'How To Exit app by double tap on back button in NativeScript Apps

Exit app on double tap on back button in Nativescript

Please help me with snippet of code

nativescript


Solution 1:[1]

Here is the solution that I have found:

var frameModule = require("ui/frame");
var application = require("application")

var activity = application.android.startActivity ||
        application.android.foregroundActivity ||
        frameModule.topmost().android.currentActivity ||
        frameModule.topmost().android.activity;
var lastPress;

activity.onBackPressed = function() {
    var timeDelay = 500
    if (lastPress + timeDelay > java.lang.System.currentTimeMillis()) {
        var startMain = new android.content.Intent(android.content.Intent.ACTION_MAIN);
        startMain.addCategory(android.content.Intent.CATEGORY_HOME);
        startMain.setFlags(android.content.Intent.FLAG_ACTIVITY_NEW_TASK);
        activity.startActivity(startMain);

        // If you want to kill the app totally, use these codes instead of above
        // activity.finish();
        // java.lang.System.exit(0);

    } else {
        frameModule.topmost().goBack();
    }
    lastPress = java.lang.System.currentTimeMillis();
}

Hope this helps.

Solution 2:[2]

for Angular

import * as application from 'tns-core-modules/application';

exitapp=0
ngOnInit() {
application.android.on(application.AndroidApplication.activityBackPressedEvent, this.handleBackButton, this);
}

handleBackButton(args: any) {
        this.ngZone.run(() => {
            args.cancel = true;
            if (this.routerExtensions.canGoBack()) {
                this.routerExtensions.back();
            }else{
                
                this.exitapp++
                this.getData.toast('Press again to exit')
                if(this.exitapp==2){
                    application.android.foregroundActivity.finish();
                }
                setTimeout(()=>{
                    this.exitapp=0
                },2000)
            }
        })
    }

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 aditya_adi_98

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