 对象，然后解析它以获取所需的信息作为字符串？

4

5 回答 5

0

有几种方法可以做到这一点。一种简单的方法是定义一个与您的数据匹配的对象结构并使用http://code.google.com/p/google-gson/来填充它：

final URL url = new URL("http://yourURL.com");
final InputStream openStream = url.openStream();
final InputStreamReader inputStreamReader = new InputStreamReader(openStream);
final SearchResult searchResult = new Gson().fromJson(inputStreamReader, SearchResult.class);

如果您的 searchResult 类包含一个字段，例如 apiVersion，它将填充来自 JSON 的数据。然后你有一个数据字段，里面有其他字段等。这是填充数据的一种非常简单的方法，但是你必须在你的对象模型中镜像 JSON 流的结构。

于 2012-11-02T09:27:23.927 回答
0

你问的问题太宽泛了..

1)在 URL 上执行http get请求。（或使用此方法）并获取响应字符串，即 JSON。这里有一些例子

2）请参阅此答案以了解如何解析 JSON。

于 2012-11-02T09:22:02.277 回答
0

简单简单....

String str = "" // your JSON string
JSONObject json = new JSONObject(str);
String apiVersion = json.getString("apiVersion");
 // continue as before for the rest of it...
于 2012-11-02T09:22:47.777 回答
0

你可以试试下面的方法。您可以传递 URL 并将响应作为结果字符串获取。公共静态字符串结果；

public static boolean connect(String url) {
        boolean flag = false;
        HttpClient httpclient = new DefaultHttpClient();

        // Prepare a request object
        HttpGet httppost = new HttpGet(url);
        Log.e("url", url);

        // Execute the request
        HttpResponse response;
        try {

            // https://api.vkontakte.ru/method/audio.search?uid=163398985&q=akoncount=100&access_token=2a4db0e223f0f5ab23f0f5ab5f23da5680223f023f1f5a3c696b018be9b17b9

            response = httpclient.execute(httppost);
            // Examine the response status
            Log.i("vkontake", response.getStatusLine().toString() + "\n"
                    + response);

            // Get hold of the response entity
            HttpEntity entity = response.getEntity();
            // If the response does not enclose an entity, there is no need
            // to worry about connection release

            if (entity != null) {

                // A Simple JSON Response Read
                InputStream instream = entity.getContent();
                Log.d("Jsomn Activity", "---- is --- " + instream);
                result = convertStreamToString(instream);
                Log.d("Jsomn Activity", "---- Result --- " + result);

                // now you have the string representation of the HTML request
                instream.close();

            } else {
                Log.d("Jsomn Activity", "---- is --- null ");
            }
            flag = true;
            net = false;
        } catch (Exception e) {
            Log.d("Jsomn Activity", "---- Catch --- " + e.toString());
            flag = false;
            e.printStackTrace();
        } finally {
            return flag;
        }

    }
于 2012-11-02T09:23:12.597 回答
-1

JSON.org站点包含不少于 20 种不同的 JSON 解析器 Java 实现。

于 2012-11-02T09:18:21.650 回答