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?

Salsify

Use HttpUrlConnection. – Clove

public void openWebURL( String inURL ) { Intent browse = new Intent( Intent.ACTION_VIEW , Uri.parse( inURL ) ); startActivity( browse ); } – Salsify

This will work perfectly dude.. so 1 up... – Sinclare

@tushar: have you tried it? i think it should work properly. do you get any error while running this code? – Clove

try this way vogella.de/articles/AndroidIntent/article.html – Willin

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.

Acquit

OR startActivity(new Intent(Intent.ACTION_VIEW,Uri.parse("http://www.google.com")) – Nodus

@Chris-Jr you have missed the last parenthesis i.e. ) – Scanlon

How is this different from opening a webview item? – Rubbish

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);
        }
    });

Wiegand

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);
}

Glorygloryofthesnow

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.

Sidetrack

Answer 5

Add this code to your OK button click listener.

startActivity(Intent(Intent.ACTION_VIEW, Uri.parse("http://www.google.com")))

Kalikalian

Answer 6

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);

Schlieren

Answer 7

No need for any Java or Kotlin code to make it a clickable link, now you just need to follow given below code. And you can also link text color change by using textColorLink.

<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:autoLink="web"
android:textColorLink="@color/white"/>

Kira

Interesting, but it doesn't allow longClicks to override the autoLink. – Mcgowan

Answer 8

The following code worked perfectly for me.

fun Context.goToUrl(url: String) {
if (url.startsWith("https://") || url.startsWith("http://")) {
    val uriUrl = Uri.parse(url)
    val launchBrowser = Intent(Intent.ACTION_VIEW, uriUrl)
    startActivity(launchBrowser)

} else {
      Toast.makeText(this, "Invalid Url", Toast.LENGTH_SHORT).show()
}

}

call it in your actvity or fragment

requireContext().goToUrl("https://"+"something")

Dollarfish

Note, I combined @Mahamud and Sojitra to arrive at this solution. – Dollarfish

Answer 9

 private fun goToUrl(url: String) {
        val uriUrl = Uri.parse(url)
        val launchBrowser = Intent(Intent.ACTION_VIEW, uriUrl)
        startActivity(launchBrowser)
    }

Yeoman

Can you please give some explanation? – Yul