如何从我的应用程序在 Android 的 web 浏览器中打开 URL？

Question

How to open an URL from code in the built-in web browser rather than within my application?

I tried this:

try {
    Intent myIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(download_link));
    startActivity(myIntent);
} catch (ActivityNotFoundException e) {
    Toast.makeText(this, "No application can handle this request."
        + " Please install a webbrowser",  Toast.LENGTH_LONG).show();
    e.printStackTrace();
}

but I got an Exception:

No activity found to handle Intent{action=android.intent.action.VIEW data =www.google.com

Answer 1

Try this:

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

That works fine for me.

As for the missing "http://" I'd just do something like this:

if (!url.startsWith("http://") && !url.startsWith("https://"))
   url = "http://" + url;

I would also probably pre-populate your EditText that the user is typing a URL in with "http://".

Answer 2

a common way to achieve this is with the next code:

String url = "http://www.stackoverflow.com";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url)); 
startActivity(i); 

that could be changed to a short code version ...

Intent intent = new Intent(Intent.ACTION_VIEW).setData(Uri.parse("http://www.stackoverflow.com"));      
startActivity(intent); 

or :

Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.stackoverflow.com")); 
startActivity(intent);

the shortest! :

startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.stackoverflow.com")));

happy coding!

Answer 3

Simple Answer

You can see the official sample from Android Developer .

/**
 * Open a web page of a specified URL
 *
 * @param url URL to open
 */
public void openWebPage(String url) {
    Uri webpage = Uri.parse(url);
    Intent intent = new Intent(Intent.ACTION_VIEW, webpage);
    if (intent.resolveActivity(getPackageManager()) != null) {
        startActivity(intent);
    }
}

How it works

Please have a look at the constructor of Intent :

public Intent (String action, Uri uri)

You can pass android.net.Uri instance to the 2nd parameter, and a new Intent is created based on the given data url.

And then, simply call startActivity(Intent intent) to start a new Activity, which is bundled with the Intent with the given URL.

Do I need the if check statement?

Yes. The docs says:

If there are no apps on the device that can receive the implicit intent, your app will crash when it calls startActivity(). To first verify that an app exists to receive the intent, call resolveActivity() on your Intent object. If the result is non-null, there is at least one app that can handle the intent and it's safe to call startActivity(). If the result is null, you should not use the intent and, if possible, you should disable the feature that invokes the intent.

Bonus

You can write in one line when creating the Intent instance like below:

Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));

Answer 4

In 2.3, I had better luck with

final Intent intent = new Intent(Intent.ACTION_VIEW).setData(Uri.parse(url));
activity.startActivity(intent);

The difference being the use of Intent.ACTION_VIEW rather than the String "android.intent.action.VIEW"

Answer 5

Try this:

Uri uri = Uri.parse("https://www.google.com");
startActivity(new Intent(Intent.ACTION_VIEW, uri));

or if you want then web browser open in your activity then do like this:

WebView webView = (WebView) findViewById(R.id.webView1);
WebSettings settings = webview.getSettings();
settings.setJavaScriptEnabled(true);
webView.loadUrl(URL);

and if you want to use zoom control in your browser then you can use:

settings.setSupportZoom(true);
settings.setBuiltInZoomControls(true);

Answer 6

If you want to show user a dialogue with all browser list, so he can choose preferred, here is sample code:

private static final String HTTPS = "https://";
private static final String HTTP = "http://";

public static void openBrowser(final Context context, String url) {

     if (!url.startsWith(HTTP) && !url.startsWith(HTTPS)) {
            url = HTTP + url;
     }

     Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
     context.startActivity(Intent.createChooser(intent, "Choose browser"));// Choose browser is arbitrary :)

}

Answer 7

Just like the solutions other have written (that work fine), I would like to answer the same thing, but with a tip that I think most would prefer to use.

