Man*_*ano -2 android android-json
我想"result"从下面的JSON响应中获取值并将其存储在本地.这是代码:
private class GetContacts extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... arg0) {
// Creating service handler class instance
ServiceHandler sh = new ServiceHandler();
// Making a request to url and getting response
String jsonStr = sh.makeServiceCall(url, ServiceHandler.GET);
if (jsonStr != null) {
try {
JSONObject jsonObj = new JSONObject(jsonStr);
//JSONArray contacts;
contacts = jsonObj.getJSONArray("response");
Log.d("Response: ", "> " + contacts);
} catch (JSONException e) {
e.printStackTrace();
}
} else {
Log.e("ServiceHandler", "Couldn't get any data from the url");
}
return null;
}
}
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我的回复 :
{"response":
[{
"name":"ajay",
"class":"7",
},
{
"rank":1
}],
"date":
{
"startdate":2/12/2012,
},
"result":"pass"
}
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您需要从json String创建一个JSON对象,然后获取并检索其数据:
JSONObject json= new JSONObject(responseString); //your response
try {
String result = json.getString("result"); //result is key for which you need to retrieve data
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
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希望能帮助到你.
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