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
Intent getWebPage = new Intent(Intent.ACTION_VIEW, Uri.parse(MyLink));
startActivity(getWebPage);
其他回答
如果您想向用户显示一个带有所有浏览器列表的对话框，以便他可以选择首选项，下面是示例代码：
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 :)
}
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();
}
该错误是由于URL无效而发生的，Android OS无法找到数据的操作视图。因此，您必须验证URL是否有效。
试试这个OmegaIntentBuilder
OmegaIntentBuilder.from(context)
.web("Your url here")
.createIntentHandler()
.failToast("You don't have app for open urls")
.startActivity();
简短甜美的Kotlin助手功能：
private fun openUrl(link: String) =
startActivity(Intent(Intent.ACTION_VIEW, Uri.parse(link)))
我认为这是最好的
openBrowser(context, "http://www.google.com")
将以下代码放入全局类
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);
}
