How to check if ANY ContentDialog is open?
So we can only have one opened content dialog at a time. This is fine. But in my app there are several possible content dialogs that can be opened, and I would like to avoid making my own variable because I can forget to add it somewhere and the whole app will crash (because trying to open second content dialog throws exception).
So my question is: How to check if any ContentDialog is open?
Note:
- I don't want to check for specific ContentDialog.
- I would like to avoid creating my own variables.
c# uwp
add a comment |
So we can only have one opened content dialog at a time. This is fine. But in my app there are several possible content dialogs that can be opened, and I would like to avoid making my own variable because I can forget to add it somewhere and the whole app will crash (because trying to open second content dialog throws exception).
So my question is: How to check if any ContentDialog is open?
Note:
- I don't want to check for specific ContentDialog.
- I would like to avoid creating my own variables.
c# uwp
add a comment |
So we can only have one opened content dialog at a time. This is fine. But in my app there are several possible content dialogs that can be opened, and I would like to avoid making my own variable because I can forget to add it somewhere and the whole app will crash (because trying to open second content dialog throws exception).
So my question is: How to check if any ContentDialog is open?
Note:
- I don't want to check for specific ContentDialog.
- I would like to avoid creating my own variables.
c# uwp
So we can only have one opened content dialog at a time. This is fine. But in my app there are several possible content dialogs that can be opened, and I would like to avoid making my own variable because I can forget to add it somewhere and the whole app will crash (because trying to open second content dialog throws exception).
So my question is: How to check if any ContentDialog is open?
Note:
- I don't want to check for specific ContentDialog.
- I would like to avoid creating my own variables.
c# uwp
c# uwp
asked Jan 1 at 14:04


MailoszMailosz
6917
6917
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add a comment |
1 Answer
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ContentDialog is shown in the PopupRoot so using VisualTreeHelper.GetOpenPopups()
will help you get it.
var openedpopups = VisualTreeHelper.GetOpenPopups(Window.Current);
foreach (var popup in openedpopups)
{
if(popup.Child is ContentDialog)
{
//some content dialog is open.
}
}
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
ContentDialog is shown in the PopupRoot so using VisualTreeHelper.GetOpenPopups()
will help you get it.
var openedpopups = VisualTreeHelper.GetOpenPopups(Window.Current);
foreach (var popup in openedpopups)
{
if(popup.Child is ContentDialog)
{
//some content dialog is open.
}
}
add a comment |
ContentDialog is shown in the PopupRoot so using VisualTreeHelper.GetOpenPopups()
will help you get it.
var openedpopups = VisualTreeHelper.GetOpenPopups(Window.Current);
foreach (var popup in openedpopups)
{
if(popup.Child is ContentDialog)
{
//some content dialog is open.
}
}
add a comment |
ContentDialog is shown in the PopupRoot so using VisualTreeHelper.GetOpenPopups()
will help you get it.
var openedpopups = VisualTreeHelper.GetOpenPopups(Window.Current);
foreach (var popup in openedpopups)
{
if(popup.Child is ContentDialog)
{
//some content dialog is open.
}
}
ContentDialog is shown in the PopupRoot so using VisualTreeHelper.GetOpenPopups()
will help you get it.
var openedpopups = VisualTreeHelper.GetOpenPopups(Window.Current);
foreach (var popup in openedpopups)
{
if(popup.Child is ContentDialog)
{
//some content dialog is open.
}
}
edited Jan 2 at 6:27


Martin Zikmund
25.7k63761
25.7k63761
answered Jan 1 at 19:06
VigneshVignesh
9142318
9142318
add a comment |
add a comment |
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