英文:
Winui 3 how to make a new app window at button click
问题
正在使用WinUI 3构建一个应用程序，使用XAML和C＃。
在运行时如何创建应用程序的新实例？在其中一个页面上，我有一个按钮，内容为“打开新窗口”。在按钮点击时，我想从我正在构建的应用程序中打开一个新的自身/窗口。在WinUI 3示例应用程序中有一个示例，但对于使用最新版本的WinUI 3并从应用程序设计器创建应用程序的人来说，它不起作用。此外，在搜索Microsoft文档时，未能找到我想要的信息。
已经在互联网上搜索到以下示例，但均未成功：
https://learn.microsoft.com/en-us/windows/apps/design/layout/app-window
https://github.com/microsoft/WinUI-Gallery/blob/main/WinUIGallery/Navigation/NavigationRootPage.xaml.cs
我遇到了一次又一次的问题。
英文:
Am building an app with Winui 3 so XAML and C#
Ok, how to make a new instance of the app at runtime. So in one of the pages I have a button with content "Open new window". At button click I want to open a new self/ window from the app I am building. In the winui 3 sample app there is an example, but that doesn't work for people who uses the newest version of winui 3 and have the app created from the app designer.
Also when looking around for Microsoft documentation didn't get me where I want to be.
Have searched the internet found the following examples that didn't work:
https://learn.microsoft.com/en-us/windows/apps/design/layout/app-window
https://github.com/microsoft/WinUI-Gallery/blob/main/WinUIGallery/Navigation/NavigationRootPage.xaml.cs
What I got was breakdown after breakdown.
答案1
得分: 2
这有点复杂，但试试看。
-
在 Assets 文件夹中创建一个带有唯一扩展名的虚拟文件，并确保将其 Build Action 属性设置为 Content。在这种情况下，我的虚拟文件是一个名为 "dummy.test" 的文本文件。
-
转到 Package.appxmanifest，Declarations 并添加一个 File Type Associations 声明，设置以下属性：
- 显示名称：Test
- 名称：test
- 文件类型：.test
-
在按钮点击事件处理程序中，启动该文件。这将启动与 "test" 扩展名关联的应用程序，本例中是您的应用程序。
private async void LaunchAppButton_Click(object sender, RoutedEventArgs e)
{
StorageFile file = await Package.Current.InstalledLocation.GetFileAsync(@"Assets\dummy.test");
_ = await Windows.System.Launcher.LaunchFileAsync(file);
}
更新
如果您想要做与 WinUI 3 Gallery 应用程序相同的事情，您可以从存储库中获取所有代码。
以下是其中的一部分：
-
CreateWindow()方法实质上只是创建一个Window，将其添加到一个集合中以跟踪它，并返回它。
因此，基本上您只需要创建一个 Window，设置您的主要内容并激活它。
Window newWindow = new();
newWindow.Content = /* 您的主要内容 */;
newWindow.Activate();
英文:
This is a bit tricky but give it a try.
-
Create a dummy file with a unique extension in the Assets folder and make sure to set its Build Action property as Content. In this case, my dummy file is a text file named "dummy.test".
-
Go to Package.appxmanifest, Declarations and add a File Type Associations declaration and set these properties:
- Display name: Test
- Name: test
- File type: .test
-
In your button click event handler, launch the file. This should launch the app associated with the "test" extension, which is your app in this case.
private async void LaunchAppButton_Click(object sender, RoutedEventArgs e)
{
StorageFile file = await Package.Current.InstalledLocation.GetFileAsync(@"Assets\dummy.test");
_ = await Windows.System.Launcher.LaunchFileAsync(file);
}
UPDATE
If you want to do the same thing that the WinUI 3 Gallery app does, you can get all the code from the repo.
Here's some part of it:
-
The
CreateWindow()method essentially just creates aWindow, adds it to a collection just to keep track of it and returns it.
So, basically you just need to create a Window, set your main content and activate it.
Window newWindow = new();
newWindow.Content = /* your main content*/
newWindow.Activate();
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