CODEWITHSUNDEEP
msbuild
wince
wince

Reputation: 531

Configure MSBuild output path

There is a Windows Forms (NET 3.5) project, foo.csproj, with localized resources files. I use MSBuild to build the project and create a deployment structure:

<MSBuild Projects="foo.csproj" Properties="Configuration=Release;OutputPath=..\deploy\foo" Targets="Build" />

It copies foo.exe and all localized DLL files to the deploy\foo folder, but I need localized DLL files to be copied into a separate folder. It should be:

Is there a way to configure MSBuild to copy EXE and DLL files to different folders?

Upvotes: 33

Views: 75924

Answers (2)

Jerome Anthony
Jerome Anthony

Reputation: 8021

Using MSBuild command line, you can specify the output path like below:

C:\Program Files (x86)\MSBuild\12.0\Bin\MSBuild.exe <path_to_project_file> /t:Build /p:OutDir=c:\custom_build_out\;Configuration=PRODUCTION;Platform=x64

Note:

  1. If you change the order of specifying the OutDir property for /p, this doesn't work.
  2. The OutDir property is for specifying a full path to an alternate directory. OutputPath is for a relative directory.
  3. It has to have a project name + build configuration name in the custom build output path as MSBuild does not append these things to the OutDir.

Upvotes: 55

JoeBilly
JoeBilly

Reputation: 3107

Resource files generation and copy is done in an internal MSBuild process during the build: GenerateSatelliteAssemblies and CopyFilesToOutputDirectory. They are copied in the output directory.

As far as I know, you can't modify this behavior.

You have to move your resources files after the build step. I would advise to use the Move task from MSBuild community tasks.

<MSBuild Projects="foo.csproj" Properties="Configuration=Release;OutputPath=..\deploy\foo" Targets="Build" />

<CreateItem Include="..\deploy\foo\**\*.resources.dll">
    <Output TaskParameter="Include" ItemName="ResourcesToMove" />
</CreateItem>

<Move SourceFiles="@(ResourcesToMove)" DestinationFiles="@(ResourcesToMove->'..\deploy\locales\%(RecursiveDir)\%(Filename)%(Extension)')"/>

Upvotes: 10

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