Question

I created a configuration named Test via Visual Studio which currently just takes all of DEBUG settings, however I employ compiler conditions to determine some specific actions if the build happens to be TEST|DEBUG|RELEASE.

However how can I get my MSBUILD script to detect the TEST configuration??

Currently I build

 <MSBuild Projects='@(SolutionsToBuild)' Properties='Configuration=$(Configuration);OutDir=$(BuildDir)\Builds\' /> 

Where @(SolutionsToBuild) is a my solution. In the Common MsBuild Project Properties it states that $(Configuration) is a common property but it always appears blank?

Does this mean that it never gets set but is simply reserved for my use or that it can ONLY detect DEBUG|RELEASE. If so what is the point in allowing the creation of different build configurations?

Answer 1

I haven’t done much with defining an MSBUILD configuration file but I have done builds of different configurations using a batch file like this

msbuild /v:n /p:Configuration=Release 'Capture.sln'  msbuild /v:n /p:Configuration=ReleaseNoUploads 'Capture.sln'  

I defined the ReleaseNoUploads configuration inside Visual Studio.

Here’s what I had to do for that (this is Visual Studio 2005):

• Open the Tools:Options menu, go to the Projects and Solutions:General option, and check Show advanced build configurations.
• From there, go to the Build:Configuration Manager menu
• In the dialog that pops up, click on the Active solution configuration pulldown and click <New…> to create a new build configuration.