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When I build a Visual Studio project, the executable is written to the output directory specified in the projects Property Page.

I have a project that has some extra files (e.g., .ini file) that are used by the program.

How can I configure the project to copy the file to the output directory so that when the program runs, it has a copy of the other file in its CWD?

I checked the Property Page of the file and there was nothing useful other than an option to exclude it from the build (which is disabled), and the custom-build-tool command is empty (plus it is a plain-text file that does not need any processing).

Synetech
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9 Answers9

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For copying a files to the output directory in Visual Studio 2003 you could use Post-Build event:

  1. Right click on the project->Properties
  2. Common Properties->Build Events

  3. Set Post-Build Event Command Line to: 

    xcopy /y $(ProjectDir)my_file.ini  $(ProjectDir)$(OutDir)

  4. OK and build!

shA.t
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Dmitry Pavlov
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    That works as well (though you don't need xcopy, copy works fine), but the catch is that it requires hard-coding the filename into the project’s build properties. Doing it the way I showed lets process each file separately and use macros for the filename. – Synetech Jun 02 '12 at 17:06
  • It depends on which version of Visual Studio you are using. This person specifically said VS 2003, which I have a project in and this helped me. The settings are terrible in 2003 - you can't even specify anything on each file's properties such as "Copy to output directory". – David Storfer Aug 05 '13 at 19:37
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    if you want the file to be copied even when the project is up-to-date and doesn't require a build, add the file to the project. – Jonathan Livni Dec 06 '13 at 22:02
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    Shouldn't it just be $(OutDir)? $(ProjectDir) is already contained within that (well, unless you modified where output going to). You want to just use the output directory. – Mary Ellen Bench Nov 23 '15 at 22:10
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    That should be copy /y "$(ProjectDir)my_file.ini" "$(OutDir)". Because of possibly directory path containing spaces. – KIM Taegyoon Feb 18 '16 at 05:46
  • FWIW, in VS 2013 I found that $(OutDir) did not work. It was necessary to use the combination $(ProjectDir)$(OutDir). This might be because $(OutDir) was given as a relative path, and the post-build step didn't process this correctly. – Tony Pulokas May 16 '23 at 23:15
38

Please try select the file in Solution Explorer. Then you should be able to see its properties in Properties window (press F4 if it is not visible). You will find there two properties:

  • "Build Action" and
  • "Copy to Output Directory"

Set "Build Action" to "Content", and then - select an appropriate value for "Copy to Output Directory" setting.

File properties window with "Build Action" and "Copy to Output Directory" settings

File properties window with "Build Action" and "Copy to Output Directory" settings

If the way above doesn't work for you, please read this post "Copy to output directory issue with .inf file". And have a look at this one then "Visual Studio: default build action for non-default file-types"

General Grievance
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Dmitry Pavlov
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    VS2003 does not seem to have that feature (for some reason). – Synetech May 31 '12 at 06:43
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    That works for some project types, not all. For example, it'll work for a C#-type project, but not for C++. – B.K. Jan 22 '16 at 22:55
  • @B.K. this is because of MS Build tasks, which might be different for different project types. You could play with adjusting it. Start digging from here https://social.msdn.microsoft.com/Forums/vstudio/en-US/af6501ed-a02c-487f-9cd1-b4bee2af3904/msbuild-custom-build-action-in-visual-studio-2010-on-c-project?forum=msbuild – Dmitry Pavlov Jan 28 '16 at 17:30
  • @DmitryPavlov That's fine; I'm just pointing out that it won't work everywhere. – B.K. Jan 28 '16 at 23:18
  • @B.K. sure, nothing works everywhere. Great that VS allows to at least extend and do almost everything developers might need. – Dmitry Pavlov Jan 29 '16 at 22:11
  • I would also like to add the you can only see this menu "Copy to Ouput Directory" on a file, but not its folder. Do this on eg. "logo.png", not the folder that contains it. Its parent folders will be created as well. – maryhadalittlelamb Mar 18 '21 at 11:09
  • You can also specify in .csproj file See https://stackoverflow.com/questions/44374074/copy-files-to-output-directory-using-csproj-dotnetcore – maryhadalittlelamb Mar 18 '21 at 11:14
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While I was searching the file’s Property Page for a build-action field, I had a thought: set the custom build step to copy the file (manually). This turned out to be easier than I thought. I had figured it would require using cmd or other external executable (xcopy, robocopy, etc.), but that is not necessary.

I set the Custom Build Step as follows:

Command Line : copy $(InputFileName) $(OutDir)
Description  : Copying foobar...
Outputs      : $(InputFileName)

Setting the outputs field (correctly) was critical in order to prevent VS from always thinking the project is out of date and requiring to be rebuilt (I’m not certain if it needs to be prefixed with $(OutDir)\).

It is reflected in the Output window as such:

Copying foobar...
        1 file(s) copied.
Compiling resources...
Linking...
Synetech
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  • fo me vs2015 it was `Command Line : copy "%(FullPath)" "$(OutDir)"` `Description : copy "%(FullPath)" "$(OutDir)"` `Outputs : %(FileName).%(Extension)` – Fantastory Sep 28 '16 at 12:50
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For VS 2017 the command Dmitry Pavlov posted would be the following:

xcopy /y "$(ProjectDir)my_file.ini"  "$(OutDir)"

Quotes are important in case there are spaces in the path to the project directory.

Kieran Chandler
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4

Expanding on Synetech's answer.

In VS2019 right click the file you want to copy in the Solution Explorer and select Properties. Then under General >> Item Type change to Copy File then hit Apply.
You now should have UI fields in the Properties Page for Destination etc.

Geordie
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3

In case this helps anyone, I needed to copy the output dll of the project i was building into another project.

xcopy /y "$(ProjectDir)$(OutDir)$(TargetName)$(TargetExt)" 
"C:\Application\MyApplicationName\bin\x86\Debug"

/y = overwrite file if already exists
$(ProjectDir) = location on your machine where the project lives
$(OutDir) = is where your current build setup outputs the build
$(TargetName) = What the project being built is set to be called. Ex: XXX of XXX.dll
$(TargetExt) = the extension of the build Ex: .dll of XXX.dll

"C:/..../x86/Debug" is the location to copy to.
vbp13
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0

You need the extra $(OutDir). Otherwise, in the rebuild/clean step it will throw away your source.

CommandLine : copy "$(SolutionDir)last-script.js" "$(TargetDir)Debug"

Outputs  : $(TargetDir)Debug\last-script.js
roland roos
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0

Improving Synetech answer :

In VS 2013 C++ project Command Line : copy %(Identity) $(OutDir) Description : Copying foobar... Outputs : %(Identity) It works , But it leads to circular dependency , i.e. it will be executed each time you demand increamental build, no meter it has been already copied.

To solve this , you can add that item at target folder, change path to $(OutDir), and use that in first added item as Output. Drawback - two items with similar name are in solution.

Also usefull xcopy with /d /y parameters in postbuild - copy only if target file date is older.

user1837919
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You could also after the unload the project (Right click on the project >> Unload Project) add the following inside an existent <ItemGroup> tag:

<Content Include="..\..\Config\db.config">
  <Link>Config\db.config</Link>
  <CopyToOutputDirectory>Always</CopyToOutputDirectory>
  <SubType>Designer</SubType>
</Content>

In this case it will grab the db.config file from 2 folders up and put it in the output folder(by default bin/Debug) after creating a Config folder with the db.config file inside

Oscar Acevedo
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