open a url on click of ok button in android

Question

I have to open a URL on Click of OK Button in a view. Can someone tell how to do this?

Answer 1

On Button click event write this:

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

that open the 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 below method, which will take your target URL as the only 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 have to check that the URL is valid or not. If URL is invalid application may crash so that you have to check URL is valid or not by this method.

Answer 5

create an intent and set an action for it while 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);