.csproj/msbuild PropertyGroup configuration for TF Build Agent

Question

How can I make a project build behave different on 1) (local) compile inside Visual Studio and 2) triggered by TFS Build Server on TF Build Agent?

I tried (both without success)

<PropertyGroup Condition="'$(TF_BUILD)' == 'true' AND '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">

<PropertyGroup Condition="'$(BuildingInsideVisualStudio)' != 'true' AND '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">

I have (in both cases) created 2 such property groups with == and != swapped on the first condition (the one that should result in different behavior) and different content in the PropertyGroup (which is what I am really aiming for).

I am using VS 2017 and TFS 2017.

edit/update:

I found the mistake in my ways, it was that I edited the AnyCPU PropertyGoup and built x86.

Answer 1

Actually, all of this works, including Daniel Mann's suggestion. I was using it on the wrong propertygroup.

Since I found this information hard to find, I hope this summary helps someone in the future.

Different property values on TF and local (VS) build can be written like

<PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
...
  <Optimize Condition="'$(BUILD_SOURCESDIRECTORY)' != ''">true</Optimize>
  <Optimize Condition="'$(TF_BUILD)' != 'true'">false</Optimize>
..redundant..
  <Optimize Condition="'$(BuildingInsideVisualStudio)' == 'false'">true</Optimize>
...
</PropertyGroup>

So on TFS Build the following conditions should be true (and false in Visual Studio):

$(BUILD_SOURCESDIRECTORY) != ''
$(TF_BUILD) == 'true'
$(BuildingInsideVisualStudio)' != 'true'