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firebase - 如何根据特定条件显示/隐藏按钮?

转载 作者:IT王子 更新时间:2023-10-29 07:20:50 27 4
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我正在尝试弄清楚如何在用户登录或注销时隐藏我的登录/注册按钮。

我已经研究了 VISIBILITY 和 GONE,两者都可以。但是,我找不到任何与我要完成的目标相匹配的教程。

以下是我要完成的一些代码片段:

从原始帖子编辑:

在 authentication.dart 中(只有函数):

loginUser(context) async {
formkey.currentState.save();
if (formkey.currentState.validate()) {
await _auth
.signInWithEmailAndPassword(email: _email, password: _password)
.catchError((e) {
home.changeState(null);
Fluttertoast.showToast(
msg: "Invalid email and/or password. Please try again",
toastLength: Toast.LENGTH_LONG,
gravity: ToastGravity.TOP,
timeInSecForIos: 5,
backgroundColor: Colors.red,
textColor: Colors.white,
fontSize: 16.0
);
}).then((newUser) {
var now = new DateTime.now();
Firestore.instance
.collection('users')
.document(newUser.uid)
.collection('userInfo')
.document('userInfo')
.setData({
'Last login': now,
}).catchError((e) {
Fluttertoast.showToast(
msg: "Update user failed: $e",
toastLength: Toast.LENGTH_LONG,
gravity: ToastGravity.TOP,
timeInSecForIos: 1,
backgroundColor: Colors.red,
textColor: Colors.white,
fontSize: 16.0
);
});
welcomeUser();
Navigator.of(context).pop();
});
}
final user = await getSignedInUser();
home.changeState(user);
}

getSignedInUser() async {

mCurrentUser = await FirebaseAuth.instance.currentUser();
if(mCurrentUser == null || mCurrentUser.isAnonymous){
return null;
}
else{
return mCurrentUser;
}
}

signOutUser () async{
await _auth.signOut();
final user = await getSignedInUser();
home.changeState(user);
}



HomeScreen home; //create instance of homescreen class, so I could call my function

在 loginScreen 中(创建登录按钮的部分):

Expanded(
child: OutlineButton(
child: Text("Login "),
onPressed: () =>
loginUser(context)),
flex: 1,
),

homeScreen类(省略无关功能):

class HomeScreen extends State<MyApp> {
FirebaseUser currentUser; //not sure what to initialize this to upon loading app because user might still be signed in.
final formkey = GlobalKey<FormState>();
@override
Widget build(BuildContext context){
return MaterialApp(
home: Scaffold(
resizeToAvoidBottomPadding: false,
appBar: AppBar(title: Text("HikeLocator"), backgroundColor: Colors.green[700],
),
body: Container(
margin: EdgeInsets.all(20.0),
child: Form(
key: formkey,
child:
Column(
children: <Widget>[
distanceFromUser(),
lengthOfTrail(),
numOfResults(),
Container(
margin: EdgeInsets.only(top: 25.0),
),
submitButton(),
// loginButton(),
//signupButton(),
//logoutButton(),
//profile(),
] + (currentUser == null ? [
loginButton(),
signupButton(),
] : [ // logged in? show the following widgets
logoutButton(),
profile(),
]),
),
)
),
),
);
}


Widget loginButton() {
return RaisedButton(
color: Color.fromRGBO(58, 66, 86, 1.0),
child: Text("Log In", style: TextStyle(color: Colors.white)),
onPressed: () {
Navigator.push(
context,
MaterialPageRoute(builder: (context) => LogInScreen()),
);
},
);
}



Widget logoutButton(){
return RaisedButton(
color: Color.fromRGBO(58, 66, 86, 1.0),
child: Text("Log Out", style: TextStyle(color: Colors.white)),
onPressed: () async {
await signOutUser();
}
);
}
Widget signupButton() {
return RaisedButton(
color: Color.fromRGBO(58, 66, 86, 1.0),
child: Text("Sign Up", style: TextStyle(color: Colors.white)),
onPressed: () {
Navigator.push(
context,
MaterialPageRoute(builder: (context) => SignUpScreen()),
);
},
);
}
changeState(value){
setState(() {
this.currentUser = value;
});
}
}


我有一个我写的函数,叫做 getSignedInUser()。如果没有,则返回 null。我是新手。上面在注释中指定的按钮显示了在某些情况下应该可见的按钮。感谢您的帮助。

最佳答案

我会选择这个:

Widget loginButton() {
return getSignedInUser() == null
? RaisedButton(
color: Color.fromRGBO(58, 66, 86, 1.0),
child: Text("Log In", style: TextStyle(color: Colors.white)),
onPressed: () {
Navigator.push(
context,
MaterialPageRoute(builder: (context) => LogInScreen()),
);
},
)
: Container();
}

如果用户已登录,则返回一个空容器,否则,返回登录按钮。您可以在返回小部件的其他函数中执行相同的操作。

您还可以将登录按钮包裹在 Opacity 小部件中,并将 opacity 设置为 0.0 以使其隐藏,但它仍会占用空间。

更新:我准备了以下示例来演示一个简单的登录页面。 getUser 函数模拟一个函数,该函数对用户进行身份验证并在登录成功时返回用户名(两秒后)。

import 'package:flutter/material.dart';

class LoginPage extends StatefulWidget {
@override
LoginPageState createState() {
return new LoginPageState();
}
}

class LoginPageState extends State<LoginPage> {
String user;

Future<String> getUser() {
return Future.delayed(
Duration(seconds: 2),
() {
return "john";
},
);
}

void login() async {
final user = await getUser();
setState(() {
this.user = user;
});
}

void logout() {
setState(() {
this.user = null;
});
}

Widget _buildLoginButton() {
return user == null
? RaisedButton(
onPressed: login,
child: Text("Login"),
)
: Container();
}

Widget _buildLogoutButton() {
return (user != null)
? RaisedButton(
onPressed: logout,
child: Text("Logout"),
)
: Container();
}

@override
Widget build(BuildContext context) {
return Scaffold(
appBar: AppBar(
title: Text("Login"),
),
body: Center(
child: Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
_buildLoginButton(),
_buildLogoutButton(),
],
),
),
);
}
}

关于firebase - 如何根据特定条件显示/隐藏按钮?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55716645/

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