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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T20:50:42+00:00 2026-05-13T20:50:42+00:00

I have three different configurations on my project, all three do not require all

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I have three different configurations on my project, all three do not require all files to be build into the application. Actually I’d prefer if I could exclude those files from the build, which would make my application a little more lightweight.

What I’m looking for is #if MYCONFIG or #if DEBUG statement but for files. I’ve already read that this can be accomplished by manually editing the csproj file, but I can’t find that anymore…and are there other ways?

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  1. Editorial Team
    Editorial Team
    2026-05-13T20:50:42+00:00Added an answer on May 13, 2026 at 8:50 pm

    There are two different ways:
    In your csproj files, you will have sections that look like this:

    <ItemGroup>
        <Compile Include="Helper.cs" />
        <Compile Include="Properties\AssemblyInfo.cs" />
    </ItemGroup>
    

    What you can do is set up a new project configuration (Build menu, Configuration Manager, select New from the Active solution configuration dropdown), then manually change the ItemGroup node to this:

    <ItemGroup Condition=" '$(Configuration)' == 'MyNewConfiguration' ">
        <Compile Include="Helper.cs" />
        <Compile Include="Properties\AssemblyInfo.cs" />
    </ItemGroup>
    

    The second way, as you referred to in your question, is to use conditional debug symbols. At the top of your file, have the statement

    #if MYDEBUGSYMBOL
    

    and at the bottom have

    #endif
    

    then you can define the debug symbols; right clickon your project file, select Properties, go to the Build tab, and enter the debug symbol in the Conditional compilation symbols textbox.

    I would probably stick with the first method.

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