In case you wish the app you start to open in a new task, indepandant of your own, instead of staying on the same stack, you can use this code:

final Intent intent=new Intent(Intent.ACTION_VIEW,Uri.parse(url));
intent.addFlags(Intent.FLAG_ACTIVITY_NO_HISTORY|Intent.FLAG_ACTIVITY_CLEAR_WHEN_TASK_RESET|Intent.FLAG_ACTIVITY_NEW_TASK|Intent.FLAG_ACTIVITY_MULTIPLE_TASK);
startActivity(intent);

There is also a way to open the URL in Chrome Custom Tabs . Example in Kotlin :

@JvmStatic
fun openWebsite(activity: Activity, websiteUrl: String, useWebBrowserAppAsFallbackIfPossible: Boolean) {
    var websiteUrl = websiteUrl
    if (TextUtils.isEmpty(websiteUrl))
        return
    if (websiteUrl.startsWith("www"))
        websiteUrl = "http://$websiteUrl"
    else if (!websiteUrl.startsWith("http"))
        websiteUrl = "http://www.$websiteUrl"
    val finalWebsiteUrl = websiteUrl
    //https://github.com/GoogleChrome/custom-tabs-client
    val webviewFallback = object : CustomTabActivityHelper.CustomTabFallback {
        override fun openUri(activity: Activity, uri: Uri?) {
            var intent: Intent
            if (useWebBrowserAppAsFallbackIfPossible) {
                intent = Intent(Intent.ACTION_VIEW, Uri.parse(finalWebsiteUrl))
                intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK or Intent.FLAG_ACTIVITY_NO_HISTORY
                        or Intent.FLAG_ACTIVITY_CLEAR_WHEN_TASK_RESET or Intent.FLAG_ACTIVITY_MULTIPLE_TASK)
                if (!CollectionUtil.isEmpty(activity.packageManager.queryIntentActivities(intent, 0))) {
                    activity.startActivity(intent)
                    return
                }
            }
            // open our own Activity to show the URL
            intent = Intent(activity, WebViewActivity::class.java)
            WebViewActivity.prepareIntent(intent, finalWebsiteUrl)
            activity.startActivity(intent)
        }
    }
    val uri = Uri.parse(finalWebsiteUrl)
    val intentBuilder = CustomTabsIntent.Builder()
    val customTabsIntent = intentBuilder.build()
    customTabsIntent.intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK or Intent.FLAG_ACTIVITY_NO_HISTORY
            or Intent.FLAG_ACTIVITY_CLEAR_WHEN_TASK_RESET or Intent.FLAG_ACTIVITY_MULTIPLE_TASK)
    CustomTabActivityHelper.openCustomTab(activity, customTabsIntent, uri, webviewFallback)
}

Answer 8

other option In Load Url in Same Application using Webview

webView = (WebView) findViewById(R.id.webView1);
webView.getSettings().setJavaScriptEnabled(true);
webView.loadUrl("http://www.google.com");

Answer 9

You can also go this way

In xml :

<?xml version="1.0" encoding="utf-8"?>
<WebView  
xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/webView1"
android:layout_width="fill_parent"
android:layout_height="fill_parent" />

In java code :

public class WebViewActivity extends Activity {

private WebView webView;

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.webview);

    webView = (WebView) findViewById(R.id.webView1);
    webView.getSettings().setJavaScriptEnabled(true);
    webView.loadUrl("http://www.google.com");

 }

}

In Manifest dont forget to add internet permission...

Answer 10

The Kotlin answer:

val browserIntent = Intent(Intent.ACTION_VIEW, uri)
ContextCompat.startActivity(context, browserIntent, null)

I have added an extension on Uri to make this even easier

myUri.openInBrowser(context)

fun Uri?.openInBrowser(context: Context) {
    this ?: return // Do nothing if uri is null

    val browserIntent = Intent(Intent.ACTION_VIEW, this)
    ContextCompat.startActivity(context, browserIntent, null)
}

As a bonus, here is a simple extension function to safely convert a String to Uri.

