Msbuild 根据生成目标更改引用路径

我有一个单元测试项目，它需要一些外部依赖项。这些依赖项有两种类型：i386（…..\External\EA\i386\Core.dll）和amd64（…..\External\EA\amd64\Core.dll）

..\..\..\..\External\EA\amd64\Core.dll
..\..\..\..\External\EA\amd64\Util.dll

MsTest是32位，我希望这些程序集的路径是……。\External\EA**i386**\Core.dll。换句话说，如何告诉msbuild选择正确的构建目标

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谢谢

只需在引用上设置一个条件，或者如下所示，在包含它们的项目组上设置一个条件

<ItemGroup
   Condition="'$(Platform)' == 'x64'">
   <Reference Include="Core">
      <HintPath>..\..\..\..\External\EA\amd64\Core.dll</HintPath>
   </Reference>
   <Reference Include="Util">
      <HintPath>..\..\..\..\External\EA\amd64\Util.dll</HintPath>
   </Reference>
</ItemGroup>
<ItemGroup
   Condition="'$(Platform)' == 'Win32'">
   <Reference Include="Core">
      <HintPath>..\..\..\..\External\EA\i386\Core.dll</HintPath>
   </Reference>
   <Reference Include="Util">
      <HintPath>..\..\..\..\External\EA\i386\Util.dll</HintPath>
   </Reference>
</ItemGroup>

..\..\..\..\External\EA\amd64\Core.dll
..\..\..\..\External\EA\amd64\Util.dll
..\..\..\..\External\EA\i386\Core.dll
..\..\..\..\External\EA\i386\Util.dll

您必须准确地发现您的项目正在使用的$（平台）的哪些值，对项目的XML进行简单的检查就会显示这些值。

请看这里：

<ItemGroup
   Condition="'$(Platform)' == 'x64'">
   <Reference Include="Core">
      <HintPath>..\..\..\..\External\EA\amd64\Core.dll</HintPath>
   </Reference>
   <Reference Include="Util">
      <HintPath>..\..\..\..\External\EA\amd64\Util.dll</HintPath>
   </Reference>
</ItemGroup>
<ItemGroup
   Condition="'$(Platform)' == 'Win32'">
   <Reference Include="Core">
      <HintPath>..\..\..\..\External\EA\i386\Core.dll</HintPath>
   </Reference>
   <Reference Include="Util">
      <HintPath>..\..\..\..\External\EA\i386\Util.dll</HintPath>
   </Reference>
</ItemGroup>