I am using a website in the ionic-5 web-view app. When I go back to the previous page in the Ionic app, Android device back button is not responding. I am unable to go back to previous page in the Ionic app. Ionic web-view app events are not in sync with Android device back button. Please help me to solve this issue.
You can handle hardware back button using platform
import { Platform } from '#ionic/angular';
this.platform.backButton.subscribeWithPriority(10, () => {
// route to previous page or component
});
Related
I've tried the two following approaches to close my ionic app but non has worked.
1.
closeApp(){
this.platform.backButton.subscribeWithPriority(999999, () => {
navigator['app'].exitApp();
})
}
In this approach, when I click on the back button, the app is disappeared, but it's in the list of the background running apps. This doesn't help me, because I need the app to be completely closed and the user be forced to reopen the app by clicking on its icon.
2.
closeApp(){
const { App } = Plugins;
App['exitApp'];
}
in this case, nothing happens, i.e., I stay on the page.
Is there any other approach or plugin which I can use?
I am working with capacitor: 3 and testing on android 12.
import { App } from '#capacitor/app';
ExitApp() {
App.exitApp();
}
try this
I have a button in my app. On click of the button, I want the notes app in the device to get launched whether iOS or Android. I there a way around this with flutter? I haven't discovered any solution yet.
Yes there is this plugin external_app_launcher will helps you to open another app from your app by providing PackageName for Android and URLscheme for IOS external_app_launcher
I am working on progressive web app with ionic 3 and angular 6 framework.
Every thing works fine but when I press my browser back button or mobile back button my application get closed.
I tried to search how to do this, I am able to disable browser back button with java script but on mobile app it's not working.
<script>
// window.onbeforeunload = function() { return "Your work will be lost.";
};
$ionicPlatform.registerBackButtonAction(function (event) {
event.preventDefault();
}, 100);
</script>
I tried to register the back button event and lot's more but none of that is worked fine for me.
Can anyone have same issue then please tell me how to resolve this.
Thanks,
Dattatray
You can use Ionic's Platform to override the physical back button on Android.
Add it into app.component.ts file as such :
import { Platform } from 'ionic-angular';
constructor(public plt: Platform) {}
platform.registerBackButtonAction(() => {
//Add your code here
},100);
Ionic Platform Service.
This is my custom view where "LondonStreet" is a button.
When I tap that button I get url and open it in Safari (it works). Then I can go back, using "Back to Wishlist" button (it also works).
The problem is when I try to test this under UITests.
itemsTable.cells.elementBoundByIndex(0).buttons["addressButton"].tap() //press the button to open link in Safari
Along with this line:
app.statusBars.buttons["Back to Wishlist"].tap() //go back, doesn't work, although it was recorded by Xcode itself.
is an error:
UI Testing Failure - Failed to get screenshot within 5s.
And also in issue Navigator
UI Testing failure - Unable to update application state promptly.
Starting in iOS 11 you can interact with other applications using the XCUIApplication(bundleIdentifier:) initializer.
To get back to your app you'd do something like:
let myApp = XCUIApplication(bundleIdentifier: "my.app.bundle.id")
let safari = XCUIApplication(bundleIdentifier: "com.apple.mobilesafari")
// Perform action in your app that opens Safari
safari.wait(for: .runningForeground, timeout: 30)
myApp.activate() // <--- Go back to your app
UI Testing cannot interact with anything outside of your application. In your scenario, the framework can no longer do anything once your app opens Safari.
To verify this, try printing out the app's hierarchy once Safari opens. You will notice that nothing in Safari nor the navigation bar will show up - you will only see your app's information.
print(XCUIApplication().debugDescription)
To open specific url in Safari on iOS 15:
safari.textFields["Address"].tap()
safari.textFields["Address"].typeText("www.urlToOpen.com")
safari.keyboards.buttons["Go"].tap()
I would not have asked the question if I owned an iphone, but so far I only have an android phone for development.
The question is: Should I use the View.viewMenuItems that are provided by flex mobile? Or should I better embed the functionality in another way? I don't know what possibilities might exist on iphone to open the menu, because it has no hardware button "menu".
By the way: How could I open the menu in the Flash-Builder mobile device emulator - there are no buttons, too....
Thanks!
you can detect the Menu Button through KeyBoardEvent in Android. In Iphone, the Home Button will exit out your application, so you should do some interface buttons to show your menu.
private function _onAddedToStage(event:FlexEvent):void
{
//removes listener
removeEventListener(Event.ADDED_TO_STAGE, _onAddedToStage);
stage.addEventListener(KeyboardEvent.KEY_DOWN, _onKeyDown);
}
private function _onKeyDown(event:KeyboardEvent):void
{
if(event.keyCode == Keyboard.MENU)
{
event.preventDefault();
_text.appendText("\nMenu Pressed");
}
}
The latest versions of Android's Honeycomb OS does not support the menu button. So if you do take advantage of Flex's ViewMenu feature beware that if you build using Flex 4.6 or newer there won't be a button on the latest Android Tablets and iOS devices.
In ADL you can choose Device > Menu to simulate clicking a hardware menu button.