I am new to React native and I am trying to open my webpages(page1,page2) inside webview using react native, and my component's webview in the below example.
Here page1 contains button on clicking that button, page2 is opening in child window in external browser.
Can somebody please tell me how to open page2 inside the webview, so that the user can get a good experience?
Example: Component 1
{
<WebView
scalesPageToFit
startInLoadingState={true}
renderLoading={() => { return <Loading/> }}
**source={{uri:"url to page1"}}**
onShouldStartLoadWithRequest={request => {
return request.url.startsWith(domain);
}}
style={styles.web}
javaScriptEnabledAndroid={true}
javaScriptEnabled={true}
originWhitelist={[domain+"*"]}
/>
}
I found the below solution but was not able to implement it in my scenario.
https://github.com/facebook/react-native/pull/6886
I found the answer, this happened because of the latest react-native-webview version.
setSupportMultipleWindows={true} becomes true and in my case, it should be false.
Related
I'm working on a react-native app using WebView.
Here's the code of web view:
<WebView
ref={webView}
onLoad={onLoad}
onMessage={() => {}}
originWhitelist={["*"]}
source={{ uri: PLATFORM_URL }}
onNavigationStateChange={onNavigationStateChange}
/>
I'd like to prevent the web-view from opening browser when some buttons are clicked.
In particular, I'm struggling on a with target blank, and in iOS it actually redirect inside the webview, but on Android it opens the browser.
I've found one possible solution, that consist on injecting javascript code that is able to remove all target-blank s.
Is this the only solution? or is there a way to intercept that the webview is going to open the web browser?
Thank you
I hope this helps although it is late. Try using props onShouldStartLoadWithRequest like this
<WebView
source={{uri:'https://stackoverflow.com/'}}
onShouldStartLoadWithRequest={request => {
if (request.url.includes('https')) {
return false;
} else return true;
}}
/>
I am using Linking from React-Native. Linking.OpenUrl seems to work with most urls but it does not seem to work with urls which have non-english characters. See the example below in Expo where I have reproduced the case. Note that if you click on the url link directly it will open properly. However, when the same link is being opened via the Linking.OpenUrl it does something to the url and lands in a 404 page.
Here is a repro in Expo:
https://snack.expo.dev/#rezahok/linking-not-working
I am using Expo 42. Any help with this would be really appreciated.
Try with below code
export default class App extends Component {
render() {
const uri = `https://www.prothomalo.com/bangladesh/district/%E0%A6%A6%E0%A6%BF%E0%A6%A8%E0%A6%BE%E0%A6%9C%E0%A6%AA%E0%A7%81%E0%A6%B0%E0%A7%87-%E0%A6%B0%E0%A7%87%E0%A6%B2%E0%A6%95%E0%A7%8D%E0%A6%B0%E0%A6%B8%E0%A6%BF%E0%A6%82%E0%A7%9F%E0%A7%87-%E0%A6%AC%E0%A7%8D%E0%A6%AF%E0%A6%95%E0%A7%8D%E0%A6%A4%E0%A6%BF%E0%A6%97%E0%A6%A4-%E0%A6%97%E0%A6%BE%E0%A7%9C%E0%A6%BF%E0%A6%95%E0%A7%87-%E0%A6%9F%E0%A7%8D%E0%A6%B0%E0%A7%87%E0%A6%A8%E0%A7%87%E0%A6%B0-%E0%A6%A7%E0%A6%BE%E0%A6%95%E0%A7%8D%E0%A6%95%E0%A6%BE-%E0%A6%A8%E0%A6%BF%E0%A6%B9%E0%A6%A4-%E0%A7%A9`
const decodedUri = decodeURIComponent(uri);
return (
<View style={styles.container}>
<Button title="Click me" onPress={ ()=>{ Linking.openURL(decodedUri)}} />
</View>
);
}
}
Inside my StackNavigator, one of the components includes web content with a long loading time. However, this screen will only be shown late in my navigation flow.
How can I use this time to render my component in the background before finally switching to it?
I couldn't find anything comparable to ReactDOM.render in React Native that would allow me to render it manually.
I am not aware of any option in react-navigation to preload a screen that is not displayed, except maybe when the screen is part of a tab navigator.
Depending on what is slowing down the rendering, you might be able to do some actions in the first screen and later pass the results to the second screen using a navigation parameter.
For instance, if you are fetching data from an api in the second screen, you could fetch this data in the first screen and pass it to the second one:
this.props.navigation.navigate('SecondScreen', { data: this.data });
If it is a component, you could also try to build it in the first screen and pass it in the same fashion:
this.props.navigation.navigate('SecondScreen', { component: this.component });
If you are rendering a WebView in the second screen, what can help is to render the WebView in the first screen too, but with no width or height. The WebView will not be displayed but the website data will be fetched and cached, making the real render more efficient:
render() {
return (
<View>
<WebView source={{ uri: 'https://github.com/facebook/react-native' }} style={{ height: 0, width: 0 }} />
{this.renderCurrentScreen()}
</View>
);
}
Actually I have a webview in react native and I am trying to open a url from that webview.Once the URL is loaded on pressing the bacǩ buttton it should navigate to previous page.How to navigate back to the webview page?
You can use reference of Webview and call goBack on that reference
<WebView
ref={ref => {
if (!ref) {
return;
}
this.webview = ref;
}}
/>
this.webview.goBack();
There's an option that you can try with is to have your webView inside a Modal that way you can control onBackPress.
it is possible to handle URL change on React Native web view.
I try to handle with onNavigationStateChange listener. But it only once time firing. When the page is load firing onNavigationStateChange. And when I navigate to another page, this event not firing.
Any idea?
With the reference to import { WebView } from 'react-native-webview' you may be using old version of webview please try updating or try following code/method to get current navigation path/url.
In your case onNavigationStateChange is not returning navigated path so you can try using onLoadProgress which return every change in url.
Instead of this
onNavigationStateChange={(state) => {
console.log("current_path",path);
}}
try this
This may get called multiple time please handle your condition accordingly.
onLoadProgress={({ path}) => {
console.log("current_path",path);
}}
This is how my WebView looks like
<WebView
source={{ uri: this.state.url }}
onLoadEnd={() => this.hideSpinner()}
onMessage={onMessage.bind(this)}
ref={WEBVIEW_REF => (this.WebViewRef = WEBVIEW_REF)}
startInLoadingState={true}
javaScriptEnabled={true}
domStorageEnabled={true}
onLoadProgress={({ nativeEvent }) => {
//your code goes here
}}
onNavigationStateChange={(state) => {
//your code goes here
}}
/>
Try upgrading your React Native (and React) version.
It should be fixed. Maybe was an old issue, because right now it's working OK.
Check this link: React Native webview get url
onNavigationStateChange={({ url, canGoBack }) => {
console.log("url>>>>>>>>",url);
}}
I load website link in react native WebView and its has button when i click that it navigate to default mobile browser.
Running application in react native android Current scenario: when i click button in website its navigate to default browser and am not getting the url i tried above suggestions.
Expected scenario: When i click that button in website it should not navigate to default browser and i want that url which is opened in default browser.
my react version: "react-native": "0.64.0",
onLoadProgress = {({ nativeEvent }) => {
//your code goes here
}}
onNavigationStateChange={navState => {
console.log("navState ", navState);
}}