How to get the last part of $(MSBuildProjectDirectory)

Question

I can't figure out how to get the last part of $(MSBuildProjectDirectory).

For example, if the value was "c:\development\projects\project_branch" then I want just the last part "project_branch".

How can I do this?

Want to deploy binaries for branches to a location that organizes the folders by branch name. So the closest thing without referring to subversion seems to be the folder name where our msbuild file is. There is a deploy task we use for automated nightly build, so the objective is to add a deploy task for branch. — Sean, Apr 21 '10 at 13:03

score 4 · Answer 1 · edited Jun 06 '22 at 18:53

4

In 4.0+ you can use Property Functions to do this in one line.

In this case for example $([System.IO.Path]::GetDirectoryName($(MSBuildProjectDirectory)))

or you could use a String function.

edited Jun 06 '22 at 18:53

Vimes

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answered Jul 17 '11 at 21:47

cheerless bog

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score 2 · Answer 2 · answered Apr 23 '10 at 00:47

<Project DefaultTargets="BuildAll" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">

    <Target Name="GetMSBuildProjectLocalDirectory">
        <CreateItem Include="$(MSBuildProjectDirectory)">
            <Output ItemName="MSBuildProjectDirectoryMeta" TaskParameter="Include"/>
        </CreateItem>
        <CreateProperty Value="%(MSBuildProjectDirectoryMeta.Filename)">
            <Output PropertyName="LocalDirectory" TaskParameter="Value"/>
        </CreateProperty>
    </Target>

    <Target Name="BuildAll" DependsOnTargets="GetMSBuildProjectLocalDirectory">
        <Message Text="$(LocalDirectory)" />
    </Target>

</Project>

score 0 · Answer 3 · answered Sep 12 '14 at 17:31

0

If you're following best practice, then your project directory will have the same name as your project file. Therefore, you should be able to use:

$(MSBuildProjectName)

answered Sep 12 '14 at 17:31

Mike Gilmore

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How to get the last part of $(MSBuildProjectDirectory)

3 Answers3