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naeron84
naeron84

Reputation: 3025

WPF show control in debug mode only

I have some useful wpf buttons to test some functionality. It would be good not to show them in release but in debug indeed.

Doing it from code is easy. But I'd prefer a declarative solution.

Upvotes: 27

Views: 9255

Answers (6)

biomiker
biomiker

Reputation: 3306

As of at latest Visual Studio 17.2.6 / WinUI3, the following simple beauty seems to work (details not related to visibility excluded):

<Page     
   xmlns:appmodel="using:Windows.ApplicationModel">

   <Button Visibility="{x:Bind Path=appmodel:Package.Current.IsDevelopmentMode}"/>
</Page>

Upvotes: -1

DanW
DanW

Reputation: 2056

This will show when the debugger is attached. First, set the namespace:

xmlns:diag="clr-namespace:System.Diagnostics;assembly=mscorlib"

then set your resource:

    <BooleanToVisibilityConverter x:Key="BoolToVisibilityConverter"/>

then use the binding: 

<MenuItem Header="onlyIfDebuggerAttached" Visibility="{Binding Source={x:Static diag:Debugger.IsAttached}, Converter={StaticResource BoolToVisibilityConverter}}" />

Upvotes: 23

Eric Ouellet
Eric Ouellet

Reputation: 11754

Based on Steven answer... You can use a static class and declare the visibility once only.

using System.Windows;

namespace HQ.Wpf.Util
{
    /* Usage:

        xmlns:wpfUtil="clr-namespace:HQ.Wpf.Util;assembly=WpfUtil"

        <Button Name="CmdTest" Click="CmdTestOnClick" Visibility="{x:Static wpfUtil:DebugVisibility.DebugOnly}">Test</Button>

    */

    public static class DebugVisibility
    {
        public static Visibility DebugOnly
        {
#if DEBUG
            get { return Visibility.Visible; }
#else
            get { return Visibility.Collapsed; }
#endif
        }

        public static Visibility ReleaseOnly
        {
#if DEBUG
            get { return Visibility.Collapsed; }
#else
            get { return Visibility.Visible; }
#endif
        }
    }
}

XAML:

<Button Name="CmdTest" Click="CmdTestOnClick" 
    Visibility="{x:Static wpfUtil:DebugVisibility.DebugOnly}">Test
</Button>

Upvotes: 1

Aaron D.
Aaron D.

Reputation: 141

Not sure what the difference is between this and Steven's approach, but I used his property as a non-static property in a non-static class, and referenced it like such:

<local:MyClass x:Key="MyClass" />
<MyControl Visibility="{Binding IsDebug, Source={StaticResource MyClass}, Mode=OneTime}" />

Upvotes: 0

Steven
Steven

Reputation: 4963

The only solution I know of is to create a static property somewhere like this:

    public static Visibility IsDebug
    {
#if DEBUG
        get { return Visibility.Visible; }
#else
        get { return Visibility.Collapsed; }
#endif
    }

Then use it in XAML like this:

<MyControl Visibility="{x:Static local:MyType.IsDebug}" />

XAML doesn't have anything for compiler flags.

Upvotes: 33

Henk Holterman
Henk Holterman

Reputation: 273219

As far as I know there is no way to use the Configuration constants (Debug, Release) from XAML.
So the best you can get is to bind the Visibility property of the buttons to a Debug property on your datacontext. But setting that property would still require some code.

Upvotes: 2

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