MSBuild: Conditional Construct (Project Reference | File Reference)

问题

I´m still trying to eleminate the need of a cobol compiler in a Project with cobol-Projects in it.

Is it possible to create following build behaviour:

If the Configuration is Debug then use ProjectReferences on ExCobol.cblproj if the Configuration is DebugVB then use FileReferences on ExCobol.dll

When Yes, How to achieve it?

I assume the use of tags in the project file will do the trick.

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

回答1:

Regarding the conditional 'how', assuming you have either

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

or

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

you want to hand-edit the .proj file to include both thusly

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

回答2:

Almost forgot this question: the answer from Brian works perfectly for me.

And to answer the second part of the question:

It indeed eliminates the need for a cobol compiler on every maschine if you setup different Configurations.

Thanks Brian.

来源：https://stackoverflow.com/questions/1872553/msbuild-conditional-construct-project-reference-file-reference