"https://stackoverflow.com".asUri()?.openInBrowser(context)

fun String?.asUri(): Uri? {
    try {
        return Uri.parse(this)
    } catch (e: Exception) {}
    return null
}

Answer 11

Webview can be used to load Url in your applicaion. URL can be provided from user in text view or you can hardcode it.

Also don't forget internet permissions in AndroidManifest.

String url="http://developer.android.com/index.html"

WebView wv=(WebView)findViewById(R.id.webView);
wv.setWebViewClient(new MyBrowser());
wv.getSettings().setLoadsImagesAutomatically(true);
wv.getSettings().setJavaScriptEnabled(true);
wv.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
wv.loadUrl(url);

private class MyBrowser extends WebViewClient {
    @Override
    public boolean shouldOverrideUrlLoading(WebView view, String url) {
        view.loadUrl(url);
        return true;
    }
}

Answer 12

Within in your try block,paste the following code,Android Intent uses directly the link within the URI(Uniform Resource Identifier) braces in order to identify the location of your link.

You can try this:

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

Answer 13

A short code version...

 if (!strUrl.startsWith("http://") && !strUrl.startsWith("https://")){
     strUrl= "http://" + strUrl;
 }


 startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(strUrl)));

Answer 14

Simple and Best Practice

Method 1:

String intentUrl="www.google.com";
Intent webIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(intentUrl));
    if(webIntent.resolveActivity(getPackageManager())!=null){
        startActivity(webIntent);    
    }else{
      /*show Error Toast 
              or 
        Open play store to download browser*/
            }

Method 2:

try{
    String intentUrl="www.google.com";
    Intent webIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(intentUrl));
        startActivity(webIntent);
    }catch (ActivityNotFoundException e){
                /*show Error Toast
                        or
                  Open play store to download browser*/
    }

Answer 15

String url = "http://www.example.com";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);

Answer 16

So I've looked for this for a long time because all the other answers were opening default app for that link, but not default browser and that's what I wanted.

I finally managed to do so:

// gathering the default browser
final Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://"));
final ResolveInfo resolveInfo = context.getPackageManager()
    .resolveActivity(intent, PackageManager.MATCH_DEFAULT_ONLY);
String defaultBrowserPackageName = resolveInfo.activityInfo.packageName;


final Intent intent2 = new Intent(Intent.ACTION_VIEW);
intent2.setData(Uri.parse(url));

if (!defaultBrowserPackageName.equals("android") {
    // android = no default browser is set 
    // (android < 6 or fresh browser install or simply no default set)
    // if it's the case (not in this block), it will just use normal way.
    intent2.setPackage(defaultBrowserPackageName);
}

context.startActivity(intent2);

BTW, you can notice context .whatever, because I've used this for a static util method, if you are doing this in an activity, it's not needed.

Answer 17

Intent getWebPage = new Intent(Intent.ACTION_VIEW, Uri.parse(MyLink));          
startActivity(getWebPage);

Answer 18

Simple, website view via intent,

Intent viewIntent = new Intent("android.intent.action.VIEW", Uri.parse("http://www.yoursite.in"));
startActivity(viewIntent);  

use this simple code toview your website in android app.

Add internet permission in manifest file,

<uses-permission android:name="android.permission.INTERNET" /> 

Answer 19

Chrome custom tabs are now available:

The first step is adding the Custom Tabs Support Library to your build.gradle file:

dependencies {
    ...
    compile 'com.android.support:customtabs:24.2.0'
}

And then, to open a chrome custom tab:

String url = "https://www.google.pt/";
CustomTabsIntent.Builder builder = new CustomTabsIntent.Builder();
CustomTabsIntent customTabsIntent = builder.build();
customTabsIntent.launchUrl(this, Uri.parse(url));

