MSBuild: conditional construct (project link: file link)

Question

I'm still trying to eliminate the need for the cobol compiler in a project with cobol-Projects in it.

Is it possible to create the following assembly behavior:

If Configuration is Debug, use ProjectReferences for ExCobol.cblproj if Configuration is DebugVB, then use FileReferences in ExCobol.dll

When yes, how to do it?

I assume that using tags in the project file will do the trick.

And does this really eliminate the need for a cobol compiler for DebugVB configuration?

Answer 1

Regarding the conditional "how," assuming that you have

<ProjectReference ...>...</ProjectReference>

or

<Reference ...>...</Reference>

.proj,

<ProjectReference Condition="'$(Configuration)'!='DebugVB'" ...>...</ProjectReference>
<Reference Condition="'$(Configuration)'=='DebugVB'" ...>...</Reference>

Answer 2

: .

:

cobol , .

.