How to make Visual Studio copy a DLL file to the output directory?

Use a post-build action in your project, and add the commands to copy the offending DLL. The post-build action are written as a batch script.

The output directory can be referenced as $(OutDir). The project directory is available as $(ProjDir). Try to use relative pathes where applicable, so that you can copy or move your project folder without breaking the post-build action.

Add builtin COPY in project.csproj file:

  <Project>
    ...
    <Target Name="AfterBuild">
      <Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Debug\bin" SkipUnchangedFiles="false" />
      <Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Release\bin" SkipUnchangedFiles="false" />
    </Target>
  </Project>

The details in the comments section above did not work for me (VS 2013) when trying to copy the output dll from one C++ project to the release and debug folder of another C# project within the same solution.

I had to add the following post build-action (right click on the project that has a .dll output) then properties -> configuration properties -> build events -> post-build event -> command line

now I added these two lines to copy the output dll into the two folders:

xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Release
xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Debug

$(OutDir) turned out to be a relative path in VS2013, so I had to combine it with $(ProjectDir) to achieve the desired effect:

xcopy /y /d  "$(ProjectDir)External\*.dll" "$(ProjectDir)$(OutDir)"

BTW, you can easily debug the scripts by adding 'echo ' at the beginning and observe the expanded text in the build output window.