Powershell 包 uri 对象

发布于 2024-09-28 09:59:30 字数 238 浏览 0 评论 0原文

我正在尝试创建一个包 ui，引用 powershell 中程序集文件内的 xaml 资源。读完这篇文章后我尝试这样做：

$resource = new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

我收到一个错误，指出它需要一个端口，因为有两个冒号。

有人可以请建议吗？

原文

I'm trying to create a pack ui referencing a xaml resource inside of an assembly file in powershell. After reading this post I tried to do this:

$resource = new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

The I get an error noting that it is expecting a port since there are two colons.

Can anyone please advice?

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风蛊 2024-10-05 09:59:30

您可以采用以下两种方法之一来解决此问题。一种是加载并初始化 WPF 基础结构：

Add-Type -AssemblyName PresentationFramework,PresentationCore
[windows.application]::current > $null # Inits the pack protocol
new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

另一种方法是手动注册包协议：

$opt = [GenericUriParserOptions]::GenericAuthority
$parser = new-object system.GenericUriParser $opt
if (![UriParser]::IsKnownScheme("pack")) { 
    [UriParser]::Register($parser,"pack",-1) 
}
new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

You can go about this one of two ways. One is to load up and init the WPF infrastructure:

Add-Type -AssemblyName PresentationFramework,PresentationCore
[windows.application]::current > $null # Inits the pack protocol
new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

The other way is to manually register the pack protocol:

$opt = [GenericUriParserOptions]::GenericAuthority
$parser = new-object system.GenericUriParser $opt
if (![UriParser]::IsKnownScheme("pack")) { 
    [UriParser]::Register($parser,"pack",-1) 
}
new-object system.uri("pack://application:,,,/WPFResource;component/test.xaml")

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