I am implementing Splash Screen in flutter along with it i am using sharedPreferences I am facing late initialisation error in it

Question

Here I have implemented splash screen and in main.dart I assigned it as home:SplashScreen(); Here I have also implemented sharedPreferences now I am facing late initialisation error for:static late SharedPreferences pref; I can't understand how resolve it and where to declare pref variable import 'package:flutter/material.dart';

import 'package:onesignal_flutter/onesignal_flutter.dart';
import 'package:shared_preferences/shared_preferences.dart';

import 'View/LoginPage.dart';
import 'View/homePageAdmin.dart';

class SplashScreen extends StatefulWidget {
  @override
  State<SplashScreen> createState() => _SplashScreenState();
}

class SharedPrefService {
  static late SharedPreferences pref;

  static Future<void> init() async {
    await SharedPrefService.init();

    SharedPrefService.pref = await SharedPreferences.getInstance();
    var usrEmail = pref.getString('email');
    String usrname = pref.getString('username').toString();
  }
}

class _SplashScreenState extends State<SplashScreen> {
  String playerId = '';

  var usrEmail = SharedPrefService.pref.getString('email');
  String usrname = SharedPrefService.pref.getString('username').toString();
  @override
  void initState() async {
    SharedPrefService.pref = await SharedPreferences.getInstance();
    super.initState();
    await SharedPrefService.init();

    initPlatformState();

    _navigateToHome();
  }

  Future<void> initPlatformState() async {
    String usrname = SharedPrefService.pref.getString('username').toString();

    OneSignal.shared.setAppId('e89acaa4-5388-4e3a-bd69-44d197bdcbd7');
    // OneSignal.shared
    //     .promptUserForPushNotificationPermission()
    //     .then((accepted) {});
    final status = await OneSignal.shared.getDeviceState();
    final String? osUserID = status?.userId;
    print('The player id from main.dart ...........${osUserID}');
    // OneSignal.shared.setNotificationOpenedHandler(
    //     (OSNotificationOpenedResult result) async {
    //   var id1 = await result.notification.additionalData!["Docid"];

    //   final int docid = int.parse(id1).toInt();

    //   navigatorKey.currentState!.push(MaterialPageRoute(
    //       builder: (context) => DocumentsDetails(docid, usrname)));
    // });

    setState(() {
      this.playerId = osUserID!;
    });
    print('this.playerid from main.dart ${this.playerId}');
  }

  _navigateToHome() async {
    String pId = this.playerId;

    await Future.delayed(Duration(milliseconds: 1500), () {});
    Navigator.pushReplacement(
        context,
        MaterialPageRoute(
          builder: (context) => usrEmail == null
              ? LoginPage()
              : homePageAdmin(
                  pId,
                  usrname,
                ),
        ));
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: Center(
        child: Container(
          child: Text(
            'Splash Screen',
            style: TextStyle(
                fontSize: 25,
                fontWeight: FontWeight.bold,
                color: Colors.blueAccent),
          ),
        ),
      ),
    );
  }
}

score 0 · Answer 1 · answered Dec 19 '22 at 10:08

0

This error occurs when you use late variable before its initialization.

Solution : making late variable to nullable variable, for example:

class SharedPrefService {
  static SharedPreferences? pref; // nullable variable

  static Future<void> init() async {
    await SharedPrefService.init();

    SharedPrefService.pref = await SharedPreferences.getInstance();
    var usrEmail = pref?.getString('email'); 
    String usrname = pref!.getString('username').toString();
  }
}

answered Dec 19 '22 at 10:08

Yashraj

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So now it will crash with a *different* error message? Does not seem to be the best solution. – nvoigt Dec 19 '22 at 10:11
can you share that error message? – Yashraj Dec 19 '22 at 10:23
Well, if it's `null` instead of not set, the program wanting to do something with it will still not work, will it? – nvoigt Dec 19 '22 at 10:38
@nvoigt you are right but my solution is for get rid of late initialize error. – Yashraj Dec 19 '22 at 11:00

score 0 · Answer 2 · answered Dec 19 '22 at 10:10

You cannot just make initState async and expect that to work. Yes, it will be async, but since it is still returning void, it cannot be waited on. So, it will be called and then your program will continue without waiting for it.

You seem to have a Future and not really an idea how to handle this inside a Flutter widget. I suggest getting a good overview:

What is a Future and how do I use it?

The solution is to use some kind of state management to handle the fact that you are doing something in the background. Your display cannot wait for it. It has to be displayed. The obvious solution is some kind of spinner or loading screen while you are waiting for the Future to resolve. There is a simple example of a FutureBuilder in the link above, or you can use more powerful solutions. Personally, I prefer bloc, but there are many others, too.

I am implementing Splash Screen in flutter along with it i am using sharedPreferences I am facing late initialisation error in it

2 Answers2