WPF: How to make binding of IsOpen property in a professional way?
Multi tool use
up vote
-1
down vote
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I need to process with IsOpen property in a code Behind.
XAML:
DataContext="{Binding RelativeSource={RelativeSource Self}}"
IsOpen="{Binding ChildWindow_IsOpen}"
Code Behind:
public bool ChildWindow_IsOpen
{
get { return (bool)GetValue(WindowProperty); }
set { SetValue(WindowProperty, value); }
}
public static readonly DependencyProperty WindowProperty = DependencyProperty.Register("ChildWindow_IsOpen", typeof(bool), typeof(MainWindow));
MainWindow:
ChildWindow childWindow = new ChildWindow();
private async void button3_OnClick(object sender, RoutedEventArgs e)
{
if (childWindow.ChildWindow_IsOpen == false)
{
await this.ShowChildWindowAsync(new ChildWindow() { IsModal = false, AllowMove = true, }, RootGrid);
childWindow.ChildWindow_IsOpen = true;
}
else if (childWindow.ChildWindow_IsOpen == true)
{
childWindow.Close();
childWindow.ChildWindow_IsOpen = false;
}
else
{
return;
}
}
So my question is how to do that in a professional way?
My code doesn't affect to ChildWindow at all.
Thanks in advance!
Update: ChildWindow's XAML is situated in MainWindow XAML. This works like a charm!
private void button1_Click(object sender, RoutedEventArgs e)
{
if (ChildWindow.IsOpen == false)
{
ChildWindow.IsOpen = true;
}
else if (ChildWindow.IsOpen == true)
{
ChildWindow.Close();
}
else
{
return;
}
c# wpf xaml binding
asked Nov 10 at 13:51
Pew
12
|
show 3 more comments
up vote
-1
down vote
favorite
I need to process with IsOpen property in a code Behind.
XAML:
DataContext="{Binding RelativeSource={RelativeSource Self}}"
IsOpen="{Binding ChildWindow_IsOpen}"
Code Behind:
public bool ChildWindow_IsOpen
{
get { return (bool)GetValue(WindowProperty); }
set { SetValue(WindowProperty, value); }
}
public static readonly DependencyProperty WindowProperty = DependencyProperty.Register("ChildWindow_IsOpen", typeof(bool), typeof(MainWindow));
MainWindow:
ChildWindow childWindow = new ChildWindow();
private async void button3_OnClick(object sender, RoutedEventArgs e)
{
if (childWindow.ChildWindow_IsOpen == false)
{
await this.ShowChildWindowAsync(new ChildWindow() { IsModal = false, AllowMove = true, }, RootGrid);
childWindow.ChildWindow_IsOpen = true;
}
else if (childWindow.ChildWindow_IsOpen == true)
{
childWindow.Close();
childWindow.ChildWindow_IsOpen = false;
}
else
{
return;
}
}
So my question is how to do that in a professional way?
My code doesn't affect to ChildWindow at all.
Thanks in advance!
Update: ChildWindow's XAML is situated in MainWindow XAML. This works like a charm!
private void button1_Click(object sender, RoutedEventArgs e)
{
if (ChildWindow.IsOpen == false)
{
ChildWindow.IsOpen = true;
}
else if (ChildWindow.IsOpen == true)
{
ChildWindow.Close();
}
else
{
return;
}
c# wpf xaml binding
asked Nov 10 at 13:51
Pew
12
What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doingchildWindow.ChildWindow_IsOpenjust fine
– MickyD
Nov 10 at 14:09
2
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45
|
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I need to process with IsOpen property in a code Behind.
XAML:
DataContext="{Binding RelativeSource={RelativeSource Self}}"
IsOpen="{Binding ChildWindow_IsOpen}"
Code Behind:
public bool ChildWindow_IsOpen
{
get { return (bool)GetValue(WindowProperty); }
set { SetValue(WindowProperty, value); }
}
public static readonly DependencyProperty WindowProperty = DependencyProperty.Register("ChildWindow_IsOpen", typeof(bool), typeof(MainWindow));
MainWindow:
ChildWindow childWindow = new ChildWindow();
private async void button3_OnClick(object sender, RoutedEventArgs e)
{
if (childWindow.ChildWindow_IsOpen == false)
{
await this.ShowChildWindowAsync(new ChildWindow() { IsModal = false, AllowMove = true, }, RootGrid);
childWindow.ChildWindow_IsOpen = true;
}
else if (childWindow.ChildWindow_IsOpen == true)
{
childWindow.Close();
childWindow.ChildWindow_IsOpen = false;
}
else
{
return;
}
}
So my question is how to do that in a professional way?
