How can I set custom MsBuild property from VSIX without writing into .csproj file?

Question

I create a VSIX extension for Visual Studio, which adds a custom drop-down menu. I need the value from this menu to apply to the .csproj property of the file without changing the file itself, like the configuration menu. For example:

<PropertyGroup>
  <Configuration Condition=" '$(Configuration)' == '' ">Debug</Configuration>
</PropertyGroup>
<PropertyGroup Condition=" '$(Configuration)' == 'Debug' ">
  <DefineConstants>DEBUG;TRACE;</DefineConstants>
</PropertyGroup>

-- default content of .csproj. When I selecting "Debug" from Visual Studio configuration drop-down menu, Configuration property value switch to Debug without changing .csproj file and '#if DEBUG'-block in code file is activated.

I have multiply projects in solution and each project have this property. If I overwrite .csproj, then these changes will be monitored by the source control, which I do not want. Help please.

Answer 1

How can I set custom MsBuild property without writing into .csproj file?

If the custom MSBuild property is a PropertyGroup in the project file, you can invoke MSBuild to pass an parameter into the project file by MSBuild command line:

<PropertyGroup>
  <CustomTestName>TestValue1</CustomTestName>
</PropertyGroup>

The command line:

msbuild /p:CustaomTestName=TestValue2

Alternatively, you can use the option "condition" in the custom property, like [Debug/Release] in project file:

<PropertyGroup>
  <CustomTestName>TestValue</CustomTestName>
</PropertyGroup>

<Target Name="CheckPropertiesHaveBeenSet">
  <Error Condition="'$(CustomTestName )'=='CustomTestNameNotSet'" Text="Something has gone wrong.. Custom Test Name not entered"/>
  <PropertyGroup>
    <CustomTestName Condition="'$(CustomTestName)'=='ChangeTestValue'">ChangeTestValue</CustomTestName >
  </PropertyGroup>
</Target>