How can I open a URL in Android's web browser from my application?
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_lin开发者_StackOverflow社区k));
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
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://".
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")));
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));
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"
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);
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? {
return try {
Uri.parse(this)
} catch (e: Exception) {
null
}
}
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 :)
}
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)
}
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");
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...
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;
}
}
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.
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);
A short code version...
if (!strUrl.startsWith("http://") && !strUrl.startsWith("https://")){
strUrl= "http://" + strUrl;
}
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(strUrl)));
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*/
}
Short & sweet Kotlin helper function:
private fun openUrl(link: String) =
startActivity(Intent(Intent.ACTION_VIEW, Uri.parse(link)))
String url = "http://www.example.com";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);
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
Intent getWebPage = new Intent(Intent.ACTION_VIEW, Uri.parse(MyLink));
startActivity(getWebPage);
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/
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
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" />
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);
}
}
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
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.
Simply go with short one to open your Url in Browser:
Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("YourUrlHere"));
startActivity(browserIntent);
Kotlin
startActivity(Intent(Intent.ACTION_VIEW).apply {
data = Uri.parse(your_link)
})
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");
e.g. :
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);
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);
}
//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);
}
}
加载中,请稍侯......
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