。为什么每次投票都是负面的..我是android新手;(。你的建议可能会帮助其他人.. 我有一个申请。登录时,我想将数据存储到共享首选项。
我正在尝试将数据从服务器保存到共享首选项
中。我已成功从服务器接收数据
我尝试过,但出现一些错误
{"Login":"Success","Login Details":[{"name":"abc","place":"abcdd","cityname":"asdf"}]}
代码是:
public class login extends AsyncTask<Void, Void, Void> {
InputStream ins;
String status,details, result, s = null, data = "", js;
int ss;
int responseCode;
@Override
protected void onPreExecute() {
super.onPreExecute();
}
@Override
protected Void doInBackground(Void... params) {
StringBuilder sb = new StringBuilder();
ArrayList al;
try {
URL url =new URL(BuildConfig.url);
String param = "username=" + uname + "&password=" + pass2;
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("POST");
connection.setConnectTimeout(15000);
connection.setReadTimeout(15000);
connection.setDoInput(true);
connection.setDoOutput(true);
OutputStream os = connection.getOutputStream();
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
bw.write(param);
bw.flush();
bw.close();
responseCode = connection.getResponseCode();
if (responseCode == HttpURLConnection.HTTP_OK) {
BufferedReader br = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line = "";
while ((line = br.readLine()) != null) {
sb.append(line + "\n");
}
}
data = sb.toString();
JSONObject json = new JSONObject(data);
status = json.getString("Login");
} catch (MalformedURLException e) {
Log.i("MalformedURLException", e.getMessage());
} catch (IOException e) {
Log.i("IOException", e.getMessage());
} catch (JSONException e) {
Log.i("JSONException", e.getMessage());
}
return null;
}
protected void onPostExecute(Void result) {
super.onPostExecute(result);
String status1 = status.trim();
if (status1.equals("Success")) {
try {
JSONObject jsonObject = new JSONObject(status1);
JSONArray array=jsonObject.getJSONArray("Login Details");
for (int i=0;i<array.length();i++){
jsonObject = array.getJSONObject(i);
name= jsonObject.getString("name");
place= jsonObject.getString("place");
cityname = jsonObject.getString("cityname");
}
} catch (JSONException e) {
e.printStackTrace();
}
SharedPreferences sharedPreferences = PreferenceManager.getDefaultSharedPreferences(getApplicationContext());
SharedPreferences.Editor editor = sharedPreferences.edit();
editor.putString("name", name);
editor.putString("place", place);
editor.apply();
Toast.makeText(Login.this,name+":"+cityname+":"+mobileno,Toast.LENGTH_SHORT).show();
}
else {
Toast.makeText(Login.this, "Username or Password is Incorrect", Toast.LENGTH_LONG).show();
}
}
}
最佳答案
您的“JSONArray”仅由一个 JSONObject 和您的 using 循环组成
try {
JSONObject jsonObject = new JSONObject(status1);
JSONArray array=jsonObject.getJSONArray("Login Details");
//"Login Details":[{"name":"abc","place":"abcdd","cityname":"asdf"}] only one object
for (int i=0;i<array.length();i++){
jsonObject = array.getJSONObject(i);
name= jsonObject.getString("name");
place= jsonObject.getString("place");
cityname = jsonObject.getString("cityname");
}
} catch (JSONException e) {
e.printStackTrace();
}
希望你能理解并且这个提示对你有用 并且请在解析和打印中使用相同的变量
关于android - 将 Web 服务数据存储到共享首选项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48031311/