For more info: https://developer.chrome.com/multidevice/android/customtabs

Answer 20

The response of MarkB is right. In my case I'm using Xamarin, and the code to use with C# and Xamarin is:

var uri = Android.Net.Uri.Parse ("http://www.xamarin.com");
var intent = new Intent (Intent.ActionView, uri);
StartActivity (intent);

This information is taked from: https://developer.xamarin.com/recipes/android/fundamentals/intent/open_a_webpage_in_the_browser_application/

Answer 21

Based on the answer by Mark B and the comments bellow:

protected void launchUrl(String url) {
    Uri uri = Uri.parse(url);

    if (uri.getScheme() == null || uri.getScheme().isEmpty()) {
        uri = Uri.parse("http://" + url);
    }

    Intent browserIntent = new Intent(Intent.ACTION_VIEW, uri);

    if (browserIntent.resolveActivity(getPackageManager()) != null) {
        startActivity(browserIntent);
    }
}

Answer 22

android.webkit.URLUtil has the method guessUrl(String) working perfectly fine (even with file:// or data:// ) since Api level 1 (Android 1.0). Use as:

String url = URLUtil.guessUrl(link);

// url.com            ->  http://url.com/     (adds http://)
// http://url         ->  http://url.com/     (adds .com)
// https://url        ->  https://url.com/    (adds .com)
// url                ->  http://www.url.com/ (adds http://www. and .com)
// http://www.url.com ->  http://www.url.com/ 
// https://url.com    ->  https://url.com/
// file://dir/to/file ->  file://dir/to/file
// data://dataline    ->  data://dataline
// content://test     ->  content://test

In the Activity call:

Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(URLUtil.guessUrl(download_link)));

if (intent.resolveActivity(getPackageManager()) != null)
    startActivity(intent);

Check the complete guessUrl code for more info.

Answer 23

Okay,I checked every answer but what app has deeplinking with same URL that user want to use?

Today I got this case and answer is browserIntent.setPackage("browser_package_name");

eg :

   Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.google.com"));
    browserIntent.setPackage("com.android.chrome"); // Whatever browser you are using
    startActivity(browserIntent);

Thank you!

Answer 24

Simply go with short one to open your Url in Browser:

Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("YourUrlHere"));
startActivity(browserIntent);

Answer 25

String url = "https://www.thandroid-mania.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();
}

That error occurred because of invalid URL, Android OS can't find action view for your data. So you have validate that the URL is valid or not.

Answer 26

Short & sweet Kotlin helper function:

private fun openUrl(link: String) =
    startActivity(Intent(Intent.ACTION_VIEW, Uri.parse(link)))

Answer 27

I think this is the best

openBrowser(context, "http://www.google.com")

Put below code into global class

    public static void openBrowser(Context context, String url) {

        if (!url.startsWith("http://") && !url.startsWith("https://"))
            url = "http://" + url;

        Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
        context.startActivity(browserIntent);
    }

Answer 28

This way uses a method, to allow you to input any String instead of having a fixed input. This does save some lines of code if used a repeated amount of times, as you only need three lines to call the method.

public Intent getWebIntent(String url) {
    //Make sure it is a valid URL before parsing the URL.
    if(!url.contains("http://") && !url.contains("https://")){
        //If it isn't, just add the HTTP protocol at the start of the URL.
        url = "http://" + url;
    }
    //create the intent
    Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url)/*And parse the valid URL. It doesn't need to be changed at this point, it we don't create an instance for it*/);
    if (intent.resolveActivity(getPackageManager()) != null) {
        //Make sure there is an app to handle this intent
        return intent;
    }
    //If there is no app, return null.
    return null;
}

Using this method makes it universally usable. IT doesn't have to be placed in a specific activity, as you can use it like this:

Intent i = getWebIntent("google.com");
if(i != null)
    startActivity();

Or if you want to start it outside an activity, you simply call startActivity on the activity instance:

Intent i = getWebIntent("google.com");
if(i != null)
    activityInstance.startActivity(i);

As seen in both of these code blocks there is a null-check. This is as it returns null if there is no app to handle the intent.

