Java 简单JSON服务器
如何在android中解析这种类型的json:Java 简单JSON服务器,java,android,json,parsing,Java,Android,Json,Parsing,如何在android中解析这种类型的json: { "chatCircles": [{ "id": "59660c155d44f20b46eab438", "name": "Hyderabad Circle", "description": "Hyderabad" }] } 我已经尝试过了,但是UI没有更新 try { JSONArray chatCircleArray = response.getJSONArray("chatCircle");
{
"chatCircles": [{
"id": "59660c155d44f20b46eab438",
"name": "Hyderabad Circle",
"description": "Hyderabad"
}]
}
我已经尝试过了,但是UI没有更新
try {
JSONArray chatCircleArray = response.getJSONArray("chatCircle");
for (int i = 0; i < chatCircleArray.length(); i++){
JSONObject geoFenceObject = chatCircleArray.getJSONObject(i);
circleId = geoFenceObject.getString("id");
String circleName = geoFenceObject.getString("name");
String location = geoFenceObject.getString("description");
Log.d(TAG, "GeoFenceObject Chat Circle Id parsed is:\t " + circleId);
System.out.println("Chat Circle id is: \t" + circleId);
} catch (JSONException e) {
e.printStackTrace();
}
试试看{
JSONArray chatcirclarray=response.getJSONArray(“聊天圈”);
对于(int i=0;i
请帮助。如何进行json解析?第一个大括号是您没有解析的JSONObject。 请尝试下面的代码
try {
JSONObject json = new JSONObject(response);
JSONArray itemArray = json.getJSONArray("chatCircles");
for (int i = 0; i < itemArray.length(); i++) {
JSONObject subJson = (JSONObject) itemArray.get(i);
String id = subJson.getString("id");
String name = subJson.getString("name");
String description = subJson.getString("description");
Log.i("TEST", "id=" + id + ", name=" + name + ", description=" + description);
}
} catch (Exception e) {
e.printStackTrace();
}
试试看{
JSONObject json=新的JSONObject(响应);
JSONArray itemArray=json.getJSONArray(“聊天圈”);
对于(int i=0;i
getJSONArray(“聊天圈”);
..?你是说getJSONArray(“聊天圈”);
?是的,非常感谢你这么做了。现在需要更加小心。谢谢大家