如何将名称从firebase显示到颤动屏幕中[重复]

Question

  FirebaseUser loggedinuser;
  FirebaseAuth _auth = FirebaseAuth.instance;

  Future<void> currentUser()async
  {
    try
    {
      FirebaseUser user = await _auth.currentUser();
      if(user != null)
      {
        setState(() {
          loggedinuser  = user;
        });
      }
    }
    catch(e){
      print(e);
    }
  }
  @override
  void initState() {
    super.initState();
    currentUser();
  }



@override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text("Logged in"),
        centerTitle: true,
        backgroundColor: Colors.green,
      ),
      body: Container(
        alignment: Alignment.center,
        padding: EdgeInsets.symmetric(horizontal: 20,vertical: 20),
        child: Text(loggedinuser.providerId)

你好，我是 FLutter 新手，正在做一个小项目来练习 但是每当我尝试显示用户名时，我都会收到一条错误消息，说 Text 小部件不能为空，但我使用 setState 初始化了登录用户 我可以看到电子邮件，但屏幕上看不到名字 感谢您的帮助

Answer 1

我认为您应该像@nvoigt 所说的那样使用 FutureBuilder。但是您也可以解决这个问题，将这一行 Text(loggedinuser.providerId) 更改为：

Text(loggedinuser != null loggedinuser.providerId : "Loading / Not Logged in")

使用 FutureBuilder 你会得到这样的东西：

class LoadUser extends StatefulWidget {
  @override
  _LoadUserState createState() => _LoadUserState();
}

class _LoadUserState extends State<LoadUser> {
  FirebaseUser loggedinuser;
  FirebaseAuth _auth = FirebaseAuth.instance;

  Future<FirebaseUser> currentUser() async {
    try {
      FirebaseUser user = await _auth.currentUser();
      if (user != null) {
        return user;
      }
      return null;
    } catch (e) {
      return null;
    }
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text("FirebaseUser"),
        centerTitle: true,
        backgroundColor: Colors.green,
      ),
      body: Container(
        alignment: Alignment.center,
        padding: EdgeInsets.symmetric(horizontal: 20, vertical: 20),
        child: FutureBuilder<FirebaseUser>(
          future: currentUser(),
          builder: (context, snapshot) {
            switch (snapshot.connectionState) {
              case ConnectionState.waiting:
              case ConnectionState.active:
                return Center(child: CircularProgressIndicator(),);
                break;
              case ConnectionState.done:
                loggedinuser = snapshot.data;
                return Text(loggedinuser.providerId);
                break;
              default:
                return Text('Not logged in :(');
                break;
            }
          },
        )
      ),
    );
  }
}

Answer 2

您的currentUser() 函数是异步的，实例化loggedinuser 需要时间。所以最初loggedinuser 是null，这就是你收到错误的原因。要提出解决方案，请将Text 小部件与FutureBuilder 包装在一起。

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text("Logged in"),
        centerTitle: true,
        backgroundColor: Colors.green,
      ),
      body: Container(
        alignment: Alignment.center,
        padding: EdgeInsets.symmetric(horizontal: 20, vertical: 20),
        child: FutureBuilder(
          future: currentUser(),
          builder: (context, snapshot) {
            if (snapshot.connectionState == ConnectionState.active) {
              return Text(loggedInUser.providerId.toString());
            } else if (snapshot.connectionState == ConnectionState.waiting) {
              return CircularProgressIndicator();
            }
          },
        ),
      ),
    );
  }