how can i implement built-in browser like facebook app - android

i read article that facebook app has started opening links with its built-in browser
( related article url : http://www.androidcentral.com/facebook-enabling-app-browser-links-some-accounts )
it's quite cool so some apps were updated to use built-in browser.
i also want to develop own built-in browser and use in my app like facebook, but i can't find how to.
( i searched 'built-in browser', 'in-app browser' etc.. in google, but can't find how to make it. )
so, how can i develop own built-in browser? is there any development guide?

You should use a webview. Look this link... If your webpage that you want load is local you must put it in assets and call it with
webView.loadUrl("file:///android_asset/index.html");
otherwise like doc
http://developer.android.com/reference/android/webkit/WebView.html

Related

what is the way to share text from website to android apps as like send intent in android sdk

Hello we are working on a mobile website. If user open this website in Google Chrome, there I have feature to share such as any article link something to android apps.
I am native android developer. I know how to share text from my native app to other apps using Intent.ACTION_SEND. We are looking to do same from mobile website. Is there any way to achieve it ?
Found a good article on Intents with Chrome here :
https://developer.chrome.com/multidevice/android/intents
Another way in my opinion would be Enabling Deep Links for App Content:
https://developer.chrome.com/multidevice/android/intents

How to open an article in android app?

Can someone tell me what steps to follow to open an article in android app?
I do not meant open via browser but in-app.
Update description:
I do not mean open the actual website article, but copy the content (along with images, video links) and show it in the app with your preferred font, background etc
You can open a web page in-app (instead of the browser) by using a WebView (see documentation here). I recommend also reading the API guide topic Building Web Apps in WebView.
you may use webview to load your html file
WebView webView = (WebView)findViewById(R.id.webView1);
webview.loadUrl("file:///android_asset/file.html");

phonegap android application webrowser

I know about phonegap but here i am want to know if it
is possible to have an android application which is not
really an application but merely a browser that opens only
a particular webpage, that is only the mobile site i already
designed, am not sure but i have seen some android apps and i know
they do something similar , below are the things i want :
The app should be installable.
Just opens a link to an existing webpage online.i.e its more like a browser
only it just opens only a specified web page
is this possible with Phonegap for Android ?
You can do that. You would want to use the inappbrowser plugin to open an external page when ondeviceready is fired. Take a look at my SO answer here for a guide on how to use the inappbrowser, and you can take a look here for the docs about ondeviceready.

Display offline html in android app with a tablet

I have to create an android app, that displays an offline html website. Does android sdk have an internal browser control? What kind of dev sw do you recommend to create this basic wrapper?
Read about WebView.
It's built-in webrowser engine and you can use it to your target.
Yes. WebView
http://developer.android.com/reference/android/webkit/WebView.html
This component allows you to specific a local url (file://) if required or even passed the complete html to the load method. Also capable of event handling for more advanced scenarios.

How do you replace the default webkit browser with another one in an android web view?

How can I change the webview in an app so it uses another engine, not the engine from the android browser?
You cannot "change the webview". If another browser elects to make its code available as a JAR exposing some sort of widget, you may be able to use that -- you would have to contact the browser developer to find out.

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