VS Post Build Event, copy one level above solution folder?

Question

I currently have

  <PropertyGroup>
    <PostBuildEvent>copy "$(TargetPath)" "$(SolutionDir)Shared.Lib\$(TargetFileName)"</PostBuildEvent>
  </PropertyGroup>

I want to do something like this, but one level above $(SolutionDir)

Hamiltonian

Answer 1

You can use ..\ to move up a directory.

 <PropertyGroup>
    <PostBuildEvent>copy "$(TargetPath)" "$(SolutionDir)..\Shared.Lib\$(TargetFileName)"</PostBuildEvent>
  </PropertyGroup>

Duhamel

what if I just want to copy one particular binary, and not every output file? – Hamiltonian

Are you saying that line currently copies more than one file? I would expect $(TargetPath) to point to the exe/library you are building for the project. – Duhamel

This answer did not work for me. This one does. The problem lies in the `..` in quote; it should be outside. – Catchings

Answer 2

Solution:

copy "$(TargetPath)" "$(SolutionDir)"..\"Shared.Lib\$(TargetFileName)"

If you have ..\ within the quotation marks, it will take it as literal instead of running the DOS command up one level.

Tertullian

thnx a lot. It worked with me as well. but with copy not xcopy – Rodrigorodrigue

Answer 3

This is not working in VS2010 .. is not resolved but becomes part of the path

Studio is running command something like this copy drive$:\a\b\bin\debug drive$:\a\b..\c

Vitalis

Make sure the path you choose exists, it didn't work for me until the path actually existed. – Hamiltonian

Answer 4

In .Net Core edit csproj file:

<Target Name="PostBuild" AfterTargets="PostBuildEvent">
  <Exec Command="copy /Y &quot;$(TargetPath)&quot; &quot;$(SolutionDir)&quot;..\&quot;lib\$(TargetFileName)&quot;" />
</Target>

/Y Suppresses prompting to confirm you want to overwrite an existing destination file.

Hydranth

Answer 5

xcopy "$(TargerDir)." "$(SolutionDir)..\Installer\bin\"

Note: "../" is used for One level up folder structure

Gordon