Question

我正在尝试为具有属性的所有项目文件调用msbuild任务。我使用硬编码配置和平台组合将msbuild任务调用四次。像

这样的东西

<MSBuild Projects="$(MSBuildProjectFile)" Targets="Build" Properties="Configuration=Debug;Platform=Win32" BuildInParallel="true"/>
<MSBuild Projects="$(MSBuildProjectFile)" Targets="Build" Properties="Configuration=Debug;Platform=x64" BuildInParallel="true"/>
<MSBuild Projects="$(MSBuildProjectFile)" Targets="Build" Properties="Configuration=Release;Platform=Win32" BuildInParallel="true"/>
<MSBuild Projects="$(MSBuildProjectFile)" Targets="Build" Properties="Configuration=Release;Platform=x64" BuildInParallel="true"/>

但我想将此属性作为ItemGroup提供类似这样的

Configuration=%(BUILD_CONFIG.Identity);Platform=%(BUILD_PLATFORM.Identity)

代码示例 MyProject.vcxproj

<?xml version="1.0" encoding="utf-8"?>
<Project DefaultTargets="BuildAll" ToolsVersion="14.0" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<Import Project="BuildAllConfiguration.vcxproj"/>
<ItemGroup Label="ProjectConfigurations">
    <ProjectConfiguration Include="Debug|Win32">
      <Configuration>Debug</Configuration>
      <Platform>Win32</Platform>
    </ProjectConfiguration>
    <ProjectConfiguration Include="Release|Win32">
      <Configuration>Release</Configuration>
      <Platform>Win32</Platform>
    </ProjectConfiguration>
    <ProjectConfiguration Include="Debug|x64">
      <Configuration>Debug</Configuration>
      <Platform>x64</Platform>
    </ProjectConfiguration>
    <ProjectConfiguration Include="Release|x64">
      <Configuration>Release</Configuration>
      <Platform>x64</Platform>
    </ProjectConfiguration>
  </ItemGroup>
  <PropertyGroup Label="Globals">
    <ProjectGuid>{E6B6F967-3BE3-428F-9288-3F838B8E726A}</ProjectGuid>
    <Keyword>Win32Proj</Keyword>
    <RootNamespace>MyProject</RootNamespace>
    <WindowsTargetPlatformVersion>8.1</WindowsTargetPlatformVersion>
  </PropertyGroup>
  <Import Project="$(VCTargetsPath)\Microsoft.Cpp.Default.props" />
  <PropertyGroup Condition="'$(Configuration)|$(Platform)'=='Debug|Win32'" Label="Configuration">
    <ConfigurationType>DynamicLibrary</ConfigurationType>
    <UseDebugLibraries>true</UseDebugLibraries>
    <PlatformToolset>v140</PlatformToolset>
    <CharacterSet>Unicode</CharacterSet>
  </PropertyGroup>
<ItemDefinitionGroup Condition="'$(Configuration)|$(Platform)'=='Debug|Win32'">
    <ClCompile>
      <PrecompiledHeader>
      </PrecompiledHeader>
      <WarningLevel>Level3</WarningLevel>
      <Optimization>Disabled</Optimization>
      <PreprocessorDefinitions>_DEBUG;_WINDOWS;_USRDLL;%(PreprocessorDefinitions)</PreprocessorDefinitions>
      <SDLCheck>true</SDLCheck>
    </ClCompile>
    <Link>
      <SubSystem>Windows</SubSystem>
      <GenerateDebugInformation>true</GenerateDebugInformation>
    </Link>
  </ItemDefinitionGroup>
... 
Similar Configuration Details for release and Platforms x64

此项目文件包含BuildAllConfiguration.vcxproj

<?xml version="1.0" encoding="utf-8"?>
<Project ToolsVersion="14.0" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
  <PropertyGroup>
    <BUILD_PLATFORMS>Win32;x64</BUILD_PLATFORMS>
    <BUILD_CONFIGURATION>Debug;Release</BUILD_CONFIGURATION>
  </PropertyGroup>
  <Target Name="BuildAll">
    <ItemGroup>      
      <CONFIGURATION Include="$(BUILD_CONFIGURATION.Split(';'))"/>
      <PLATFORM Include="$(BUILD_PLATFORMS.Split(';'))"/>
      <ProjectToBuild Include="$(MSBuildProjectFile)">
        <Properties>Configuration=%(CONFIGURATION.Identity);Platform=%(PLATFORM.Identity)</Properties>
        <Targets>Build</Targets>
      </ProjectToBuild>
    </ItemGroup>
    <Message Text="MSBUILD TASK input @(ProjectToBuild)"/>
    <MSBuild Projects="@(ProjectToBuild)" />
 </Target>
</Project>

这个项目将使用目标构建和属性调用MyProject.vcxproj，但是没有很好地完成。我的期望是属性如下

Properties=Configuration=Debug;Platform=Win32 
Properties=Configuration=Release;Platform=Win32  
Properties=Configuration=Debug;Platform=x64 
Properties=Configuration=Release;Platform=x64

而是按以下方式传递属性

Properties=Configuration=Debug;Platform=
Properties=Configuration=Release;Platform=
Properties=Configuration=;Platform=Win32
Properties=Configuration=;Platform=x64

Answer 1

你需要一个跨产品，如果你搜索它，你应该找到很多answers，但我想如果你不知道它被称为可能很难找。像这样：

<Target Name="BuildAll">
  <ItemGroup>
    <CONFIGURATION Include="$(BUILD_CONFIGURATION.Split(';'))"/>
    <PLATFORM Include="$(BUILD_PLATFORMS.Split(';'))"/>

    <!-- cross product of both -->
    <ConfigAndPlatform Include="@(CONFIGURATION)">
      <Platform>%(PLATFORM.Identity)</Platform>
    </ConfigAndPlatform>

    <ProjectToBuild Include="$(MSBuildProjectFile)"/>
  </ItemGroup>
  <MSBuild Projects="@(ProjectToBuild)" Properties="Configuration=%(ConfigAndPlatform.Identity);Platform=%(ConfigAndPlatform.Platform)" />
</Target>

有些注意事项：首都会让事情变得更难读，也许不会使用它们？此外，如果您将配置/平台放在ItemGroup而不是PropertyGroup中，则不需要额外的拆分逻辑：

<ItemGroup>
  <Configuration Include="Debug;Release"/>
  <Platform Include="Win32;x64"/>
<ItemGroup>

将属性传递给MSBUILD任务

1 个答案: