I'm attempting to write an automated build for one of our products, and I've hit up against a wall for some of our VC++ projects: I need to be able to set the output path to where the assemblies will be copied once its done.
Here is a makeshift msbuild file:
<Project DefaultTargets="Build"
xmlns="http://schemas.microsoft.com/developer/msbuild/2003"
ToolsVersion="3.5">
<Target Name="Build">
<VCBuild Projects="C:\src\SomeProject\SomeProject.vcproj"
ToolPath="C:\Program Files\Microsoft Visual Studio 9.0\VC\vcpackages"
Configuration="Debug" />
</Target>
</Project>
Stijn's Answer:
I thought I'd use this space to clarify how I personally used Stijn's answer to solve this. He has some code in his MSBuild file that writes the vsprops file for him. I decided to take a simpler approach and just write the file manually.
I created this file, called build.vsprops (my output path is V:)
<?xml version="1.0"?>
<VisualStudioPropertySheet ProjectType="Visual C++"
Version="8.00"
Name="Overrides"
OutputDirectory="V:\">
<Tool Name="VCCLCompilerTool"
AdditionalUsingDirectories="V:\" />
</VisualStudioPropertySheet>
Then I edited my MSBuild file to add the Override parameter:
<Project DefaultTargets="Build"
xmlns="http://schemas.microsoft.com/developer/msbuild/2003"
ToolsVersion="3.5">
<Target Name="Build">
<VCBuild Projects="C:\src\SomeProject\SomeProject.vcproj"
ToolPath="C:\Program Files\Microsoft Visual Studio 9.0\VC\vcpackages"
Configuration="Debug"
Override="$(MSBuildProjectDirectory)\build.vsprops" />
</Target>
</Project>
have a look at the Override parameter for the VCBuild task. Basically you specify a property sheet which you can use to override whatever property you want (it has the same effect as adding a property sheet to the top of the list in a project within VS). You could even generate the override file using the WriteLinesToFile task.
Example:
<PropertyGroup>
<VCOverridesFile Condition=" '$(VCOverridesFile)'=='' ">overrides.vsprops</VCOverridesFile>
<VCOverridesOpen>%3C?xml version=%221.0%22?%3E%0D%0A%3CVisualStudioPropertySheet ProjectType=%22Visual C++%22 Version=%228.00%22 Name=%22My Overrides%22%3E</VCOverridesOpen>
<VCOverridesClose>%3C/VisualStudioPropertySheet%3E</VCOverridesClose>
<MyOutPath><Tool Name="VCLinkerTool" OutputFile ="c:\my.exe"/></MyOutPath>
</PropertyGroup>
<Target Name="WriteOverridesFile">
<WriteLinesToFile
File="$(VCOverridesFile)"
Lines="$(VCOverridesOpen);$(AdditionalVCOverrides);$(VCOverridesClose)"
Overwrite="true" />
</Target>
Then pass $(VCOverridesFile) to the Override property and make sure your VCBuild Task DependsOnTarget WriteOverridesFile.
Doing it the dirty way you can pass output directory path through command line arguments of msbuild.
msbuild yourProject /p:OutDir=yourPath
Although I suspect, there should be the better way to accomplish the task. The main idea is to set 'OutDir' property in such a way that it will not be overriden by your SomeProject.vcproj
if you are using Azure DevOps and needs to create a YAML do build a .net framework (vintage[old])
- task: VSBuild#1
inputs:
solution: '**\*.sln'
msbuildArgs: '/p:DeployOnBuild=true /p:SkipInvalidConfigurations=false /p:OutDir="$(System.DefaultWorkingDirectory)\publish_output"'
platform: 'Any CPU'
configuration: 'Release'
Related
We can define a target in a csproj file and then specify that target when we call msbuild from the command line. That looks like this:
my.csproj
<Target Name="CopyFiles">
<Copy
SourceFiles="#(MySourceFiles)"
DestinationFolder="c:\MyProject\Destination" />
</Target>
msbuild
msbuild my.csproj /t:CopyFiles
The CopyFiles targets asks msbuild to run the Copy task.
What if we don't want to edit the csproj file. How can we define a target just from the command line? Alternatively, using only the command line, how can we ask msbuild to run just one or maybe two tasks?
Pseudo-Code
msbuild my.csproj /t:"Copy SourceFiles=#(MySourceFiles) DestinationFolder=..."
Based on the MSBuild Command-Line Reference, this isn't possible exactly as you describe it, i.e. MSBuild will not take Target definitions from the command-line input. They have to be defined in a file somewhere.
But, you can have a script (maybe even a .bat?) that does something like:
Create a new, essentially empty, project file.
Import the .csproj, with <Import Project="foo.csproj" />
Add the targets
Call MSBuild with the new project file with the targets you want. Multiple targets can be specified by multiple /t: switches. Properties can be specified with /p:
The final, programmatically/script generated project file might look something like:
<?xml version="1.0" encoding="utf-8"?>
<Project ToolsVersion="12.0" DefaultTargets="Build" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<Import Project="foo.csproj" />
<Target Name="CopyFiles">
<Copy
SourceFiles="#(MySourceFiles)"
DestinationFolder="$(Destination)" />
</Target>
</Project>
The actual MSBuild command called might then be:
msbuild temp.proj /t:CopyFiles /p:"Destination=c:\MyProject\Destination"
I am a beginner with MSBuild. So far I have been able to create a custom task called 'MakeTextFile' which creates a text file in C:\ based on the contents property you pass it. This works running from a command line prompt.
I have also included this in my .targets file (under the project tag):
<ItemGroup>
<AvailableItemName Include="CreateTextFileAction" />
</ItemGroup>
When I use the Import tag on my client applications .csproj I can now set items build actions to 'CreateTextFileAction', however the action never triggers (as no text file on C:\ is created)
How do I get all the file paths of items that were marked with my build action 'CreateTextFileAction' and pass them onto my custom task?
For reference, my .targets file:
<?xml version="1.0" encoding="utf-8" ?>
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<ItemGroup>
<AvailableItemName Include="CreateTextFileAction" />
</ItemGroup>
<UsingTask AssemblyFile="CustomMSBuildTask.dll" TaskName="CustomMSBuildTask.MakeTextFile" />
<Target Name="MyTarget">
<MakeTextFile Contents="TODO HOW DO I GRAB MARKED FILES?" />
</Target>
</Project>
A csproj file has a defined set of targets. The three main entry points are Build, Rebuild and Clean. These targets each have a set of dependencies. If you write your own targets to be part of the standard csproj build you need to find a suitable injection point within these dependencies.
For ease of use there are two standard targets for you to override called BeforeBuild and AfterBuild. If you define this in the csproj file (after the import of the csharp targets file) and call your custom task in there then it should work (or at least move further along).
<Target Name="BeforeBuild">
<MakeTextFile Contents="TODO HOW DO I GRAB MARKED FILES?" />
</Target>
I am executing MSBuild from a batch file. The MSBuild script is in a different directory than the directory I want MSBuild to consider the working directory when running the script. When invoking MSBuild.exe, how do I change its working directory?
Edit: More details
Let's say I have an MSBuild script located on some other server. I want to run a command thusly:
msbuild.exe \\my_server\c$\My\Path\To\Scripts\TestScript.msbuild
I run that command with my command prompt at c:\temp. Let's say my TestScript.msbuild has a task to create a file. The file has no path just a filename. I would expect that the file gets created inside c:\temp. But it doesn't it gets created next to the msbuild file that is sitting on the server. This is the behavior I want to change.
Edit #2
Here is the script I'm using in my test:
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<ItemGroup>
<Files Include="HelloWorld.txt" />
</ItemGroup>
<Target Name="TouchFiles">
<Touch Files="#(Files)" AlwaysCreate="True" />
</Target>
</Project>
I am going into a command shell CDing into c:\temp and then executing the script. With or without the /p:OutDir switch that #Nick Nieslanik mentions, the HelloWorld.txt file appears in the folder where the *.msbuild file is and not c:\temp.
I ran across this while looking for a solution to my problem. Here's my solution (build script):
<?xml version="1.0" encoding="utf-8"?>
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<Target Name="Default">
<Exec Command="build.bat" WorkingDirectory="..\[your dir]\" />
</Target>
</Project>
I believe that's more what you were originally looking for?
My problem was that my batch file called another that it expected to be in the same directory, but since my ms build script was being run elsewhere, the batch file failed to find the second batch file.
#jkohlhepp - I see now. You are doing the opposite of what I described in my comment to some degree.
MSBuild common targets use the MSBuildProjectDirectory to determine the output folder unless you override that. So in your case, you could run
msbuild.exe \\my_server\c$\My\Pat\To\Scripts\TestScript.msbuild /p:OutDir=c:\temp
to force the output to be dropped in that location.
EDIT:
Given the project file above, you'd need to edit it to do something like the following for this to work:
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<PropertyGroup>
<OutDir Condition=" '$(OutDir)' == '' ">bin\debug\</OutDir>
</PropertyGroup>
<ItemGroup>
<!-- Without prefacing files with paths, they are assumed relative to the proj file -->
<FilesToCreate Include="$(OutDir)HelloWorld.txt" />
</ItemGroup>
<Target Name="TouchFiles">
<Touch Files="#(FilesToCreate)" AlwaysCreate="True" />
</Target>
</Project>
In current versions of MSBuild the well-known property MSBuildStartupDirectory can be used in the msbuild file to retrieve the absolute path of the folder where MSBuild is called.
https://learn.microsoft.com/en-us/visualstudio/msbuild/msbuild-reserved-and-well-known-properties?view=vs-2019
This option perhaps did not exist in msbuild around the time when the question was asked. I didn't want to spend too much time investigating it.
I have Windows application in csproj in my solution, and I want generate Publish using command line (bat, cmd).
My script is (I put \r\n for better reading):
SET MSBUILD="%SystemRoot%\Microsoft.NET\Framework\v3.5\MSBuild.exe"
SET CARWIN="..\..\Security.CarWin.csproj"
rem msbuild para publish
%MSBUILD% /target:rebuild;publish %CARWIN%
/p:ApplicationVersion="1.0.0.0"
/p:Configuration=release
/p:PublishUrl="C:\ClickOnce\CarWin.WebInstall\Publicacion\"
/p:InstallUrl="http://desserver/carwinclickonce/Publicacion/"
/p:PublishDir="C:\ClickOnce\CarWin.WebInstall\Publicacion\"
note: I'll try too using /target:publish
But in path PublishDir or PublishUrl (C:\ClickOnce\CarWin.WebInstall\Publicacion) not generates any files.
I have seen many posts in this site and google but I not found any solution.
Use PublishDir instead of PublishUrl when running from command line.
msbuild /target:publish /p:Configuration=Release;PublishDir=c:\playground\
You can also change version, like ApplicationRevision=666;MinimumRequiredVersion=1.1
Take a look at this Stack Overflow question. Basically the PublishUrl property is ignored when running ClickOnce from the command line. But you can easily add the behaviour with an additional MSBuild-task.
I've created an additional MSBuild-File, for example a build.csproj. This contains a publish-task. This task first invokes the regular MS-Build of the target-project. Afterwards it copies the result to the publish-directory. Now I invoke the 'build.csproj' instead of the reguar project-file from the command-line:
<?xml version="1.0" encoding="utf-8"?>
<Project ToolsVersion="3.5" DefaultTargets="Publish" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<PropertyGroup>
<!-- project name-->
<ProjectName>MyExampleProject</ProjectName>
<!--properties for the project-build-->
<DefaultBuildProperties>Configuration=Release</DefaultBuildProperties>
<!-- location of the click-once stuff, relative to the project -->
<ProjectPublishLocation>.\bin\Release\app.publish</ProjectPublishLocation>
<!-- Location you want to copy the click-once-deployment. Here an windows-share-->
<ProjectClickOnceFolder>\\TargetServer\deployments</ProjectClickOnceFolder>
</PropertyGroup>
<Target Name="Publish" DependsOnTargets="Clean">
<Message Text="Publish-Build started for build no $(ApplicationRevision)" />
<!-- run the original build of the project -->
<MSBuild Projects="./$(ProjectName).csproj"
Properties="$(DefaultBuildProperties)"
Targets="Publish"/>
<!-- define the files required for click-once-->
<ItemGroup>
<SetupFiles Include="$(ProjectPublishLocation)\*.*"/>
<UpdateFiles Include="$(ProjectPublishLocation)\Application Files\**\*.*"/>
</ItemGroup>
<!-- and copy them -->
<Copy
SourceFiles="#(SetupFiles)"
DestinationFolder="$(ProjectClickOnceFolder)\"/>
<Copy
SourceFiles="#(UpdateFiles)"
DestinationFolder="$(ProjectClickOnceFolder)\Application Files\%(RecursiveDir)"/>
</Target>
<Target Name="Clean">
<Message Text="Clean project" />
<MSBuild Projects="./$(ProjectName).csproj"
Properties="$(DefaultBuildProperties)"
Targets="Clean"/>
</Target>
</Project>
I don't know if this is a problem, but I noticed that you pass the /target parameter twice?
