webview

问题 I have a simple android WebView application which is displaying a website featuring 'copy' buttons using the following code:- navigator.clipboard.writeText('Text to be copied') .then(() => { console.log('Text copied to clipboard'); }) .catch(err => { // This can happen if the user denies clipboard permissions: console.error('Could not copy text: ', err); }); this is working in all desktop and mobile browsers and an iOS WebView application however I cant it fails and catches the following

问题 I have a simple android WebView application which is displaying a website featuring 'copy' buttons using the following code:- navigator.clipboard.writeText('Text to be copied') .then(() => { console.log('Text copied to clipboard'); }) .catch(err => { // This can happen if the user denies clipboard permissions: console.error('Could not copy text: ', err); }); this is working in all desktop and mobile browsers and an iOS WebView application however I cant it fails and catches the following

问题 Is there a way to tell if a form is valid from Java? Basically, a method that returns true if all input fields constrains are satisfied and false otherwise. Strangely enough, org.w3c.dom.html.HTMLFormElement doesn't actually have a checkValidity() method. EDIT: Even more strangely, the implementation com.sun.webkit.dom.HTMLFormElementImpl does support the method checkValidity() . Unfortunately, the package com.sun.webkit is not accessible directly and thus the method is unavailable. 回答1: DOM

问题 Is there a way to tell if a form is valid from Java? Basically, a method that returns true if all input fields constrains are satisfied and false otherwise. Strangely enough, org.w3c.dom.html.HTMLFormElement doesn't actually have a checkValidity() method. EDIT: Even more strangely, the implementation com.sun.webkit.dom.HTMLFormElementImpl does support the method checkValidity() . Unfortunately, the package com.sun.webkit is not accessible directly and thus the method is unavailable. 回答1: DOM

问题 Is there a way to tell if a form is valid from Java? Basically, a method that returns true if all input fields constrains are satisfied and false otherwise. Strangely enough, org.w3c.dom.html.HTMLFormElement doesn't actually have a checkValidity() method. EDIT: Even more strangely, the implementation com.sun.webkit.dom.HTMLFormElementImpl does support the method checkValidity() . Unfortunately, the package com.sun.webkit is not accessible directly and thus the method is unavailable. 回答1: DOM

问题 I have a web view that I want to get the URL when the user long clicks on links. I know I can do this: // Register the context menu for web view registerForContextMenu(webView); And: @Override public void onCreateContextMenu(@NonNull ContextMenu menu, @NonNull View v, ContextMenu.ContextMenuInfo menuInfo) { super.onCreateContextMenu(menu, v, menuInfo); final WebView.HitTestResult result = webView.getHitTestResult(); Log.i("url is ===" , "result.getExtra()") } But this way, In some search

问题 Tools used D3 js: v4.11.0 react-native: v0.60.3 react-native-webview: v9.3.0 Previously I was using react-native 0.59.8 and I was using WebView from react-native, in this version WebView was working fine, SVG was also loading perfectly, but after upgrading react native to 0.60.3, I also have to take WebView from react-native-webview, React Native WebView :- <WebView scrollEnabled = {true} originWhitelist={["*"]} javaScriptEnabled={true} source={{ uri: isAndroid ? "file:///android_asset/widget

问题 Tools used D3 js: v4.11.0 react-native: v0.60.3 react-native-webview: v9.3.0 Previously I was using react-native 0.59.8 and I was using WebView from react-native, in this version WebView was working fine, SVG was also loading perfectly, but after upgrading react native to 0.60.3, I also have to take WebView from react-native-webview, React Native WebView :- <WebView scrollEnabled = {true} originWhitelist={["*"]} javaScriptEnabled={true} source={{ uri: isAndroid ? "file:///android_asset/widget

问题 hello i have a webview android app. i want that my website open in webview only and if somone click on a url that is not related to my website then open in browser. Example: if my website is little.com and if some one clcik a link that is facebook.com or google.com then facebbok or google open in browser. androidmainfest.xml <?xml version="1.0" encoding="utf-8"?> <manifest xmlns:android="http://schemas.android.com/apk/res/android" package="com.little.example"> <uses-permission android:name=

问题 It is possible to achieve this currently in Android? I only can find deprecated questions about old methods (CookieSynchManager) which not seems to work for this actually. It is possible to achieve it? can't find anything also on the android developers guide. Thank you 回答1: Having the same problem...I solved reading the doc here: https://developer.android.com/reference/android/webkit/CookieSyncManager For future readers: to force the Cookie sync process you can manually call the flush()