This method defaults to HTTP if there is no protocol defined, as there are websites who don't have an SSL certificate(what you need for an HTTPS connection) and those will stop working if you attempt to use HTTPS and it isn't there. Any website can still force over to HTTPS, so those sides lands you at HTTPS either way

---

Because this method uses outside resources to display the page, there is no need for you to declare the INternet permission. The app that displays the webpage has to do that

Answer 29

//OnClick Listener

  @Override
      public void onClick(View v) {
        String webUrl = news.getNewsURL();
        if(webUrl!="")
        Utils.intentWebURL(mContext, webUrl);
      }

//Your Util Method

public static void intentWebURL(Context context, String url) {
        if (!url.startsWith("http://") && !url.startsWith("https://")) {
            url = "http://" + url;
        }
        boolean flag = isURL(url);
        if (flag) {
            Intent browserIntent = new Intent(Intent.ACTION_VIEW,
                    Uri.parse(url));
            context.startActivity(browserIntent);
        }

    }

Answer 30

Kotlin

startActivity(Intent(Intent.ACTION_VIEW).apply {
            data = Uri.parse(your_link)
        })

Answer 31

From Anko library method

fun Context.browse(url: String, newTask: Boolean = false): Boolean {
    try {
        val intent = Intent(Intent.ACTION_VIEW)
        intent.data = Uri.parse(url)
        if (newTask) {
            intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK)
        }
        startActivity(intent)
        return true
    } catch (e: ActivityNotFoundException) {
        e.printStackTrace()
        return false
    }
}

Answer 32

A new and better way to open link from URL in Android 11.

  try {
        val intent = Intent(ACTION_VIEW, Uri.parse(url)).apply {
            // The URL should either launch directly in a non-browser app
            // (if it’s the default), or in the disambiguation dialog
            addCategory(CATEGORY_BROWSABLE)
            flags = FLAG_ACTIVITY_NEW_TASK or FLAG_ACTIVITY_REQUIRE_NON_BROWSER or
                    FLAG_ACTIVITY_REQUIRE_DEFAULT
        }
        startActivity(intent)
    } catch (e: ActivityNotFoundException) {
        // Only browser apps are available, or a browser is the default app for this intent
        // This code executes in one of the following cases:
        // 1. Only browser apps can handle the intent.
        // 2. The user has set a browser app as the default app.
        // 3. The user hasn't set any app as the default for handling this URL.
        openInCustomTabs(url)
    }

References:

https://medium.com/androiddevelopers/package-visibility-in-android-11-cc857f221cd9 and https://developer.android.com/training/package-visibility/use-cases#avoid-a-disambiguation-dialog

Answer 33

Check whether your url is correct. For me there was an unwanted space before url.

Answer 34

Basic Introduction:

https:// is using that one into the "code" so that no one in between can read them. This keeps your information safe from hackers.

http:// is using just sharing purpose, it's not secured.

About Your Problem:
XML designing:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:orientation="vertical"
    tools:context="com.example.sridhar.sharedpreferencesstackoverflow.MainActivity">
   <LinearLayout
       android:orientation="horizontal"
       android:background="#228b22"
       android:layout_weight="1"
       android:layout_width="match_parent"
       android:layout_height="0dp">
      <Button
          android:id="@+id/normal_search"
          android:text="secure Search"
          android:onClick="secure"
          android:layout_weight="1"
          android:layout_width="0dp"
          android:layout_height="wrap_content" />
      <Button
          android:id="@+id/secure_search"
          android:text="Normal Search"
          android:onClick="normal"
          android:layout_weight="1"
          android:layout_width="0dp"
          android:layout_height="wrap_content" />
   </LinearLayout>

   <LinearLayout
       android:layout_weight="9"
       android:id="@+id/button_container"
       android:layout_width="match_parent"
       android:layout_height="0dp"
       android:orientation="horizontal">

