在Powershell中运行的XAML文本框中捕获Enter键

<Window x:Class="WpfApp.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="MainWindow" Height="350" Width="525">
    <Grid>
        <TextBox x:Name="textBox" KeyDown="TextBox_KeyDown" />
        <Button Content="Submit" Click="Button_Click" />
    </Grid>
</Window>

Add-Type -AssemblyName PresentationFramework

# 加载XAML文件
$reader = [System.Xml.XmlReader]::Create("Path\to\your\XAML\File.xaml")
$window = [Windows.Markup.XamlReader]::Load($reader)

# 获取文本框和按钮对象
$textBox = $window.FindName("textBox")
$button = $window.FindName("button")

# 定义Enter键的事件处理函数
$textBox_KeyDown = {
    param($sender, $e)
    if ($e.Key -eq "Enter") {
        # 在这里处理Enter键的逻辑
        Write-Host "Enter键被按下"
    }
}

# 绑定事件处理函数
$textBox.Add_KeyDown($textBox_KeyDown)

# 显示窗口
$window.ShowDialog() | Out-Null