从第一个用户控件的磁贴单击显示第二个用户控件 [英] Showing Second User control from first user control's tile click
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问题描述
I have created a Wpf application with mvvm architecture. In main window I have shown a first user control with tile in it. Now I need when I click on a tile of the first user control then it will hide and open second user control in the same Main window and if i close second usercontrol it has to go back to first user control. (Similar to Windows 8 when we click on tile it will open and if we close it will go back to main window)
Can any one tell me ho to achieve this.
i have tried in this way:
on click of tile event i am trying to display second user control.
private void Tile_Click(object sender, EventArgs e)
{
SecondUserControl.Visibility = Visibility.Visible;
}
but not working :(
推荐答案
在MVVM中如果你想要隐藏和显示你需要设置2个属性的用户控件。
In MVVM if you would like to hide and show user controls you need to set up 2 properties.
public bool IsFirstVisible {get; set;}
public bool IsSecondVisible {get; set;}
public void OnTileClick()
{
if (IsFirstVisble)
{
IsSecondVisible = true;
IsFirstVisible = false;
}
else
{
IsSecondVisible = false;
IsFirstVisible = true;
}
}
现在,在您的磁贴上,您将切换用户控件的可见性被展示。接下来,您需要使用BoolToVisibility Converter在xaml上反映这一点。
Now, on your tile click, you will toggle the visibility on which user control is being shown. Next you need to reflect this on the xaml with a BoolToVisibility Converter like so.
<window.resources>
<booleantovisibilityconverter x:key="boolToVis" />
</window.resources>
<local:firstusercontrol visibility="{Binding IsFirstVisible, Mode=TwoWay, UpdateSourceTrigger=PropertyChanged, Converter = {StaticResource boolToVis}}" />
<local:secondusercontrol visibility="{Binding IsSecondVisible, Mode=TwoWay, UpdateSourceTrigger=PropertyChanged, Converter = {StaticResource boolToVis}}" />;
这应该反映属性为真时的可见状态,以及当pr属性时的崩溃状态operty是假的。
This should reflect a visible state when the property is true, and a collapsed state when the property is false.
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