      <WebView
          android:id="@+id/webView1"
          android:layout_width="match_parent"
          android:layout_height="match_parent" />

   </LinearLayout>
</LinearLayout>

Activity Designing:

public class MainActivity extends Activity {
    //securely open the browser
    public String Url_secure="https://www.stackoverflow.com";
    //normal purpouse
    public String Url_normal="https://www.stackoverflow.com";

    WebView webView;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        webView=(WebView)findViewById(R.id.webView1);

    }
    public void secure(View view){
        webView.setWebViewClient(new SecureSearch());
        webView.getSettings().setLoadsImagesAutomatically(true);
        webView.getSettings().setJavaScriptEnabled(true);
        webView.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
        webView.loadUrl(Url_secure);
    }
    public void normal(View view){
        webView.setWebViewClient(new NormalSearch());
        webView.getSettings().setLoadsImagesAutomatically(true);
        webView.getSettings().setJavaScriptEnabled(true);
        webView.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
        webView.loadUrl(Url_normal);

    }
    public class SecureSearch extends WebViewClient{
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String Url_secure) {
            view.loadUrl(Url_secure);
            return true;
        }
    }
    public class NormalSearch extends WebViewClient{
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String Url_normal) {
            view.loadUrl(Url_normal);
            return true;
        }
    }
}

Android Manifest.Xml permissions:

<uses-permission android:name="android.permission.INTERNET"/>

You face Problems when implementing this:

1. getting The Manifest permissions
2. excess space's between url
3. Check your url's correct or not

Answer 35

If you want to do this with XML not programmatically you can use on your TextView:

android:autoLink="web"
android:linksClickable="true"

Answer 36

Try this..Worked for me!

    public void webLaunch(View view) {
            WebView myWebView = (WebView) findViewById(R.id.webview);
            myWebView.setVisibility(View.VISIBLE);
            View view1=findViewById(R.id.recharge);
            view1.setVisibility(View.GONE);
            myWebView.getSettings().setJavaScriptEnabled(true);
            myWebView.loadUrl("<your link>");

        }

xml code :-

 <WebView  xmlns:android="http://schemas.android.com/apk/res/android"
        android:id="@+id/webview"
        android:visibility="gone"
        android:layout_width="fill_parent"
        android:layout_height="fill_parent"
        />

--------- OR------------------

String url = "";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);

Answer 37

Try this one OmegaIntentBuilder

OmegaIntentBuilder.from(context)
                .web("Your url here")
                .createIntentHandler()
                .failToast("You don't have app for open urls")
                .startActivity();

Answer 38

dataWebView.setWebViewClient(new VbLinksWebClient() {
     @Override
     public void onPageFinished(WebView webView, String url) {
           super.onPageFinished(webView, url);
     }
});




public class VbLinksWebClient extends WebViewClient
{
    @Override
    public boolean shouldOverrideUrlLoading(WebView view, String url)
    {
        view.getContext().startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(url.trim())));
        return true;
    }
}

Answer 39

try this code

AndroidManifest.xml

<manifest xmlns:android="http://schemas.android.com/apk/res/android"
  xmlns:tools="http://schemas.android.com/tools"
   package="com.example.myapplication5">

    <uses-permission android:name="android.permission.INTERNET" />

    <application
    android:usesCleartextTraffic="true"
    android:allowBackup="true"
    .....
     />
     <activity android:name=".MainActivity"
        android:screenOrientation="portrait"
        tools:ignore="LockedOrientationActivity">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
</application>
 </manifest>

MainActivity.java

import android.app.Activity;
import android.content.res.Resources;
import android.os.Bundle;
import android.view.View;
import android.view.Window;
import android.webkit.WebSettings;
import android.webkit.WebView;
import android.webkit.WebViewClient;
import android.widget.Toast;

public class MainActivity extends Activity {
    private WebView mWebview;
    String link = "";// global variable
    Resources res;// global variable

