Question

假设我有属性$(Foo)，它被定义为某个函数的结果，返回字符串值$(Bar)。是否有可能以某种方式扩展它，以便$(Foo)将扩展为$(Bar)的值？

给出示例项目：

<?xml version="1.0" encoding="utf-8"?>
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
    <PropertyGroup>
        <Bar>Bar Value</Bar>
        <Foo>$([System.String]::Concat("$(","Bar",")"))</Foo>
        <Baz>$(Foo)</Baz>
        <Qux>$(Bar)</Qux>
    </PropertyGroup>
    <Target Name="Test">
        <Message Text="Foo == $(Foo)" />
        <Message Text="Baz == $(Baz)" />
        <Message Text="Qux == $(Qux)" />
    </Target>
</Project>

这就是我所拥有的：

S:\>msbuild Test.proj /t:Test /nologo                 
Build started 18.09.2013 17:52:14.                    
Project "S:\Test.proj" on node 1 (Test target(s)).    
Test:                                                 
  Foo == $(Bar)                                       
  Baz == $(Bar)                                       
  Qux == Bar Value                                    
Done Building Project "S:\Test.proj" (Test target(s)).


Build succeeded.                                      
    0 Warning(s)                                      
    0 Error(s)

因此，$(Qux)（直接定义为$(Bar)）会正确展开，但$(Foo)和$(Baz)不会。是否有可能扩展它们？

S:\>msbuild /version                                     
Microsoft (R) Build Engine version 4.0.30319.17929       
[Microsoft .NET Framework, version 4.0.30319.18052]      
Copyright (C) Microsoft Corporation. All rights reserved.

4.0.30319.17929

Answer 1

你想模拟像$（$（Foo）这样的东西，这是msbuild的无效语法。但您可以通过使用项目和在目标中动态创建项目来模拟此行为。由于Property and Item evaluation order，您无法在“全局范围”中执行此操作。

所以你必须在一些目标中做到这一点。以下是通过InitialTargets设置属性的示例。

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

<PropertyGroup>
    <Bar>Bar Value</Bar>
    <!-- A value of Foo property specifies the name of the property it takes value from. -->
    <Foo>Bar</Foo>
    <Baz></Baz>
    <Qux>$(Bar)</Qux>
</PropertyGroup>

<Target Name="Test">
    <Message Text="Foo == $(Foo)" />
    <Message Text="Baz == $(Baz)" />
    <Message Text="Qux == $(Qux)" />
</Target>

<Target Name="MyPropertiesSetup">
    <ItemGroup>
        <_Foo Include="$(Foo)" />
        <_Baz Include="%(_Foo.Identity)" />
    </ItemGroup>
    <PropertyGroup>
        <Foo>$(%(_Foo.Identity))</Foo>
        <Baz>$(%(_Baz.Identity))</Baz>
    </PropertyGroup>
</Target>

</Project>

如果你可以使用item作为值而不是属性，那么有更优雅的方法（MSBuild Trickery＃68 - @（$（CanYouDoThis））：

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

<PropertyGroup>
    <Foo>Bar</Foo>
    <Foo2>Bar2</Foo2>
</PropertyGroup>

<ItemGroup>
   <Bar Include="Bar Value" />
   <Bar2 Include="Bar2 Value" />
</ItemGroup>

<Target Name="Test">
    <Message Text="Foo == '@($(Foo))'" />
    <Message Text="Foo2 == '@($(Foo2))'" />
</Target>

</Project>

展开包含其他属性名称的MSBuild属性

1 个答案: