如何从内置web浏览器而不是应用程序中的代码打开URL？
我试过了：
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();
}
但我有个例外：
No activity found to handle Intent{action=android.intent.action.VIEW data =www.google.com
当前回答
Chrome自定义选项卡现在可用：
第一步是将自定义选项卡支持库添加到build.gradle文件中：
dependencies {
...
compile 'com.android.support:customtabs:24.2.0'
}
然后，要打开chrome自定义选项卡：
String url = "https://www.google.pt/";
CustomTabsIntent.Builder builder = new CustomTabsIntent.Builder();
CustomTabsIntent customTabsIntent = builder.build();
customTabsIntent.launchUrl(this, Uri.parse(url));
有关详细信息：https://developer.chrome.com/multidevice/android/customtabs
其他回答
试试这个OmegaIntentBuilder
OmegaIntentBuilder.from(context)
.web("Your url here")
.createIntentHandler()
.failToast("You don't have app for open urls")
.startActivity();
android.webkit.URLUtil自Api级别1（android 1.0）以来，guessUrl（String）方法工作得非常好（即使使用file://或data://）。用作：
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
在“活动”调用中：
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(URLUtil.guessUrl(download_link)));
if (intent.resolveActivity(getPackageManager()) != null)
startActivity(intent);
有关详细信息，请查看完整的guessUrl代码。
在Android 11中打开URL链接的新的更好方法。
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)
}
参考文献：
https://medium.com/androiddevelopers/package-visibility-in-android-11-cc857f221cd9和https://developer.android.com/training/package-visibility/use-cases#avoid-a-消除歧义对话框
Kotlin回答：
val browserIntent = Intent(Intent.ACTION_VIEW, uri)
ContextCompat.startActivity(context, browserIntent, null)
我在Uri上添加了一个扩展，以使这更加容易
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)
}
另外，这里有一个简单的扩展函数，可以将字符串安全地转换为Uri。
"https://stackoverflow.com".asUri()?.openInBrowser(context)
fun String?.asUri(): Uri? {
return try {
Uri.parse(this)
} catch (e: Exception) {
null
}
}
Kotlin开发人员可以使用
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)
}
