Set content files to “copy local : always” in a nu

2019-01-02 20:48发布

I generate a nuget package from a project with this command in the post-build event. the variable %conf% is set to the right configuration (debug or release) and %1 is the project name (e.g. "MyCompany.MyProject").

nuget pack -Prop Configuration=%conf% "%1.csproj" -exclude *.sql -IncludeReferencedProjects

This package is for our own usage only, it will never be published on nuget. It ends in our private repository.

In the project, there is a file that is set to generate action : content and copy local : always. (My visual studio is in french, so I'm not 100% sure of the traduction). Let's name it importantfile.xml.

In the generated package, I end up with this structure :

- content
    - importantfile.xml
- lib
    -net45 (.NetFramework,Version=v4.5)
        -MyCompany.MyProject.dll

Which is fine, I want importantfile.xml to be deployed in the package, because, well, this file is important!

When I install the package in another project, importantfile.xml is deployed at the root of the project. That's OK. But it is not set to copy local : always.

I need importantfile.xml to be copy local : always in this project where I install my package.

How can I achieve that?

Notes :

I can set copy local : always on the file just after installing the package, that's no big deal. I would live with it if later updates of the package would let this property as-is, wich is not the case. When updating the package, copy local is resetted to never (as stated here).

There's a nuspec file in the project's folder, here it is :

<?xml version="1.0"?>
<package >
  <metadata>
    <id>$id$</id>
    <version>$version$</version>
    <title>$title$</title>
    <authors>$author$</authors>
    <owners>$author$</owners>
    <requireLicenseAcceptance>false</requireLicenseAcceptance>
    <description>$description$</description>
    <copyright>Copyright 2014</copyright>
    <tags>some random tags</tags>
  </metadata>
</package>

5条回答
若你有天会懂
2楼-- · 2019-01-02 20:51

I have made this which copies files from my build folder to the output folder (bin/debug or bin/release). Works like a charm for me.

Nuspec file:

<package>
  <files>
    <file src="\bin\Release\*.dll" target="lib" />
    <file src="\bin\Release\x64\*.dll" target="build\x64" />
    <file src="\bin\Release\x86\*.dll" target="build\x86" />
    <file src="MyProject.targets" target="build\" />    
  </files>
</package>

MyProject.targets

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
  <ItemGroup>
    <NativeLibs Include="$(MSBuildThisFileDirectory)**\*.dll" />
    <None Include="@(NativeLibs)">
      <Link>%(RecursiveDir)%(FileName)%(Extension)</Link>
      <CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
    </None>
  </ItemGroup>
</Project>
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余生请多指教
3楼-- · 2019-01-02 20:52

You can use PowerShell and the Install.ps1 hook provided by NuGet.

See the documentation.

Via PowerShell you have to 'search' for the content element which includes your importantfile.xml in an attribute. When the script found it, it has to add <CopyToOutputDirectory>Always</CopyToOutputDirectory> as a child element.

    <Content Include="importantfile.xml">
      <CopyToOutputDirectory>Always</CopyToOutputDirectory>
    </Content>

You can find some PowerShell snippets here. Just take a look at the .ps1 files.

You could try the following (not tested). The file has to be named Install.ps1 and copied into the tools folder:

param($installPath, $toolsPath, $package, $project)

# Load project XML.
$doc = New-Object System.Xml.XmlDocument
$doc.Load($project.FullName)
$namespace = 'http://schemas.microsoft.com/developer/msbuild/2003'

# Find the node containing the file. The tag "Content" may be replace by "None" depending of the case, check your .csproj file.
$xmlNode = Select-Xml "//msb:Project/msb:ItemGroup/msb:Content[@Include='importantfile.xml']" $doc -Namespace @{msb = $namespace}


#check if the node exists.
if($xmlNode -ne $null)
{
    $nodeName = "CopyToOutputDirectory"

    #Check if the property already exists, just in case.
    $property = $xmlNode.Node.SelectSingleNode($nodeName)
    if($property -eq $null)
    {
        $property = $doc.CreateElement($nodeName, $namespace)
        $property.AppendChild($doc.CreateTextNode("Always"))
        $xmlNode.Node.AppendChild($property)

        # Save changes.
        $doc.Save($project.FullName)
    }
}

You should also check if everything is removed completely when uninstalling the package.

Note by Jonhhy5

When updating the package via update-package, Visual Studio warns that the project is modified "outside the environnment". That's caused by $doc.Save($project.FullName). If I click reload before the command is fully terminated, it sometimes causes errors. The trick is to leave the dialog there until the process finishes, and then reload the projects.

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骚的不知所云
4楼-- · 2019-01-02 21:00

I have written a little tool called NuGetLib to automatically add files to the nuget package after build.

  1. create a tools folder with your Install.ps1 script
  2. build your nugetPackage
  3. add the tools folder to the built nugetPackage

https://stackoverflow.com/a/47134733/6229375

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笑指拈花
5楼-- · 2019-01-02 21:04

Instead of using a PowerShell script another approach is to use an MSBuild targets or props file with the same name as the package id:

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
  <ItemGroup>
    <None Include="$(MSBuildThisFileDirectory)importantfile.xml">
      <Link>importantfile.xml</Link>
      <CopyToOutputDirectory>Always</CopyToOutputDirectory>
    </None>
  </ItemGroup>
</Project>

In the nuspec file then, instead of adding the required files to the Content directory, add them to the Build directory along with the targets file.

  • Build
    • importantfile.xml
    • MyPackage.targets
  • lib
    • net45
      • MyAssembly.dll

If you require different content for different architectures then you can add architecture folders under Build also each with their own targets file.

Benefits to using a targets file over the PowerShell script with NuGet Content directory:

  • required content files aren't shown in the project in Visual Studio
  • content files are linked to rather than copied into the directory of each project which references the NuGet package (preventing there being multiple copies and keeping behaviour the same as for assemblies / libraries from NuGet packages)
  • PowerShell scripts only work in Visual Studio and aren't run when NuGet is run from the commandline (build servers, other IDEs and other OS), this approach will work everywhere
  • PowerShell install scripts are not supported in NuGet 3.x project.json system.
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梦该遗忘
6楼-- · 2019-01-02 21:11

I know you guys got a working solution to this but it didn't work for me so I'm going to share what I pulled out of the NLog.config NuGet package install.ps1 (github source here).

NOTE: this is not my code, this is the content of the install.ps1 from the NLog.config nuget package just sharing the knowledge.

It seems a little more straight forward to me and just hoping to help others that will likely stumble upon this.

You can find the accepted int values for BuildAction here and the accepted values for CopyToOutputDirectory here.

if the link breaks again enter image description here

Fields prjBuildActionCompile 1
The file is compiled.

prjBuildActionContent 2
The file is included in the Content project output group (see Deploying Applications, Services, and Components)

prjBuildActionEmbeddedResource 3
The file is included in the main generated assembly or in a satellite assembly as a resource.

prjBuildActionNone 0
No action is taken.

param($installPath, $toolsPath, $package, $project)

$configItem = $project.ProjectItems.Item("NLog.config")

# set 'Copy To Output Directory' to 'Copy if newer'
$copyToOutput = $configItem.Properties.Item("CopyToOutputDirectory")

# Copy Always Always copyToOutput.Value = 1
# Copy if Newer copyToOutput.Value = 2  
$copyToOutput.Value = 2

# set 'Build Action' to 'Content'
$buildAction = $configItem.Properties.Item("BuildAction")
$buildAction.Value = 2
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