Open url of ok button in android

Question

I need to open the url on the OK button in the view. Can someone tell how to do this?

Answer 1

In Button click an event:

 Uri uri = Uri.parse("http://www.google.com"); // missing 'http://' will cause crashed Intent intent = new Intent(Intent.ACTION_VIEW, uri); startActivity(intent); 

which will open your URL.

Answer 2

  Button imageLogo = (Button)findViewById(R.id.iv_logo); imageLogo.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { // TODO Auto-generated method stub String url = "http://www.gobloggerslive.com"; Intent i = new Intent(Intent.ACTION_VIEW); i.setData(Uri.parse(url)); startActivity(i); } }); 

Answer 3

You can use the method below that will display your destination URL as a single input (don't forget http: //)

 void GoToURL(String url){ Uri uri = Uri.parse(url); Intent intent= new Intent(Intent.ACTION_VIEW,uri); startActivity(intent); } 

Answer 4

 String url = "https://www.murait.com/"; if (url.startsWith("https://") || url.startsWith("http://")) { Uri uri = Uri.parse(url); Intent intent = new Intent(Intent.ACTION_VIEW, uri); startActivity(intent); }else{ Toast.makeText(mContext, "Invalid Url", Toast.LENGTH_SHORT).show(); } 

You need to make sure the URL is valid or not. If the URL is invalid, the application may crash, so you should check if the URL is valid or not in this way.

Answer 5

create an intent and set an action for it when passing the url to the intent

 yourbtn.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { String theurl = "http://google.com"; Uri urlstr = Uri.parse(theurl); Intent urlintent = new Intent(); urlintent.setData(urlstr); urlintent.setAction(Intent.ACTION_VIEW); startActivity(urlintent);