My code doesn't affect to ChildWindow at all.
Thanks in advance!
Update: ChildWindow's XAML is situated in MainWindow XAML. This works like a charm!
private void button1_Click(object sender, RoutedEventArgs e)
{
if (ChildWindow.IsOpen == false)
{
ChildWindow.IsOpen = true;
}
else if (ChildWindow.IsOpen == true)
{
ChildWindow.Close();
}
else
{
return;
}
c# wpf xaml binding
asked Nov 10 at 13:51
Pew
12
I need to process with IsOpen property in a code Behind.
XAML:
DataContext="{Binding RelativeSource={RelativeSource Self}}"
IsOpen="{Binding ChildWindow_IsOpen}"
Code Behind:
public bool ChildWindow_IsOpen
{
get { return (bool)GetValue(WindowProperty); }
set { SetValue(WindowProperty, value); }
}
public static readonly DependencyProperty WindowProperty = DependencyProperty.Register("ChildWindow_IsOpen", typeof(bool), typeof(MainWindow));
MainWindow:
ChildWindow childWindow = new ChildWindow();
private async void button3_OnClick(object sender, RoutedEventArgs e)
{
if (childWindow.ChildWindow_IsOpen == false)
{
await this.ShowChildWindowAsync(new ChildWindow() { IsModal = false, AllowMove = true, }, RootGrid);
childWindow.ChildWindow_IsOpen = true;
}
else if (childWindow.ChildWindow_IsOpen == true)
{
childWindow.Close();
childWindow.ChildWindow_IsOpen = false;
}
else
{
return;
}
}
So my question is how to do that in a professional way?
My code doesn't affect to ChildWindow at all.
Thanks in advance!
Update: ChildWindow's XAML is situated in MainWindow XAML. This works like a charm!
private void button1_Click(object sender, RoutedEventArgs e)
{
if (ChildWindow.IsOpen == false)
{
ChildWindow.IsOpen = true;
}
else if (ChildWindow.IsOpen == true)
{
ChildWindow.Close();
}
else
{
return;
}
c# wpf xaml binding
c# wpf xaml binding
asked Nov 10 at 13:51
Pew
12
asked Nov 10 at 13:51
Pew
12
edited Nov 10 at 15:05
asked Nov 10 at 13:51
Pew
12
asked Nov 10 at 13:51
Pew
12
asked Nov 10 at 13:51
Pew
12
12
What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doingchildWindow.ChildWindow_IsOpenjust fine
– MickyD
Nov 10 at 14:09
2
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45
|
show 3 more comments
What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doingchildWindow.ChildWindow_IsOpenjust fine
– MickyD
Nov 10 at 14:09
2
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45
What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doing
childWindow.ChildWindow_IsOpen just fine– MickyD
Nov 10 at 14:09
What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doing
childWindow.ChildWindow_IsOpen just fine– MickyD
Nov 10 at 14:09
2
2
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45
|
show 3 more comments
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What do you mean by "professional"? What do you mean "My code doesn't affect to ChildWindow"? You seem to be doing
childWindow.ChildWindow_IsOpenjust fine– MickyD
Nov 10 at 14:09
2
The “professional” way would be to use MVVM
– Dave M
Nov 10 at 14:13
@DaveM Exactly. WPF/UWP and XAML were designed with MVVM in mind. While you can use any other approach and pattern, doing so missed about 90% of it's power and runs into issues at every other corner. This does not look like proper MVVM, so I asume the issue is there.
– Christopher
Nov 10 at 14:17
@MickyD, first button click should open ChildWindow, second one should close ChildWindow. However it doesn't happen.
– Pew
Nov 10 at 14:17
@Christopher Since the OP wants to open/close windows, MVVM has questionable benefit for this particular problem. If you intend to open a view from the viewmodel (including by indirect means of injected window'ing services into the VM) then that is a violation of the MVVM best practices
– MickyD
Nov 10 at 14:45