android java:foreach通过json数组返回值?

android java:foreach通过json数组返回值?,java,android,arrays,json,for-loop,Java,Android,Arrays,Json,For Loop,我在java代码中执行for-each循环时遇到问题。我可以得到单一的json结果,但是如何在代码中使用for-each循环 有人能帮我吗 public JSONObject feedTimeline(String username) throws ClientProtocolException, IOException, JSONException{ StringBuilder url = new StringBuilder(URL); url.append(username)

我在java代码中执行for-each循环时遇到问题。我可以得到单一的json结果,但是如何在代码中使用for-each循环

有人能帮我吗

public JSONObject feedTimeline(String username) throws ClientProtocolException, IOException, JSONException{
    StringBuilder url = new StringBuilder(URL);
    url.append(username);

    HttpGet get = new HttpGet(url.toString());
    HttpResponse response = client.execute(get);
    int status = response.getStatusLine().getStatusCode();
    if(status == 200){
        HttpEntity e = response.getEntity();
        String data = EntityUtils.toString(e);
        JSONArray timeline = new JSONArray(data);
        for (int i = 0; i < timeline.length(); i++) {
        JSONObject value= timeline.getJSONObject(i); //no error if this i is 0 and without for each loop
        return value; //getting errors because of this return tweets
        }


    }else{
        Toast.makeText(Feed.this,"error",Toast.LENGTH_SHORT);
        return null;
    }
}


public class Read extends AsyncTask<String, Integer, String>{

    @Override
    protected String doInBackground(String... params) {
        // TODO Auto-generated method stub
        try {
            json = feedTimeline("name");
            return json.getString(params[0]); //this would need to change I assume?
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return null;
    }
公共JSONObject feedTimeline(字符串用户名)抛出ClientProtocolException、IOException、JSONException{ StringBuilder url=新的StringBuilder(url); url.append(用户名); HttpGet=newhttpget(url.toString()); HttpResponse response=client.execute(get); int status=response.getStatusLine().getStatusCode(); 如果(状态==200){ HttpEntity e=response.getEntity(); 字符串数据=EntityUtils.toString(e); JSONArray时间线=新JSONArray(数据); 对于(int i=0;i 我得到了一个JSONObject feedTimeline的错误…如果我有for循环。但是如果我将其作为for循环输出,而不是在
JSONObject value=timeline.getJSONObject(I)
中使用
I
,并使用一个类似
0
1
的数字值,那么它会输出

另外,我相信在类Read中,
return json.getString(params[0])
也需要在as-for循环中工作?我对JAVA非常陌生,我正在尝试自己学习所有东西

提前感谢!

公共JSONObject feedTimeline(字符串用户名)抛出ClientProtocolException、IOException、JSONException{
    public JSONObject feedTimeline(String username) throws ClientProtocolException, IOException, JSONException{
        StringBuilder url = new StringBuilder(URL);
        url.append(username);

        HttpGet get = new HttpGet(url.toString());
        HttpResponse response = client.execute(get);
        int status = response.getStatusLine().getStatusCode();
        if(status == 200){
            HttpEntity e = response.getEntity();
            String data = EntityUtils.toString(e);
            JSONArray timeline = new JSONArray(data);
            for (int i = 0; i < timeline.length(); i++) {
            JSONObject value= timeline.getJSONObject(i); //no error if this i is 0 and without for each loop
            return value; //getting errors because of this return tweets
            }


        }else{
            Toast.makeText(Feed.this,"error",Toast.LENGTH_SHORT);      
        }
// changed the position of return statement. This should work now.
          return null;
    }
StringBuilder url=新的StringBuilder(url); url.append(用户名); HttpGet=newhttpget(url.toString()); HttpResponse response=client.execute(get); int status=response.getStatusLine().getStatusCode(); 如果(状态==200){ HttpEntity e=response.getEntity(); 字符串数据=EntityUtils.toString(e); JSONArray时间线=新JSONArray(数据); 对于(int i=0;i
String value=timeline.getJSONString(“您的字符串名称”)嗨,对不起,我对Java还是有点陌生,您能解释一下您的意思吗?我正在获取twitter信息,所以它只是
公共静态字符串URL=”http://api.twitter.com/1/statuses/user_timeline.json?screen_name="
发布您的错误,我明天会看到。
公共JSONObject feedTimeline(字符串用户名)…上有一个错误。
行,它只是说这个方法应该返回JSONObject类型的结果