我正在开发一个基于 React Native 的应用程序,它有一个登录选项。我正在使用 firebase 身份验证。我设法从 firebase 捕获当前用户 uid。我的问题是我无法保持登录状态。每次我打开应用程序时,它都会要求我一次又一次地登录。
这是我的代码:
我知道我应该使用异步,但我不知道该怎么做。
import React from 'react';
import { StyleSheet,
Text,
View,
TouchableOpacity,
AsyncStorage,
} from 'react-native';
import {RkTextInput, RkButton } from 'react-native-ui-kitten';
import {Actions} from 'react-native-router-flux';
import { createSwitchNavigator, createAppContainer } from 'react-navigation';
import Profile from "../Profile/Profile";
import * as firebase from 'firebase';
export default class Login extends React.Component {
constructor(props){
super(props)
this.state=({
email:'savadks1919@gmail.com',
password:'123123',
userId:'',
errorMessage: null
})
}
signup() {
Actions.signup()
}
Home() {
Actions.home()
}
handleLogin = (email, password) => {
firebase.auth().signInWithEmailAndPassword(email, password).then(
this.Home,
this.state=({userId:firebase.auth().currentUser.uid})
).catch(function(error) {
var errorCode = error.code;
var errorMessage = error.message;
if (errorCode === 'auth/wrong-password') {
alert('Wrong password.');
} else {
alert(errorMessage);
}
console.log(error);
});
}
render() {
return (
<View style={styles.container}>
<Text style={styles.titleText}>Taams</Text>
<Text style={styles.edition}>Developer's Edition</Text>
<Text style={styles.titleText}>Login.js</Text>
<Text>Alpha 0.0.0.1</Text>
{/*-----UserName Input-------*/}
<RkTextInput
rkType= 'rounded'
labelStyle= {{color: 'black', fontWeight: 'bold'}}
placeholder='UserName'
//--------------value Handler----------------//
onChangeText={(email) => this.setState({email})}
//---------------------------------//
selectionColor="#000000"
keyboardType="email-address"
onSubmitEditing={() => { this.password.focusInput(); }}
inputStyle={{
color: 'black',
fontWeight: 'bold',
}}/>
{/*-----Password-------*/}
<RkTextInput
secureTextEntry={true}
rkType= 'rounded'
placeholder='Password'
//--------------value Handler----------------//
onChangeText={(password) => this.setState({password})}
//---------------------------------//
ref={(input) => { this.password = input; }}
inputStyle={{
color: 'black',
fontWeight: 'bold',
}}/>
<RkButton onPress = {()=>this.handleLogin(this.state.email,this.state.password)}>
<Text style={styles.LoginButtonText}>Login</Text>
</RkButton>
<View style={styles.signupTextCont}>
<Text style={styles.signupText}>Don't have an account yet?</Text>
<TouchableOpacity onPress={this.signup}><Text style={styles.signinButton}>SignUp</Text></TouchableOpacity>
</View>
</View>
);
}
}
const styles = StyleSheet.create({
container: {
flex: 1,
backgroundColor: '#fff',
alignItems: 'center',
justifyContent: 'center',
},
signupTextCont:{
flexGrow: 0,
alignItems:'center',
justifyContent:'flex-end',
marginVertical:15
},
signupText:{
color:'rgba(64,64,64,0.6)',
fontSize:16
},
signinButton:{
color:'#000000',
fontSize:16,
fontWeight:'500'
},
titleText: {
fontSize: 20,
fontWeight: 'bold',
},
edition: {
fontSize: 15,
},
TextInput: {
width: 300,
height:50,
borderColor: 'grey',
borderWidth: 1,
},
LoginButtonText: {
fontSize: 20,
fontWeight: 'bold',
color: 'white',
},
});
最佳答案
在 AsyncStorage 中存储从 Firebase 登录请求获取的 userId,例如
Home(userId) {
(async ()=>{
await AsyncStorage.setItem("userId",userId)
Actions.home()
})()
}
然后下次打开应用程序时检查 AsyncStorage,例如
redirectToHome() {
(async ()=>{
let userId = await AsyncStorage.getItem("userId",userId)
if(userId){
Actions.home()
}
else {
//redirect to login page
}
})()
}
注销时清除 AsyncStorage 就像
AsyncStorage.clear()
关于javascript - 如何在关闭和打开后在 react native 应用程序上保持登录状态,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54364609/