您好我需要触发按钮才能打开并关闭弹出窗口,只要在弹出窗口外发生鼠标点击,弹出窗口就会关闭。
这是我的XAML:
<Button x:Name="About"
Height="50"
Margin="0,-30,5,0"
HorizontalAlignment="Right"
Style="{StaticResource AboutButtonStyle}" />
<Popup HorizontalOffset="-300"
IsOpen="{Binding IsAboutPopupOpen, Mode=TwoWay}"
Placement="RelativePoint"
PlacementTarget="{Binding ElementName=About}"
StaysOpen="False"
VerticalOffset="-125">
<Border Padding="10"
Background="White"
BorderBrush="{StaticResource SeparatorColorBrush}"
BorderThickness="1">
<TextBlock>Some Text</TextBlock>
</Border>
</Popup>
在我的AboutViewModel中,我实现了一个属性IsAboutPopupOpen
:
private bool isAboutPopupOpen;
public bool IsAboutPopupOpen
{
get
{
return this.isAboutPopupOpen;
}
set
{
if (value != this.isAboutPopupOpen)
{
this.isAboutPopupOpen = value;
this.NotifyOfPropertyChange(() => IsAboutPopupOpen);
}
}
}
public void About()
{
IsAboutPopupOpen = true;
}
问题是,当弹出窗口打开时,我单击“关于”按钮,PopUp将关闭并再次打开。它应该关闭。除此之外,行为是正确的。
我已经找到了一个简单的解决方案,似乎无法找到它。这应该是一个常见的问题。哦,我正在使用Caliburn.Micro,但这不重要,我不会想。感谢
答案 0 :(得分:0)
<ToggleButton
x:Name="ShowAboutPopup"
Grid.Row="1"
Height="50"
Margin="0,-30,5,0"
HorizontalAlignment="Right"
Grid.ZIndex="1000"
Style="{StaticResource AboutButtonStyle}" />
然后我决定通过IsOpen属性将我的ToggleButton连接到弹出窗口,并将StaysOpen设置为false:
<Popup
Grid.Row="1"
Closed="AboutPopup_OnClosed"
HorizontalOffset="-300"
IsOpen="{Binding ElementName=ShowAboutPopup, Path=IsChecked, Mode=OneWay}"
Placement="RelativePoint"
PlacementTarget="{Binding ElementName=ShowAboutPopup}"
StaysOpen="False"
VerticalOffset="-125">
...
</Popup>
最后我在代码隐藏中实现了AboutPopup_OnClosed:
private void AboutPopup_OnClosed(object sender, EventArgs e)
{
if (!object.Equals(this.ShowAboutPopup, Mouse.DirectlyOver))
{
this.ShowAboutPopup.IsChecked = false;
}
}
希望这有助于其他可能与此斗争的人。最重要的是,您通常应该使用切换按钮来打开和关闭弹出窗口。