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
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
其他回答
短代码版本。。。
if (!strUrl.startsWith("http://") && !strUrl.startsWith("https://")){
strUrl= "http://" + strUrl;
}
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse(strUrl)));
简单和最佳实践
方法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*/
}
方法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*/
}
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)
}
MarkB的反应是正确的。在我的例子中，我使用的是Xamarin，与C#和Xamarin一起使用的代码是：
var uri = Android.Net.Uri.Parse ("http://www.xamarin.com");
var intent = new Intent (Intent.ActionView, uri);
StartActivity (intent);
该信息来源于：https://developer.xamarin.com/recipes/android/fundamentals/intent/open_a_webpage_in_the_browser_application/
kolin扩展函数。
fun Activity.openWebPage(url: String?) = url?.let {
val intent = Intent(Intent.ACTION_VIEW, Uri.parse(it))
if (intent.resolveActivity(packageManager) != null) startActivity(intent)
}
