Question

我正在为我正在计划的应用程序进行测试注册活动，并且我希望在用户成功注册后让用户注册该对象。这是我的一些代码：

public void signUp(View v){
String firstNameString = FirstName.getText().toString();
String familyNameString = LastName.getText().toString();
String emailString = email.getText().toString();

if (firstNameString.length()<2){
    Context context = getApplicationContext();
    CharSequence text = "Your first name must be at least 2 characters";
    int duration = Toast.LENGTH_SHORT;

    Toast toast = Toast.makeText(context, text, duration);
    toast.show();
}

if (familyNameString.length()<2){
    Context context = getApplicationContext();
    CharSequence text = "Your family name must be at least 2 characters";
    int duration = Toast.LENGTH_SHORT;

    Toast toast = Toast.makeText(context, text, duration);
    toast.show();
}
if (firstNameString.isEmpty()){
    Context context = getApplicationContext();
    CharSequence text = "Please Enter a First Name";
    int duration = Toast.LENGTH_SHORT;

    Toast toast = Toast.makeText(context, text, duration);
    toast.show();
}
if (familyNameString.isEmpty()){
    Context context = getApplicationContext();
    CharSequence text = "Please Enter a First Name";
    int duration = Toast.LENGTH_SHORT;

    Toast toast = Toast.makeText(context, text, duration);
    toast.show();
}
else {
    ParseUser user = new ParseUser();
    user.setUsername(firstNameString + familyNameString);
    user.setEmail(emailString);
    user.setPassword("null");

user.signUpInBackground(new SignUpCallback() {
    public void done(ParseException e) {
        if (e == null) {
        Context context = getApplicationContext();
            CharSequence success = "Sign up Successfull";
            int duration = Toast.LENGTH_LONG;

            Toast successfullTostConformation = Toast.makeText(context,success,duration);
            successfullTostConformation.show();
        }
        else {
            // Sign up didn't succeed. Look at the ParseException
            // to figure out what went wrong
            Context context = getApplicationContext();
            CharSequence success = "Sign up failed. Check your internet connection";
            int duration = Toast.LENGTH_LONG;

            Toast successfullTostConformation = Toast.makeText(context,success,duration);
            successfullTostConformation.show();
        }
    }
});
        }



}

这是我注册按钮的onclick监听器。因此，正如我所说，我如何获取用户的对象ID，将其存储在字符串中，然后在本地保存。我已尝试过第一步：

String userID = user.getObjectID

但我在user.getObjectID();下有错误 从内部类中访问变量'user'，需要声明为final

编辑：哦。我必须将String userID = user.getObjectID置于ParseUser user = new ParseUser();之下。感谢任何试图提供帮助的人

Answer 1

喔。我必须将String userID = user.getObjectID置于ParseUser user = new ParseUser();之下。感谢任何试图提供帮助的人

Android - 如何从Parse获取用户的对象ID并将其存储在字符串

1 个答案: