c# - 用户做出选择后保持 WPF 弹出窗口打开

Question

当用户从主网格中选择记录时，我会出现一个小的弹出窗口。我想在用户做出选择后保持此窗口打开，直到他或她单击取消。这是我的小代码。如果我注释掉 this.close() 方法，则控制永远不会返回以执行其余代码。

private void ClickSubmit(object sender, RoutedEventArgs e)
{
    try
    {
        IsCancelled = false;
        // Figure out to keep this little window open after clicking submit !!
        this.Close();
        return;
    }
    catch (Exception ex)
    {
        ex.ToString();
    }
}

private void ClickCancel(object sender, RoutedEventArgs e)
{
    try
    {
        Trace.WriteLine("Popup exit !");
        IsCancelled = true;
        this.Close();
        return;
    }
    catch (Exception ex)
    {
        ex.ToString();
    }

}

OrdRec or = lvOrders.SelectedItem as OrdRec;

PxQtyPopup pop = new PxQtyPopup(or);
//pop.ShowDialog();
pop.Show();

if (!pop.IsCancelled) // check whether price/qty change submitted or popup window was cancelled
    (DCtxt as OrdViewModel).ModifyOrder(or, int.Parse(pop.Qty), decimal.Parse(pop.Px));

Answer 1

将 IsOpen 绑定(bind)到合适的 bool 值，这将打开或关闭弹出窗口。下面的示例使用 ToggleButton 的 OpenState( bool 值)来打开/关闭弹出窗口。

<ToggleButton Height="30" Width="60"             
                x:Name="OpenState">
    <StackPanel>
        <TextBlock Text="{Binding IsChecked, ElementName=OpenState}"
                    FontSize="18"
                    VerticalAlignment="Center"
                    HorizontalAlignment="Center">
        </TextBlock>

        <Popup Name="myPopup"
                IsOpen="{Binding IsChecked, ElementName=OpenState}">


            <TextBlock Name="myPopupText"
                       Background="LightBlue"
                       Foreground="Blue">Popup Text</TextBlock>

        </Popup>

    </StackPanel>
</ToggleButton>