MSBuild OutputPath property and absolute paths

Question

I'm trying to set the OutputPath value to an absolute path:

<OutputPath>c:\Projects\xxx\Deployment</OutputPath>

But I get this error:

Error   17  The expression "[System.IO.Path]::GetFullPath(D:\Projects\xxx\trunk\xxx.Web.Deployment\c:\Projects\xxx\Deployment\)" cannot be evaluated. The given path's format is not supported.     1   1   xxx.Web.Deployment

Is there a way to use an absolute path with the OutputPath property? I've tried experimenting with the BaseOutputPath property:

<PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Deployment|AnyCPU'">
  <BaseOutputPath>C:\Projects\xxx\</BaseOutputPath>
  <OutputPath>.\Deployment</OutputPath>
  <EnableUpdateable>true</EnableUpdateable>
  <UseMerge>true</UseMerge>
  <SingleAssemblyName>xxx.Web.Deployment</SingleAssemblyName>

But it seems to get ignored. What are BaseOutputPath and BaseIntermediateOutputPath used for?

niels · Accepted Answer · 2011-09-23 21:05:36Z

5

I'm not sure whether you can do what you're talking about, but you can add something similar to the following:

<PropertyGroup>  
    <CentralisedBinariesFolderLocation>c:\wherever</CentralisedBinariesFolderLocation>
</PropertyGroup>  

<Target Name="AfterBuild">
    <Exec Command="xcopy /Y /S /F /R &quot;$(TargetPath)&quot; &quot;$(CentralisedBinariesFolderLocation)&quot;" />
</Target>

Which will copy it to the relevant location after the build.

edited Sep 23, 2011 at 21:05

niels

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answered Jun 7, 2010 at 12:37

Paul Michaels

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2 Comments

Frederik Vig Over a year ago

Think this is the way to go as well. I have one small problem though, $(TargetPath) returns nothing. Is there another variable I can use to get the deployment path?

Frederik Vig Over a year ago

Got it working with: <Exec Command="xcopy /Y /S /F /R $(MSBuildProjectDirectory)\Deployment $(CentralisedBinariesFolderLocation)" />. Not very elegant, but worked in my case. Please let me know if there is a better way! :)

Julien Hoarau · Accepted Answer · 2010-06-07 12:51:41Z

3

Try using OutDir instead of OutputPath :

<PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Deployment|AnyCPU'">
  <OutDir>C:\Projects\xxx\$(Configuration)</OutDir>
  <EnableUpdateable>true</EnableUpdateable>
  <UseMerge>true</UseMerge>
  <SingleAssemblyName>xxx.Web.Deployment</SingleAssemblyName>
</PropertyGroup>

answered Jun 7, 2010 at 12:51

Julien Hoarau

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1 Comment

Richard Dingwall Over a year ago

I tried this and it output to C:\Projects\xxx\$(Configuration)\_PublishedWebsites\MyApp.Website.

Henrik · Accepted Answer · 2010-10-22 15:58:38Z

Copy the .target and .dll files from the installer directory
Modify the lines at the top that look like <UsingTask TaskName="GetProjectProperties" AssemblyFile="../../ ..lallal/VisualStudio/v10.0/Microsoft.Web.Publishing.Tasks.dll"/> and c*opy those .target and .dll files to your vendors folder next to the copied Microsoft.WebDeployment.targets-file your are editing*. Set the attr, AssemblyFile="Microsoft.Web.Publishing.Tasks.dll"
Add the line <EnablePackageProcessLoggingAndAssert Condition="'$(EnablePackageProcessLoggingAndAssert)' == ''">True</EnablePackageProcessLoggingAndAssert> to the initial PropertyGroup.
Set the OutputPath as you wish in the actual file/other tagets/other build-proj file.
Edit line ~290 to <WebPublishPipelineProjectDirectory Condition="'$(WebPublishPipelineProjectDirectory)'==''">$(OutputPath)</WebPublishPipelineProjectDirectory>

tiago · Accepted Answer · 2012-11-21 04:27:00Z

0

Instead of all the steps in the October answer, is it not possible just to define WebPublishPipelineProjectDirectory with the same path as OutputPath?

I tried it in my CI solution (using CruiseControl) and it seemed to work.

Does anyone know of any side effects that are not apparent to me from doing this?

edited Nov 21, 2012 at 4:27

tiago

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answered May 27, 2011 at 19:26

Tom Smellie

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