如何将此 json [Blogger] 解析为 android
how to parse this json [Blogger] to android
我在将此博主 JSON 文件解析为 android 时遇到问题
blogger JSON structure
我想在字符串变量中显示标题请帮助我
我试过这个代码
@Override
protected Boolean doInBackground(String... urls) {
try {
//------------------>>
HttpGet httppost = new HttpGet(urls[0]);
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response = httpclient.execute(httppost);
// StatusLine stat = response.getStatusLine();
int status = response.getStatusLine().getStatusCode();
if (status == 200) {
HttpEntity entity = response.getEntity();
String data = EntityUtils.toString(entity);
JSONObject jsono = new JSONObject(data);
JSONArray jarray = jsono.getJSONObject("feed").getJSONArray("entry");
for (int i = 0; i < jarray.length(); i++) {
JSONObject object = jarray.getJSONObject(i);
Actors actor = new Actors();
actor.setName(object.getString("title"));
actorsList.add(actor);
}
return true;
}
//------------------>>
} catch (ParseException e1) {
e1.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return false;
}
我必须将上面嵌套的 JSON 数组的数据解析到我的应用程序中。我很困惑如何从中获取值。
entry
不在根对象中,而是在 feed
对象中,所以使用
//...
JSONObject jsono = new JSONObject(data);
JSONArray jarray = jsono.getJSONObject("feed").getJSONArray("entry");
//...
我在将此博主 JSON 文件解析为 android 时遇到问题 blogger JSON structure
我想在字符串变量中显示标题请帮助我
我试过这个代码
@Override
protected Boolean doInBackground(String... urls) {
try {
//------------------>>
HttpGet httppost = new HttpGet(urls[0]);
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response = httpclient.execute(httppost);
// StatusLine stat = response.getStatusLine();
int status = response.getStatusLine().getStatusCode();
if (status == 200) {
HttpEntity entity = response.getEntity();
String data = EntityUtils.toString(entity);
JSONObject jsono = new JSONObject(data);
JSONArray jarray = jsono.getJSONObject("feed").getJSONArray("entry");
for (int i = 0; i < jarray.length(); i++) {
JSONObject object = jarray.getJSONObject(i);
Actors actor = new Actors();
actor.setName(object.getString("title"));
actorsList.add(actor);
}
return true;
}
//------------------>>
} catch (ParseException e1) {
e1.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return false;
}
我必须将上面嵌套的 JSON 数组的数据解析到我的应用程序中。我很困惑如何从中获取值。
entry
不在根对象中,而是在 feed
对象中,所以使用
//...
JSONObject jsono = new JSONObject(data);
JSONArray jarray = jsono.getJSONObject("feed").getJSONArray("entry");
//...