使用MSBuild生成多个配置 [英] Using MSBuild to Build Multiple Configurations

问题描述

我正在尝试编辑我的项目文件，以使我拥有一个可以一次构建多个构建配置的项目.我已经使用批处理方法和MSBuild任务完成了此操作(请参见下文).

I'm trying to edit my project file to enable me to have a project that builds multiple build configs at once. I've done this using a batching approach and using the MSBuild task (see below).

如果运行脚本，则会出现此错误:

If I run the script, I get an this error:

错误103 OutputPath属性是 没有为项目设置 "ThisMSBuildProjectFile.csproj". 请检查以确保您 指定了以下内容的有效组合 为此的配置和平台 项目.配置='调试' Platform ='AnyCPU'.

Error 103 The OutputPath property is not set for project "ThisMSBuildProjectFile.csproj". Please check to make sure that you have specified a valid combination of Configuration and Platform for this project. Configuration='Debug' Platform='AnyCPU'.

如果我从MSBuild任务添加或省略OutputPath，则会得到此信息.如果使用VS2010调试器来逐步执行脚本并调用MSBuild任务-调试器将再次进入该文件，然后进入OutputPath，所以afaik，它应该拾取该值，不是吗?

I get this if I add or omit the OutputPath from the MSBuild task. If used the VS2010 debugger to step through the script and the MSBuild Task is called - the debugger steps into the file again and then steps into OutputPath, so afaik, it should pick that value up, no?

对此的任何帮助将不胜感激-这使我发疯.谢谢保罗.

Any help for this would be greatly appreciated - it's driving me crazy. Thanks, Paul.

ThisMSBuildProjectFile.csproj(取出多余的东西):

ThisMSBuildProjectFile.csproj (surplus stuff taken out):

<?xml version="1.0" encoding="utf-8"?>
<Project ToolsVersion="4.0" xmlns="http://schemas.microsoft.com/developer/msbuild/2003" DefaultTargets="Build">

  <!-- Only Import normal targets if not building multiple projects -->
  <Import Project="$(MSBuildToolsPath)\Microsoft.CSharp.targets" Condition="'$(Configuration)|$(Platform)' != 'AllBuild|AnyCPU' "/>

  <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == '' ">
    <DebugType>pdbonly</DebugType>
    <Optimize>true</Optimize>
    <OutputPath>C:\Folder\Etc\Output\$(Configuration)\</OutputPath>
    <OutDir>C:\Folder\Etc\Output\$(Configuration)\</OutDir>
    <BaseOutputPath>C:\Folder\Etc\Output\$(Configuration)\</BaseOutputPath>
    <DefineConstants>TRACE</DefineConstants>
    <ErrorReport>prompt</ErrorReport>
    <WarningLevel>4</WarningLevel>
  </PropertyGroup>

  <!-- Common -->
  <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
    <Platform>AnyCPU</Platform>
    <!-- Repeated properties from above here (including, of course, OutputPath) -->  
   </PropertyGroup>
  <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Release|AnyCPU' ">
    <!-- Repeated properties from above here (including, of course, OutputPath) --> 
  </PropertyGroup>

  <ItemGroup>
    <Projects Include="C:\Folder\Etc\ThisMSBuildProjectFile.csproj" />
  </ItemGroup>

   <!-- Call this project file again, but with a different configuration - if this was working, this would call multiple  build configs -->
  <Target Name="Build" Condition="'$(Configuration)|$(Platform)' == 'AllBuild|AnyCPU' ">
    <Message Text="hm!"/>
    <!-- Tried thiswith and without the OutputPath property - makes no difference. -->
   <MSBuild  Projects="@(Projects)" Properties="Configuration=Debug;OutputPath=C:\Folder\Etc\Output\" ToolsVersion="4.0" Condition="'$(Configuration)|$(Platform)' == 'AllBuild|AnyCPU' "/>
 </Target>

   <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'AllBuild|AnyCPU' ">
    <!-- Repeated properties from above here (including, of course, OutputPath) --> 
  </PropertyGroup>

