react-native-webview

问题 React Native WebView App not exit on pressing back button after setting Go back functionality on back button pressed. I want go back functionality on pressing back button when webview is not on home page and when webview is on home page then exit the app. export default class WebView extends Component { constructor (props) { super(props); this.WEBVIEW_REF = React.createRef(); } componentDidMount() { BackHandler.addEventListener('hardwareBackPress', this.handleBackButton); }

问题 I'm implementing in App WebView for my App. I've to open some info pages and I've to get some data based on the click of any particular place(which contains a different type of data) in webview. But in iOS, while loading any URL onShouldStartLoadWithRequest calling automatically which leads opening different URLs in the HTML content. But it is working as expected in Android. <WebView originWhitelist={["*"]} style={style} source={source} showsVerticalScrollIndicator=

问题 I'm implementing in App WebView for my App. I've to open some info pages and I've to get some data based on the click of any particular place(which contains a different type of data) in webview. But in iOS, while loading any URL onShouldStartLoadWithRequest calling automatically which leads opening different URLs in the HTML content. But it is working as expected in Android. <WebView originWhitelist={["*"]} style={style} source={source} showsVerticalScrollIndicator=

问题 I want to intercept clicks on a link in my webview in react-native and perform a custom action instead of navigating to the target of the link as described in the official guide. Here's what I do: import React from 'react'; import {View, Linking} from 'react-native'; import AsyncStorage from '@react-native-community/async-storage'; import {WebView} from 'react-native-webview'; export default class WebViewScreen extends React.Component { static navigationOptions = { title: 'Produck', };

问题 I want to intercept clicks on a link in my webview in react-native and perform a custom action instead of navigating to the target of the link as described in the official guide. Here's what I do: import React from 'react'; import {View, Linking} from 'react-native'; import AsyncStorage from '@react-native-community/async-storage'; import {WebView} from 'react-native-webview'; export default class WebViewScreen extends React.Component { static navigationOptions = { title: 'Produck', };

来源： https://stackoverflow.com/questions/61832232/invariant-violation-requirenativecomponent-rncwebview-was-not-found-in-the-u

来源： https://stackoverflow.com/questions/64680403/webview-freezed-after-using-the-stoploading-of-react-native-webview

来源： https://stackoverflow.com/questions/58155621/react-native-webview-for-android-not-displaying-pdf-and-word-files

问题 I've created a React-native app (tested in android) using WebView with the initial URL of my company's website. I want the user to be able to get redirected to a spesific website's product page (for example https://www.test.com/product/hoodie-with-zipper/) WHEN they clicked the OneSignal push notification with the intended launch URL included in the sent notification. The problem is, I want the user to open the URL with my WebView App, not the browsers. The features I've implemented and

问题 I'm getting this error with npm react-native-webview@9.0.1 when I try to generate a QR code with react-native-qrcode-generator I'm using react-native with an expo managed workflow. And the thing is it works on iOS, and i only get the error on Android I searched for a solution myself and I tried installing react-native-get-random-values but that also didn't work. Any thoughts? 回答1: I just had the same issue on android. Works fine on iOS. I solved it with: Uninstall the existing one npm