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
实现这一点的常见方法是使用以下代码:
String url = "http://www.stackoverflow.com";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);
可以更改为短代码版本。。。
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);
最短!:
startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.stackoverflow.com")));
Kotlin溶液
所有答案都是在该url的默认应用程序中打开该url。我想总是在浏览器中打开任何url。我需要一些kotlin的解决方案,并实现了下面的代码。
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)
}