如何解决为什么存储在变量中的JSX代码从不渲染?
理想情况下,该代码应检查用户是否已登录应用程序,如果是,请导航至主屏幕,否则,该屏幕应显示包含登录按钮和注册按钮的组件。
import React,{ useEffect,useState } from "react";
import { View,ActivityIndicator } from "react-native";
import WelcomeButtons from "../components/welcomeButtons";
import Firebase from "../fireConfig";
const LoadingScreen = (props) => {
let content = <ActivityIndicator />;
useEffect(() => {
Firebase.auth().onAuthStateChanged(function (user) {
if (user) {
console.log("User Signed in");
props.navigation.navigate("Home");
} else {
content = (
<WelcomeButtons
onSignUpPress={() => props.navigation.navigate("Register")}
onSignInPress={() => props.navigation.navigate("Login")}
/>
);
console.log("User not signed in");
}
});
});
return <View>{content}</View>;
};
export default LoadingScreen;
当用户未登录时,活动指示器将加载,但welcomeButtons组件将永远不会呈现。另外,当用户未登录时,“用户未登录”将两次登录到控制台。
解决方法
您正在异步将组件设置为变量,因此当组件返回JSX表达式时,内容仍包含<ActivityIndicator />
元素。要实现此行为,您应该创建一个状态变量,并根据其值返回JSX,就像这样
import React,{ useEffect,useState } from "react";
import { View,ActivityIndicator } from "react-native";
import WelcomeButtons from "../components/welcomeButtons";
import Firebase from "../fireConfig";
const LoadingScreen = (props) => {
let content = <ActivityIndicator />;
const [signedIn,setSignedIn] = useState(false); //false is the default value
useEffect(() => {
Firebase.auth().onAuthStateChanged(function (user) {
if (user) {
console.log("User Signed in");
props.navigation.navigate("Home");
setSignedIn(false);
} else {
setSignedIn(true);
console.log("User not signed in");
}
});
});
if (signedIn) {
content = (
<WelcomeButtons
onSignUpPress={() => props.navigation.navigate("Register")}
onSignInPress={() => props.navigation.navigate("Login")}
/>
);
}
return <View>{content}</View>;
};
export default LoadingScreen;
要了解有关状态的更多信息,可以在此处查看React的官方文档:
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