    @Override


    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        requestWindowFeature(Window.FEATURE_NO_TITLE);
        setContentView(R.layout.home);

        loadWebPage();
    }

    public void loadWebPage()
    {
        mWebview = (WebView) findViewById(R.id.webView);
        WebSettings webSettings = mWebview.getSettings();
        webSettings.setJavaScriptEnabled(true);
        webSettings.setUseWideViewPort(true);
        webSettings.setLoadWithOverviewMode(true);
        final Activity activity = this;
        mWebview.setWebViewClient(new WebViewClient() {
            public void onReceivedError(WebView view, int errorCode, String description, String failingUrl) {
                Toast.makeText(activity, description, Toast.LENGTH_SHORT).show();
            }
        });
        mWebview.loadUrl("http://www.google.com");

    }

    public void reLoad(View v)
    {
        loadWebPage();
    }
}

Layout.xml

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:orientation="vertical">

    <TextView
        android:id="@+id/textView"
        android:layout_width="335dp"
        android:layout_height="47dp"
        android:layout_alignParentStart="true"
        android:layout_alignParentTop="true"
        android:layout_marginStart="9dp"
        android:layout_marginTop="8dp"
        android:paddingLeft="10dp"
        android:paddingTop="5dp"
        android:text="URL : https://ktmmovie.co/"
        android:textSize="18dp"
        android:layout_marginLeft="9dp"
        android:layout_alignParentLeft="true" />

    <com.google.android.material.floatingactionbutton.FloatingActionButton
        android:id="@+id/floatingActionButton2"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_alignParentTop="true"
        android:layout_alignParentEnd="true"
        android:layout_marginStart="7dp"
        android:layout_marginLeft="7dp"
        android:layout_marginEnd="8dp"
        android:layout_toEndOf="@+id/textView"
        android:layout_toRightOf="@+id/textView"
        android:clickable="true"
        android:src="@android:drawable/ic_popup_sync"
        android:layout_marginRight="8dp"
        android:layout_alignParentRight="true"
        android:onClick="reLoad"/>

    <WebView
        android:id="@+id/webView"
        android:layout_width="401dp"
        android:layout_height="665dp"
        android:layout_below="@+id/textView"
        android:layout_alignParentStart="true"
        android:layout_alignParentLeft="true"
        android:layout_alignParentBottom="true"
        android:layout_centerHorizontal="true"
        android:layout_marginStart="3dp"
        android:layout_marginLeft="3dp"
        android:layout_marginTop="3dp"
        android:layout_marginBottom="7dp" />


</RelativeLayout>

Answer 40

Kotlin Developers can use this

var webpage = Uri.parse(url)
    if (!url.startsWith("http://") && !url.startsWith("https://")) {
        webpage = Uri.parse("http://$url")
    }
    val intent = Intent(Intent.ACTION_VIEW, webpage)
    if (intent.resolveActivity(packageManager) != null) {
        startActivity(intent)
    }

Answer 41

Kotlin Solution

All the answers were opening the url in default app for that url. I wanted to always open any url in the browser. I needed some solution in kotlin and implemented the code below.

fun getPackageNameForUrl(context: Context, url: String): String? {
    val intent = Intent(Intent.ACTION_VIEW, Uri.parse(url))
    val resolveInfo = context.packageManager.resolveActivity(intent, PackageManager.MATCH_DEFAULT_ONLY)
    return resolveInfo?.activityInfo?.packageName
}

fun openInBrowser(context: Context, url: String) {
    val intent = Intent(Intent.ACTION_VIEW, Uri.parse(url))
    val packageName = getPackageNameForUrl(context, "http://")
    packageName?.takeIf { 
        it == "android" 
    }?.let { intent.setPackage(defaultBrowserPackageName); }
    startActivity(context, intent, null)
}