如何防止用户在React native中按两次按钮?
即。用户无法在可触摸的高亮显示屏上快速点按两次
答案 0 :(得分:17)
https://snack.expo.io/@patwoz/withpreventdoubleclick
使用此HOC扩展可触摸组件,如TouchableHighlight,Button ...
import debounce from 'lodash.debounce'; // 4.0.8
const withPreventDoubleClick = (WrappedComponent) => {
class PreventDoubleClick extends React.PureComponent {
debouncedOnPress = () => {
this.props.onPress && this.props.onPress();
}
onPress = debounce(this.debouncedOnPress, 300, { leading: true, trailing: false });
render() {
return <WrappedComponent {...this.props} onPress={this.onPress} />;
}
}
PreventDoubleClick.displayName = `withPreventDoubleClick(${WrappedComponent.displayName ||WrappedComponent.name})`
return PreventDoubleClick;
}
<强>用法强>
import { Button } from 'react-native';
import withPreventDoubleClick from './withPreventDoubleClick';
const ButtonEx = withPreventDoubleClick(Button);
<ButtonEx onPress={this.onButtonClick} title="Click here" />
答案 1 :(得分:5)
使用属性Button.disabled
import React, { Component } from 'react';
import { AppRegistry, StyleSheet, View, Button } from 'react-native';
export default class App extends Component {
state={
disabled:false,
}
pressButton() {
this.setState({
disabled: true,
});
// enable after 5 second
setTimeout(()=>{
this.setState({
disabled: false,
});
}, 5000)
}
render() {
return (
<Button
onPress={() => this.pressButton()}
title="Learn More"
color="#841584"
disabled={this.state.disabled}
accessibilityLabel="Learn more about this purple button"
/>
);
}
}
// skip this line if using Create React Native App
AppRegistry.registerComponent('AwesomeProject', () => App);
&#13;
答案 2 :(得分:3)
如果您使用反应导航,请使用此格式导航到另一页。
this.props.navigation.navigate({key:"any",routeName:"YourRoute",params:{param1:value,param2:value}})
StackNavigator将阻止将具有相同键的路由再次推入堆栈。
您可以编写任何独特的内容,因为key
和params
属性是可选的,如果您想将参数传递到另一个屏幕。
答案 3 :(得分:2)
我通过参考上面的答案来使用它。 &#39;禁用&#39;不必是一个州。
import React, { Component } from 'react';
import { TouchableHighlight } from 'react-native';
class PreventDoubleTap extends Component {
disabled = false;
onPress = (...args) => {
if(this.disabled) return;
this.disabled = true;
setTimeout(()=>{
this.disabled = false;
}, 500);
this.props.onPress && this.props.onPress(...args);
}
}
export class ButtonHighLight extends PreventDoubleTap {
render() {
return (
<TouchableHighlight
{...this.props}
onPress={this.onPress}
underlayColor="#f7f7f7"
/>
);
}
}
它可以是其他可触摸的组件,如TouchableOpacity。
答案 4 :(得分:2)
公认的解决方案效果很好,但是它必须包装整个组件并导入lodash以实现所需的行为。 我编写了一个自定义的React钩子,使它仅可以包装您的回调:
useTimeBlockedCallback.js
import { useRef } from 'react'
export default (callback, timeBlocked = 1000) => {
const isBlockedRef = useRef(false)
const unblockTimeout = useRef(false)
return (...callbackArgs) => {
if (!isBlockedRef.current) {
callback(...callbackArgs)
}
clearTimeout(unblockTimeout.current)
unblockTimeout.current = setTimeout(() => isBlockedRef.current = false, timeBlocked)
isBlockedRef.current = true
}
}
用法:
yourComponent.js
import React from 'react'
import { View, Text } from 'react-native'
import useTimeBlockedCallback from '../hooks/useTimeBlockedCallback'
export default () => {
const callbackWithNoArgs = useTimeBlockedCallback(() => {
console.log('Do stuff here, like opening a new scene for instance.')
})
const callbackWithArgs = useTimeBlockedCallback((text) => {
console.log(text + ' will be logged once every 1000ms tops')
})
return (
<View>
<Text onPress={callbackWithNoArgs}>Touch me without double tap</Text>
<Text onPress={() => callbackWithArgs('Hello world')}>Log hello world</Text>
</View>
)
}
默认情况下,回调在默认情况下被阻止1000毫秒,但您可以使用钩子的第二个参数对其进行更改。
答案 5 :(得分:1)
这是我的简单钩子。
import { useRef } from 'react';
const BOUNCE_RATE = 2000;
export const useDebounce = () => {
const busy = useRef(false);
const debounce = async (callback: Function) => {
setTimeout(() => {
busy.current = false;
}, BOUNCE_RATE);
if (!busy.current) {
busy.current = true;
callback();
}
};
return { debounce };
};
这可以在您喜欢的任何地方使用。即使它不是用于按钮。
const { debounce } = useDebounce();
<Button onPress={() => debounce(onPressReload)}>
Tap Me again and adain!
</Button>
答案 6 :(得分:0)
您还可以在等待某些异步操作时显示加载gif。只需确保使用onPress
标记async () => {}
,以便await
&#39; d。
import React from 'react';
import {View, Button, ActivityIndicator} from 'react-native';
class Btn extends React.Component {
constructor(props) {
super(props);
this.state = {
isLoading: false
}
}
async setIsLoading(isLoading) {
const p = new Promise((resolve) => {
this.setState({isLoading}, resolve);
});
return p;
}
render() {
const {onPress, ...p} = this.props;
if (this.state.isLoading) {
return <View style={{marginTop: 2, marginBottom: 2}}>
<ActivityIndicator
size="large"
/>
</View>;
}
return <Button
{...p}
onPress={async () => {
await this.setIsLoading(true);
await onPress();
await this.setIsLoading(false);
}}
/>
}
}
export default Btn;
答案 7 :(得分:0)
我有一个使用runAfterInteractions的非常简单的解决方案:
_GoCategoria(_categoria,_tipo){
if (loading === false){
loading = true;
this.props.navigation.navigate("Categoria", {categoria: _categoria, tipo: _tipo});
}
InteractionManager.runAfterInteractions(() => {
loading = false;
});
};
答案 8 :(得分:0)
我的包装器组件的实现。
import React, { useState, useEffect } from 'react';
import { TouchableHighlight } from 'react-native';
export default ButtonOneTap = ({ onPress, disabled, children, ...props }) => {
const [isDisabled, toggleDisable] = useState(disabled);
const [timerId, setTimerId] = useState(null);
useEffect(() => {
toggleDisable(disabled);
},[disabled]);
useEffect(() => {
return () => {
toggleDisable(disabled);
clearTimeout(timerId);
}
})
const handleOnPress = () => {
toggleDisable(true);
onPress();
setTimerId(setTimeout(() => {
toggleDisable(false)
}, 1000))
}
return (
<TouchableHighlight onPress={handleOnPress} {...props} disabled={isDisabled} >
{children}
</TouchableHighlight>
)
}
答案 9 :(得分:0)
同意接受但很简单的方法,我们可以使用以下方法
import debounce from 'lodash/debounce';
componentDidMount() {
this.onPressMethod= debounce(this.onPressMethod.bind(this), 500);
}
onPressMethod=()=> {
//what you actually want on button press
}
render() {
return (
<Button
onPress={() => this.onPressMethod()}
title="Your Button Name"
/>
);
}