我在React Native中调用函数时遇到问题。我只是想更改“上下文”的值。这是一些代码,首先是“上下文”的脚本:
//LogContext.js
import React, { useState } from 'react'
export const LogContext = React.createContext({
set: "en",
login: "false"
})
export const LogContextProvider = (props) => {
const setLog = (login) => {
setState({set: "jp", login: login})
}
const initState = {
set: "en",
login: "false"
}
const [state, setState] = useState(initState)
return (
<LogContext.Provider value={state}>
{props.children}
</LogContext.Provider>
)
}
和'app.js'代码:
//app.js
import React, { useState, useContext } from 'react';
import { Button, Text, TextInput, View } from 'react-native';
import { NavigationContainer } from '@react-navigation/native';
import { createStackNavigator } from '@react-navigation/stack';
import { LogContextProvider, LogContext } from './LogContext'
function HomeScreen({ navigation }) {
const state = useContext(LogContext);
return (
<>
<View style={{ flex: 1, alignItems: 'center', justifyContent: 'center' }}>
<Text>Passed config: {JSON.stringify({state})}</Text>
<Text>Home Screen</Text>
</View>
{state.login === 'false' ? (
<Button
title="Go to Login"
onPress={() => navigation.navigate('Login')}
/>
) : (
<Button title="Stuff" onPress={() => navigation.navigate('DoStuff')} />
)}
</>
);
}
function LoginScreen({ navigation }) {
const state = useContext(LogContext);
//do stuff to login here...
state.setLog('true'); //not functional...
return (
<LogContext.Provider value={'true'}> //value={'true'} also not functional...
<View style={{ flex: 1, alignItems: 'center', justifyContent: 'center' }}>
<Text>Login Screen</Text>
<Button title="Go to Home" onPress={() => navigation.navigate('Home')} />
</View>
</LogContext.Provider>
);
}
function StuffScreen({ navigation }) {
//do other stuff here...
}
const Stack = createStackNavigator();
function App() {
return (
<NavigationContainer>
<Stack.Navigator>
<Stack.Screen name="Home" component={HomeScreen} />
<Stack.Screen name="Login" component={LoginScreen} />
<Stack.Screen name="DoStuff" component={StuffScreen} />
</Stack.Navigator>
</NavigationContainer>
);
}
export default App;
显然,我对React Native不太熟悉。任何有关如何调用“ setLog()”函数以启用“ Context”全局变量的值更新的建议都将不胜感激。我先谢谢你。
我正尝试修改我的“ App()”函数,以按照另一位用户的建议将导航器包装在提供程序中。但是,以下内容完全无法正常工作...建议:
const Stack = createStackNavigator();
function App() {
const [data, setData] = useState({
set: 'en',
login: 'false',
});
const state = { data, setData };
return (
<LogContext.Provider value={state}>
<NavigationContainer>
{state.data.login === 'true' ? (
<Stack.Navigator>
<Stack.Screen name="BroadCast" component={VideoScreen} />
<Stack.Screen name="Logout" component={LogoutScreen} />
</Stack.Navigator>
) : (
<Stack.Navigator>
<Stack.Screen name="Login" component={LoginScreen} />
<Stack.Screen name="Details" component={DetailsScreen} />
<Stack.Screen name="Home" component={HomeScreen} />
</Stack.Navigator>
)}
</NavigationContainer>
</LogContext.Provider>
);
}
答案 0 :(得分:0)
您遇到的问题是您的上下文中没有set函数,我看不到需要单独的LogContext提供程序函数。
您可以简单地在app.js或任何根函数中执行该操作。下面的例子做到了。您可以看到如何将状态值与用于设置值的函数一起传递,并且可以通过提供程序内部的Login组件对其进行修改。如果使用单独的提供程序,则会有些混乱。下面是一个工作示例,其中没有导航部分,可以使您有所了解。
const LogContext = createContext({
data: {
set: 'en',
login: 'false',
},
});
export default function App() {
const [data, setData] = useState({
set: 'en',
login: 'false',
});
const state = { data, setData };
return (
<LogContext.Provider value={state}>
<View style={{ flex: 1 }}>
<Text>{JSON.stringify(state.data)}</Text>
<Login />
</View>
</LogContext.Provider>
);
}
const Login = () => {
const state = React.useContext(LogContext);
return (
<View>
<Button
onPress={() => state.setData({ set: 'bb', login: 'true' })}
title="Update"
/>
</View>
);
};
要修改代码,应将主导航器包装在LogContext.Provider内,并保持在那里的状态,这将帮助您完成其余工作。 请随时提出任何进一步的说明:)