颤振如何像变量一样使用Future返回值 [英] Flutter how to use Future return value as if variable

问题描述

我想获取 Future 返回值，并将其用作变量。
我有这个 Future 函数

I want to get Future return value and use it like variable. I have this Future function

  Future<User> _fetchUserInfo(String id) async {
    User fetchedUser;
    await Firestore.instance
        .collection('user')
        .document(id)
        .get()
        .then((snapshot) {
      final User user = User(snapshot);
      fetchedUser = user;
    });
    return fetchedUser;
  }

我想得到这样的价值

final user = _fetchUserInfo(id);

但是当我尝试使用这种方式

However when I tried to use like this

new Text(user.userName);

Dart无法识别为 User 类。它说动态。

如何获得返回值并使用它？

首先我做错了吗？
任何帮助表示赞赏！

Dart doesn't recognize as User class. It says dynamic.
How Can I get return value and use it?
Am I doing wrong way first of all? Any help is appreciated!

推荐答案

您可以简化代码：

Future<User> _fetchUserInfo(String id) async {
    User fetchedUser;
    var snapshot = await Firestore.instance
        .collection('user')
        .document(id)
        .get();
    return User(snapshot);
  }

您还需要async / await来获取值

you also need async/await to get the value

void foo() async {
  final user = await _fetchUserInfo(id);
}

这篇关于颤振如何像变量一样使用Future返回值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！