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

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>

To do it with the GUI, first add the file(s) to the project: right-click the project, select "Add...", then "Existing Item", then browse to the file or files you want to add and click "Add". Next, tell Visual Studio to copy the file when you build: right-click the file you want to copy, select "Properties". You'll see a list of properties, including "Item Type". Change the "Item Type" to "Copy File". Hit OK and you're done.

Here's the file properties dialog:

Looking in the *.vcxproj file, the steps above add something like this:

<ItemGroup>
    <CopyFileToFolders Include="libs\a.dll" />
    <CopyFileToFolders Include="libs\a.dll" />
</ItemGroup>

I couldn't find any official documentation for <CopyFileToFolders>, but clearly it's supported or the GUI wouldn't use it. But, if you're doing it by hand and an undocumented item type makes you uncomfortable you can always use the well known but slightly more verbose <Content> type:

<ItemGroup>
    <Content Include="libs\a.dll" >
        <CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
    </Content>
    <Content Include="libs\b.dll" >
        <CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
    </Content>
</ItemGroup>

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

You can also refer to a relative path, the next example will find the DLL in a folder located one level above the project folder. If you have multiple projects that use the DLL in a single solution, this places the source of the DLL in a common area reachable when you set any of them as the Startup Project.

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

The /y option copies without confirmation. The /d option checks to see if a file exists in the target and if it does only copies if the source has a newer timestamp than the target.

I found that in at least newer versions of Visual Studio, such as VS2109, $(ProjDir) is undefined and had to use $(ProjectDir) instead.

Leaving out a target folder in xcopy should default to the output directory. That is important to understand reason $(OutDir) alone is not helpful.

$(OutDir), at least in recent versions of Visual Studio, is defined as a relative path to the output folder, such as bin/x86/Debug. Using it alone as the target will create a new set of folders starting from the project output folder. Ex: … bin/x86/Debug/bin/x86/Debug.

Combining it with the project folder should get you to the proper place. Ex: $(ProjectDir)$(OutDir).

However $(TargetDir) will provide the output directory in one step.

Microsoft's list of MSBuild macros for current and previous versions of Visual Studio

I had a similar question. In my project, there were couple of external DLLs. So I created a new folder in the project called "lib" and copied all the external dlls to this folder.

1. Add a reference to these DLLs.
2. Go to Project References>dll properties and change the following properties enter image description here