如何从Android中的发布请求响应中获取特定的json密钥值?

时间:2017-12-07 04:05:32

标签: java android json parsing android-json

我使用HttpURLConnection来发布请求后跟响应:

请求后回复:

{
    "LatestData": [{
        "ExtraData": null,
        "ID": 0,
        "season": false,
        "latest": 0,
        "url": "http://www.awebsite.com/images/12.jpg"
    }]
}

如何获取网址的价值?我试过跟随,但Android Studio一直给我错误:

String newURL = sb.getJSONObject("LatestData").getString("url");
String newURL = sb.getJSONArray("LatestData").getJSONObject(0).getString("url");

Android Studio错误:

error: cannot find symbol method getJSONObject(String)
error: cannot find symbol method getJSONArray(String)
你可以帮助我获取url的值,让我知道我需要导入到android studio的库,以便getsonObject有效吗?谢谢

android代码:

 if (myURLConnection.getResponseCode() == 200) {
                br = new BufferedReader(new InputStreamReader(myURLConnection.getInputStream(), "utf-8"));
                while (true) {
                    line = br.readLine();
                    if (line != null) {
                        sb.append(line + "\n");


                        //String newURL = sb.getJSONObject("LatestData").getString("url");
                        String newURL =sb.getJSONArray("LatestData").getJSONObject(0).getString("url");
                        mWebView.loadUrl("javascript:MyFunction('" +newURL + "');");
                    } else {
                        br.close();
                        return sb.toString();
                    }
                }
            }

6 个答案:

答案 0 :(得分:2)

您需要将sb转换为JSONObject才能访问属性:

JSONOjbect jsonResponse = new JSONObject(new String(sb));

然后:

try {
    JSONArray jsonArray = jsonResponse.getJSONArray("LatestData");
    if (jsonArray != null && jsonArray.length() > 0) {
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject dataOjbect = jsonArray.getJSONObject(i);
            String url = dataOjbect.getString("url");
        }
    }
} catch (Exception e) {
    e.printStackTrace();
}

答案 1 :(得分:1)

try {
            JSONObject jsonObject = new JSONObject(sb);
            String url = jsonObject.optJSONArray("LatestData").getJSONObject(0).optString("url");
        } catch (JSONException e) {
            e.printStackTrace();
        }

答案 2 :(得分:0)

你的方法应该是这样的。

try {
        List<String> urlist = new ArrayList<>();
        JSONObject jsonObject = new JSONObject(sb.toString());
        JSONArray jsonArray =  jsonObject.getJSONArray("LatestData");
        for (int count = 0; count<jsonArray.length(); count++){
            JSONObject jsonInner =  jsonArray.getJSONObject(count);
            String url = jsonInner.getString("url");
            //store your url in some list
            urlist.add(url);
        }

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

答案 3 :(得分:0)

使用loadJSONArray

尝试
String url = loadJSONArray(sb)..getJSONObject(0).getString("url");

答案 4 :(得分:0)

你可以试试这个:

StringBuilder sb = new StringBuilder();
line = br.readLine();
while(line != null){
    sb.append(line);
}
// if it's fail here then this means there is an issue in parsing JSONObject
JSONObject jsonObj = new JSONObject(sb.toString());        
JSONArray latestData = jsonObj.getJSONArray("LatestData");
JSONObject jsonObject = latestData.getJSONObject(0);
String url = jsonObject.getString("url");

此外,我强烈建议您使用GSON进行Retrofit ..您可以按照以下文章进行操作:http://www.vogella.com/tutorials/Retrofit/article.html

答案 5 :(得分:0)

使用jsonschema2pojo为JSON数据生成pojo类