如何从msbuild项目文件本身内部使用不同的参数两次调用同一msbuild目标 [英] How to invoke the same msbuild target twice with different parameters from within msbuild project file itself

问题描述

我有以下几段msbuild代码:

I have the following piece of msbuild code:

  <PropertyGroup>
    <DirA>C:\DirA\</DirA>
    <DirB>C:\DirB\</DirB>
  </PropertyGroup>

  <Target Name="CopyToDirA"
          Condition="Exists('$(DirA)') AND '@(FilesToCopy)' != ''"
          Inputs="@(FilesToCopy)"
          Outputs="@(FilesToCopy -> '$(DirA)%(Filename)%(Extension)')">
    <Copy SourceFiles="@(FilesToCopy)" DestinationFolder="$(DirA)" />
  </Target>

  <Target Name="CopyToDirB"
          Condition="Exists('$(DirB)') AND '@(FilesToCopy)' != ''"
          Inputs="@(FilesToCopy)"
          Outputs="@(FilesToCopy -> '$(DirB)%(Filename)%(Extension)')">
    <Copy SourceFiles="@(FilesToCopy)" DestinationFolder="$(DirB)" />
  </Target>

  <Target Name="CopyFiles" DependsOnTargets="CopyToDirA;CopyToDirB"/>

因此，调用目标CopyFiles会将相关文件复制到$(DirA)和$(DirB)(如果它们尚不存在且是最新的).

So invoking the target CopyFiles copies the relevant files to $(DirA) and $(DirB), provided they are not already there and up-to-date.

但是目标CopyToDirA和CopyToDirB看起来相同，只是一个副本复制到$(DirA)，另一个副本复制到$(DirB).

But the targets CopyToDirA and CopyToDirB look identical except one copies to $(DirA) and the other - to $(DirB). Is it possible to unify them into one target first invoked with $(DirA) and then with $(DirB)?

谢谢.

推荐答案

您应该能够生成包含Dirs，然后包含％的ItemGroup.

You should be able to generate an ItemGroup containing the Dirs and then % on that.

<ItemGroup>
    <Dirs Include="C:\DirA\;C:\DirB\">
</ItemGroup>
<Target Name="CopyFiles"
    Condition="Exists('%(Dirs)') AND '@(FilesToCopy)' != ''"
    Inputs="@(FilesToCopy)"
    Outputs="@(FilesToCopy -> '%(Dirs)%(Filename)%(Extension)')">
    <Copy SourceFiles="@(FilesToCopy)" DestinationFolder="%(Dirs)" />
</Target>

或者您可以进行2次显式调用:

Or you can do 2 explicit calls:

<Target Name="CopyFiles">
    <MsBuild Projects="$(MSBuildProjectFullPath)" Targets="CopyASetOfFiles" Properties="FilesToCopy=@(FilesToCopy);DestDir=$(DirA)" />
    <MsBuild Projects="$(MSBuildProjectFullPath)" Targets="CopyASetOfFiles" Properties="FilesToCopy=@(FilesToCopy);DestDir=$(DirB)" />
</Target>

<Target Name="CopyASetOfFiles"
    Condition="Exists('$(DestDir)') AND '@(FilesToCopy)' != ''"
    Inputs="@(FilesToCopy)"
    Outputs="@(FilesToCopy -> '$(DestDir)%(Filename)%(Extension)')">
    <Copy SourceFiles="@(FilesToCopy)" DestinationFolder="$(DestDir)" />
</Target>

我没有测试过任何一种语法，但是对第二种语法比较有信心.

I haven't tested either syntax, but am relatively more confident of the second.

(答案是，如果有的话，在我桌上的Sayed Hashimi书中-您必须等到第一个:

(The answer, if there is one, is in my Sayed Hashimi book on my desk - you'll have to wait until the first of:

1. 拿书
2. 我很无聊
3. Sayed找到了这篇文章，并给出了一个经过测试的出色答案)

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