Question

我想在用户成功登录 flutter 后保留用户令牌。我正在尝试将设备令牌保存到 SharedPreferences，以便我可以在其他小部件中使用它，但错误始终是：无效参数（值） : 不能为空。如何保存 ID 和访问令牌？

我的代码：

root = tk.Tk()

    figure1 = plt.Figure(figsize=(2,2), dpi=100)

    ax1 = figure1.add_subplot(221)
    ax1.plot(df1['year'], df1['personal'], color='red')
    scatter1 = FigureCanvasTkAgg(figure1, root)
    scatter1.get_tk_widget().pack()
    ax1.legend([''])
    ax1.set_xlabel('valeur de personals')
    ax1.set_title('ev de personal ')

    figure2 = plt.Figure(figsize=(2,2), dpi=100)
    ax2 = figure2.add_subplot(222)
    scatter2 = FigureCanvasTkAgg(figure2, root)
    scatter2.get_tk_widget().pack(side=tk.RIGHT)
    ax2.legend([''])
    ax2.set_xlabel('valeur BSA')
    ax2.set_title('Evolutiion des valeurs BSA depuis 1990 ')
    ax2.plot(df2['year'], df2['value'], color='red')

    figure3 = plt.Figure(figsize=(2,2), dpi=100)
    ax3 = figure3.add_subplot(223)
    #the same code for the reste 


    root.mainloop()

还有用户模型：

Future<User> login(String username, String password) async {
    await checkInternet();

    Map<String, String> headers = {
      'Content-type': 'application/json',
      'Accept': 'application/json',
      // 'Authorization': 'Bearer $token'
    };
    Map<String, String> body = {'username': username, 'password': password};

    var response = await http.post(Uri.parse(ApiUtil.AUTH_LOGIN),
        headers: headers, body: jsonEncode(body));
    switch (response.statusCode) {
      case 200:
        var body = jsonDecode(response.body);
        // print(body);
        var data = body['user'];
        // print(data);
        User user = User.fromJson(data);
        await _saveUser(user.id, user.access_token);
        return user;
      case 404:
        throw ResourceNotFound('User');
        break;
      case 422:
        throw UnProcessedEntity();
        break;
      case 401:
        throw LoginFailed();
      default:
        return null;
        break;
    }
  }

  Future<void> _saveUser(int id, String access_token) async {
    SharedPreferences sharedPreferences = await SharedPreferences.getInstance();
    sharedPreferences.setInt('id', id);
    sharedPreferences.setString('access_token', access_token);
  }
}

Answer 1

首先检查您要保存的值是否不为空且 setInt 或 setString 是 async 函数，因此您需要在它们之前添加 await

Future<void> _saveUser(int id, String access_token) async {
    SharedPreferences sharedPreferences = await SharedPreferences.getInstance();
    await sharedPreferences.setInt('id', id);
    await sharedPreferences.setString('access_token', access_token);
  }

如果您已经注意到，您的 User 类不会返回 .fromJson

class User {
  int id;
  String name;
  String email;
  String username;
  String phone;
  String access_token;
  String activeboxe_id;
  String adress;
  DateTime email_verifed_at;

  User(this.name, this.email, this.username, this.phone, this.activeboxe_id,
      this.adress, this.email_verifed_at,
      [this.access_token, this.id]);
  
// return a User object here
  factory User.fromJson(Map<String, dynamic> jsonObject) => User({
    this.id = jsonObject['id'];
    this.name = jsonObject['name'];
    this.email = jsonObject['email'];
    this.phone = jsonObject['phone'];
    this.access_token = jsonObject['access_token'];
    this.activeboxe_id = jsonObject['activeboxe_id'];
    this.adress = jsonObject['adress'];
    this.email_verifed_at = jsonObject['email_verifed_at'];
    this.username = jsonObject['username'];
  });
}

如何使用共享首选项在颤振中保存令牌？

1 个答案: