Configure MSBuild output path

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.

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)')"/>