MSBuild Condition IsDebug

Question

How can I determine if a project is build in Debug (or Release) mode within an MSBuild .targets file and use this information as a condition for another property?

Something like:

<OutDir Condition="IsDebug">bin\Debug\$(SomeOtherProperty)\</OutDir>
<OutDir Condition="!IsDebug">bin\Release\$(SomeOtherProperty)\</OutDir>

Is there such thing as Debug/Release mode, or are they just conventional names for different sets of configuration properties' values?

Perspex

Answer 1

Debug/Release or whatever are just conventional values for the Configuration property.

So, as long the project that includes/calls your .targets file adheres to the convention; you can check for debug mode as follows:

<OutDir>bin\Release\$(SomeOtherProperty)\</OutDir>
<OutDir Condition=" '$(Configuration)' == 'Debug' ">bin\Debug\$(SomeOtherProperty)\</OutDir>

or you could just use that variable directly:

<OutDir>bin\$(Configuration)\$(SomeOtherProperty)\</OutDir>

Gassman

Well the problem is that I have 8 configurations, half debug and half release. Their names of course are not 'Debug' or 'Release' but conventionally i have named them 'Debug <something>' 'Debug <somethingElse>' etc. So instead of a == check I would probably do a Contains check right? – Perspex

Exactly. Depending on the complexity of your targets file, it may be useful to either introduce the IsDebug property from your example, or an ConfigurationType (or similar) property having the value Debug or Release. – Inexpugnable