React 应用程序在每次单击时都运行得非常慢。我正在使用react本机v0.40.0,下面是我的项目的依赖项。
{
"analytics-react-native": "^1.1.0",
"apisauce": "^0.7.0",
"babel-preset-es2015": "^6.18.0",
"es6-promise": "^4.0.5",
"flow-bin": "^0.36.0",
"geolib": "^2.0.22",
"immutable": "^3.8.1",
"intl": "^1.2.5",
"isomorphic-fetch": "^2.2.1",
"lodash": "^4.17.4",
"lodash.range": "^3.2.0",
"prop-types": "^15.5.10",
"raven-js": "^3.13.1",
"react": "^15.4.2",
"react-native": "^0.40.0",
"react-native-apple-healthkit-rn0.40": "^0.3.2",
"react-native-blur": "^2.0.0",
"react-native-button": "^1.7.1",
"react-native-checkbox": "^2.0.0",
"react-native-code-push": "^1.17.3-beta",
"react-native-datepicker": "^1.4.4",
"react-native-device-info": "^0.10.1",
"react-native-easy-toast": "^1.0.6",
"react-native-fbsdk": "^0.5.0",
"react-native-geocoder": "^0.4.5",
"react-native-gifted-chat": "^0.1.3",
"react-native-global-props": "^1.1.1",
"react-native-image-crop-picker": "^0.15.1",
"react-native-image-picker": "^0.25.1",
"react-native-image-slider": "^1.1.5",
"react-native-keyboard-aware-scroll-view": "^0.2.7",
"react-native-maps": "0.15.2",
"react-native-modal-dropdown": "^0.4.4",
"react-native-popup-menu": "^0.7.2",
"react-native-push-notification": "^2.2.1",
"react-native-radio-buttons": "^0.14.0",
"react-native-router-flux": "3.38.0",
"react-native-segmented-android": "^1.0.4",
"react-native-snap-carousel": "2.1.4",
"react-native-stars": "^1.1.0",
"react-native-swipeout": "^2.2.2",
"react-native-swiper": "^1.5.4",
"react-native-tableview-simple": "0.16.5",
"react-native-vector-icons": "^4.0.0",
"react-native-video": "^1.0.0",
"react-native-zendesk-chat": "^0.2.1",
"react-redux": "^4.4.6",
"recompose": "^0.20.2",
"redux": "^3.5.2",
"redux-thunk": "^2.0.1"
}我在android中使用stacktrace进行了分析,发现mqt_js是每次点击UI都要花费更多时间的原因之一。您可以检查堆栈跟踪报告这里。
有人能帮我解决这个性能问题吗?
发布于 2018-04-02 17:36:38
以下是一些优化的建议-本机反应:
response.json()或JSON.stringify)阻塞js线程,因此所有JS动画和处理程序(例如onScroll和onPress,当然还有render方法)都会受到这种影响。尝试从后端加载您需要显示的数据。useNativeDriver: true参数),并尽量不使用onScroll。另外,考虑在动画中使用反应本土化。render方法,并尽可能少地调用它。当组件的道具或状态发生更改时,调用它。PureComponent和React.memo的工作原理,并在必要时尝试使用它们。但是要记住,如果不能正确使用,它们也可以减慢应用程序的速度。StyleSheet,它将它们兑现并替换为样式id (整数),因此样式只传递一次,然后使用样式id。坏:
// style object is created on every render
render() {
return <View style={{flex:1}}/>
}好:
render() {
return <View style={styles.flex}/>
}
// style is created once
const styles = StyleSheet.create({
flex: { flex: 1 }
});坏:
// onPress handler is created on every render
render() {
return <TouchableOpacity onPress={() => this.props.navigator.navigate('SignIn')}/>
}好:
render() {
return <TouchableOpacity onPress={this.onPressSignIn}/>
}
// onPressSignIn is created once
onPressSignIn = () => {
this.props.navigator.navigate('SignIn');
}Object和Set而不是Array。当需要从服务器/数据库加载大量数据、为服务器保留排序和其他繁重的计算时,请使用分页。例如,如果您经常需要通过id获取对象,最好使用:
let items = {
"123": { id: "123", ... },
"224": { id: "224", ... }
};
let item = items["123"];而不是通常的数组:
let items = [
0: { id: "123", ... },
1: { id: "224", ... }
];
let item = items.find(x => x.id === "123");发布于 2018-03-29 11:00:41
这是一个非常广泛的基于的和观点的问题,但我将尝试根据列出的profiler突出most common points和建议。
查看堆栈跟踪,主要问题在于包名(即UI Thread )中的com.fitspot.app.debug。
正如前面提到的,这里。
为了显示一个框架,我们所有的UI工作都需要在16 of周期结束前完成。
一旦边界间隔设置为16ms,您就可以看到,在一个周期内,mqt_js或JS Thread比16ms花费的时间要长得多,这意味着您的JS Thread在不断地运行。
在当前的profiler中,不清楚在JS Thread中执行哪些进程,因此很明显,问题主要在于JS Code而不是UI Thread。
有多种方法可以使react-native应用程序更快,这在页面中有很好的记录。下面是一个相同的basic要点。
dev=true中提供的,您可以在整个应用程序中禁用它们以获得更好的性能。console.log语句,因为它会导致JS线程上的bottleneck。您可以使用此插件删除console*文件中提到的所有这里语句,如
{ "env":{“生产”:{ "plugins":“转换-删除-控制台”}componentize您的项目结构,并使用Pure Components,以仅依赖props和state,使用immutable data structures进行更快的比较。slower navigation transitions,您可能需要检查navigation library代码,因为大多数情况下,它们都有用于default transitions的timeout。作为一种解决办法,您可以考虑构建自己的transitioner。Animations,可以考虑设置nativeDriver=true,这将减少JS thread的负载。这是一个解释得很好的示例。JS Thead和Main Thread操作,在本页面中有很好的解释。requiring/importing模块,这是不必要的,只导入classes所需,而不是whole component。external libraries来制作simple UI components,因为它们的性能比react-native的原生元素要慢得多。您可以考虑使用styled-components将UI组件化。发布于 2018-03-26 15:35:31
- add `initialNumToRender={number}` prop to Flatlist, as it will show only those components which are visible on screen and detach the other components
做这些改变,你就会发现你的表现比以前好多了。
https://stackoverflow.com/questions/49492135
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