  <!-- Project files -->
  <ItemGroup>
    <Reference Include="System" />
    <Reference Include="System.Core" />
  </ItemGroup>
  <ItemGroup>
    <Compile Include="Properties\AssemblyInfo.cs" />
    <Compile Include="Blah\Blah.cs" />
  </ItemGroup>

推荐答案

重要的是要意识到，当您使用"MSBuild"任务时，将启动一个新的子MSBuild进程.这意味着您在父MSBuild流程中定义的所有项目和属性将不会自动传递给子MSBuild流程/从子MSBuild流程可见除非您明确地通过以下方式传递它们: MSBuild元素上的Properties属性(如<MSbuild Properties="..." />中一样).

It is important to realize that when you use a "MSBuild" task, a new child MSBuild process will be started. The implication of this is that any items and properties you define in the parent MSBuild process will not be automatically passed to/visible from the child MSBuild process unless you explicitely pass them via Properties attribute on MSBuild element (as in <MSbuild Properties="..." />).

为回答您的问题，我编写了以下独立的示例，该示例针对所有指定的配置运行子MSBuild项目:

To answer your question, I wrote the following self-contained example that runs a child MSBuild project for all the specified configurations:

1. 首先，为您的MSBuild实验创建一个目录(例如，我使用C:\temp\msbuildtest)

在此目录中，创建第一个文件main.proj:

In this directory, create the first file, main.proj:

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" DefaultTargets="Build" ToolsVersion="4.0">
    <ItemGroup>
        <ConfigList Condition=" '@(ConfigList)' == '' and $(Config) != '' " Include="$(Config.Split('+'))" /><!-- parse all requested configurations into a list -->
        <ConfigList Condition=" '@(ConfigList)' == '' " Include="Debug" /><!-- if no configurations were specified, default to Debug -->
    </ItemGroup>
    <!--

    Build the child project for each requested configuration. -->
    <Target Name="Build">
        <MSBuild Projects="$(MSBuildProjectDirectory)\child.proj" Properties="Configuration=%(ConfigList.Identity);OutputPath=$(MSBuildProjectDirectory)\bin\%(ConfigList.Identity)" Targets="Build" />
    </Target>
</Project>

在同一目录中，创建第二个文件child.proj(在您的情况下，这将是您要构建的实际C#项目，但是由于我试图说明我的观点，所以我使用一个简单的子项目，而不是运行C#编译器，而只是打印属性值:-))

In the same directory, create the second file, child.proj (in your case this would be the actual C# project you're trying to build, but because I'm trying to illustrate my point, I am using a simple child project that instead of running C# compiler just prints values of properties :-) )

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" DefaultTargets="Build" ToolsVersion="4.0">
    <Target Name="Build">
        <Message Text="Building configuration $(Configuration) with output path $(OutputPath)" Importance="High" />
    </Target>
</Project>

现在您可以运行该示例.首先是默认值，如果您未明确指定要配置的配置:

Now you can run the example. First the default, if you don't explicitly specify configurations to build:

C:\WINDOWS\Microsoft.NET\Framework\v4.0.30319\msbuild main.proj
> (cut the noise)
> Build:
>   Building configuration Debug with output path C:\temp_c\d\bin\Debug

然后显式指定多个配置:

And then explicitly specified multiple configurations:

C:\WINDOWS\Microsoft.NET\Framework\v4.0.30319\msbuild main.proj /property:Config=Debug+Release+Staging+Production
> (cut the noise)
> Build:
>   Building configuration Debug with output path C:\temp_c\d\bin\Debug
> Build:
>   Building configuration Release with output path C:\temp_c\d\bin\Release
> Build:
>   Building configuration Staging with output path C:\temp_c\d\bin\Staging
> Build:
>   Building configuration Production with output path C:\temp_c\d\bin\Production

您应该能够使这种技术适应您的情况.

You should be able to adapt this technique to your situation.

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