you could you use a semi-colon delimited example:
/target:rebuild;publish
MSDN Documentation on command line parameters and MSBuild
If that also does not work you could perhaps try to debug it by passing
/verbosity:diag
I'm trying to make a batch file to publish the few ClickOnce application we have in one click. I'm using msbuild for that, and as an example the below command line shows how I'm doing it:
msbuild
MyApp.sln
/t:Publish
/p:Configuration=Release
/p:PublishUrl="C:\Apps\"
/v:normal > Log.txt
(wrapped for easier reading)
when I run the above command it builds and publish the application in the release directory, i.e. bin\release! Any idea why msbuild doesn't respect PublishUrl property in my example above?
PS: I tried also different combinations including remove 'Configuration', use 'Rebuild' and 'PublishOnly' as targets, and remove the the quotation marks but without any success.
You are setting the wrong property. Try PublishDir instead.
You can pass it into MSBuild as you are or you can set it in the project file (or maybe the sln file too, not sure I always use the project file.) like this
<PropertyGroup>
<PublishDir>C:\Dev\Release\$(BuildEnvironment)\</PublishDir>
</PropertyGroup>
I've just done a few blog posts on MsBuild and ClickOnce stuff, check it out you 'should' find them useful...
Some features are done by Visual-Studio and not by the MSBuild-script. So the click-once-deployment behaves differently when it's executed from the command-line.
The ApplicationRevision isn't increased with every build. This works only when is exectued from Visual Studio
In in somecases, the PublishUrl isn't used. Quote from MSDN:
For example, you could set the PublishURL to an FTP path and set the InstallURL to a Web URL. In this case, the PublishURL is only used in the IDE to transfer the files, but not used in the command-line builds. Finally, you can use UpdateUrl if you want to publish a ClickOnce application that updates itself from a separate location from which it is installed.
I've created a special MSBuild-file which does this things. It runs the publish-target and copies then the files to the right location.
An example of the build-file, as requested by alhambraeidos. It basically runs the regular VisualStudio-build and then copies the click-once data to the real release folder. Note that removed some project-specific stuff, so it's maybe broken. Furthermore it doesn't increase the build-number. Thats done by our Continues-Build-Server:
<?xml version="1.0" encoding="utf-8"?>
<Project ToolsVersion="4.0" DefaultTargets="Publish" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<PropertyGroup>
<!-- the folder of the project to build -->
<ProjLocation>..\YourProjectFolder</ProjLocation>
<ProjLocationReleaseDir>$(ProjLocation)\bin\Release</ProjLocationReleaseDir>
<ProjPublishLocation>$(ProjLocationReleaseDir)\app.publish</ProjPublishLocation>
<!-- This is the web-folder, which provides the artefacts for click-once. After this
build the project is actually deployed on the server -->
<DeploymentFolder>D:\server\releases\</DeploymentFolder>
</PropertyGroup>
<Target Name="Publish" DependsOnTargets="Clean">
<Message Text="Publish-Build started for build no $(ApplicationRevision)" />
<MSBuild Projects="$(ProjLocation)/YourProject.csproj" Properties="Configuration=Release" Targets="Publish"/>
<ItemGroup>
<SchoolPlannerSetupFiles Include="$(ProjPublishLocation)\*.*"/>
<SchoolPlannerUpdateFiles Include="$(ProjPublishLocation)\Application Files\**\*.*"/>
</ItemGroup>
<Copy
SourceFiles="#(SchoolPlannerSetupFiles)"
DestinationFolder="$(DeploymentFolder)\"
/>
<Copy
SourceFiles="#(SchoolPlannerUpdateFiles)"
DestinationFolder="$(DeploymentFolder)\Application Files\%(RecursiveDir)"
/>
<CallTarget Targets="RestoreLog"/>
</Target>
<Target Name="Clean">
<Message Text="Clean project:" />
<MSBuild Projects="$(ProjLocation)/YourProject.csproj" Properties="Configuration=Release" Targets="Clean"/>
</Target>
</Project>
I'll put in my 2 cents, this syntax seems to work (right or wrong):
/p:publishUrl="C:\\_\\Projects\\Samples\\artifacts\\Web\\"
For me, the soultion was to escape the path.
Instead of:
/p:PublishUrl="C:\Apps\"
Put:
/p:PublishUrl="C:\\Apps\\"