【发布时间】:2018-10-23 13:52:07
【问题描述】:
如果用户登录与否,我正在尝试使用反应导航身份验证流程来管理登录屏幕。但现在我陷入了 AsyncStorage。因此,当用户未登录时,我认为 componentWillMount 将等到用户输入凭据,点击登录按钮,从 API 调用接收 user_id,然后重试。对我来说,现在它在开始时调用什么很好,但随后我必须退出应用程序并返回以获取仪表板。有什么解决办法吗?
这是我在 App.js 中创建路由的代码。我也在底部加载 redux 地图。
export const createRootNavigator = (signedIn = false) => {
return SwitchNavigator(
{
SignedIn: {
screen: SignedIn
},
SignedOut: {
screen: SignedOut
}
},
{
initialRouteName: signedIn ? "SignedIn" : "SignedOut"
}
);
};
class App extends Component {
constructor(props) {
super(props);
this.state = {
signedIn: false,
checkedSignIn: false
};
}
async componentWillMount() {
await isSignedIn()
.then(res => this.setState({ signedIn: res, checkedSignIn: true }))
.catch(err => alert("An error occurred"));
}
render() {
const { checkedSignIn, signedIn } = this.state;
// If we haven't checked AsyncStorage yet, don't render anything (better ways to do this)
if (!checkedSignIn) {
return null;
}
const Layout = createRootNavigator(signedIn);
return (
<SafeAreaView style={styles.safeArea}>
<View style={{flex: 1, backgroundColor: '#ffffff'}}>
<StatusBar barStyle="light-content"/>
<Layout />
<AlertContainer/>
</View>
</SafeAreaView>
)
}
};
这是我正在等待 user_key 的 Auth.js。
export let USER_KEY = 'myKey';
export const onSignIn = async () => { await AsyncStorage.setItem(USER_KEY, 'true') };
export const onSignOut = async () => { await AsyncStorage.removeItem(USER_KEY) };
export const isSignedIn = () => {
return new Promise((resolve, reject) => {
AsyncStorage.getItem(USER_KEY)
.then(res => {
if (res !== null) {
// console.log('true')
resolve(true);
} else {
resolve(false);
// console.log('false')
}
})
.catch(err => reject(err));
});
};
【问题讨论】:
标签: react-native async-await react-navigation