Copy a single file in MSBuild without using Exec or ItemGroup

Question

Is there any way to do this? I just need to copy a single file and thought there may be some syntax for the SourceFiles parameter of the Copy task that means you don't need to define an ItemGroup beforehand, I'd rather stick with ItemGroup than use Exec though.

score 21 · Accepted Answer · answered Nov 14 '13 at 10:05

21

Copy files also takes a straight propertygroup as input:

<PropertyGroup>
  <SourceFile>Some file</SourceFile>
</PropertyGroup>
<Copy SourceFiles="$(SourceFile)" DestinationFolder="c:\"/>

Or even just a string

<Copy SourceFiles="Pathtofile" DestinationFolder="c:\"/>

answered Nov 14 '13 at 10:05

James Woolfenden

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This helped me in automating the creation of web.config file for a WCF Library Project hosted in IIS. :) – Achilles Apr 15 '15 at 10:41
For copying a single file if you want to specify the full path including a new filename, there's a DestinationFile attribute you would use instead of DestinationFolder. – AaronLS Feb 22 '22 at 09:28

granadaCoder · Answer 2 · 2013-11-14T15:34:25.693

Just put the single file name as the value for "SourceFiles". Easy-Peezey.

<?xml version="1.0" encoding="utf-8"?>
<Project DefaultTargets="AllTargetsWrapper" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">

    <PropertyGroup>
        <WorkingCheckout>.</WorkingCheckout>
    </PropertyGroup>

    <Target Name="AllTargetsWrapper">
        <CallTarget Targets="CopyItTarget" />
    </Target>

    <Target Name="CopyItTarget">
        <Copy SourceFiles="c:\windows\system.ini" DestinationFolder="$(WorkingCheckout)\"/>     
        <Error Condition="!Exists('$(WorkingCheckout)\system.ini')" Text="No Copy Is Bad And Sad" />
    </Target>

</Project>

This is actually covered by the second part of James' answer above, thanks though! — Shaun, Apr 17 '15 at 14:56

score 2 · Answer 3 · answered Feb 06 '17 at 21:05

For what it's worth, I needed to do the same thing, and wanted to put some version information in the file name. Here is how I did it for a project in $(SolutionDir) that references an executable created by another project in another solution that I can easily express the path to:

  <Target Name="AfterBuild">
    <GetAssemblyIdentity AssemblyFiles="$(SolutionDir)..\bin\$(Configuration)\SomeExectuable.exe">
      <Output TaskParameter="Assemblies" ItemName="AssemblyVersions" />
    </GetAssemblyIdentity>
    <CreateProperty Value="$(TargetDir)$(TargetName)-%(AssemblyVersions.Version)$(TargetExt)">
      <Output TaskParameter="Value" PropertyName="NewTargetPath" />
    </CreateProperty>
    <Copy SourceFiles="$(TargetPath)" DestinationFiles="$(NewTargetPath)" />
  </Target>

Copy a single file in MSBuild without using Exec or ItemGroup

3 Answers3