MSBuild：条件构造（项目参考|文件参考） [英] MSBuild: Conditional Construct (Project Reference | File Reference)

问题描述

我可能会创建以下构建行为：

如果配置是调试，那么在ExCobol.cblproj
上使用ProjectReferences如果配置是DebugVB，那么在ExCobol.dll上使用FileReferences

当是的时候，如何实现呢？

我假设在项目文件中使用标签将
做到。

这是真的消除了需要一个cobol编译器的
DebugVB配置？

解决方案

，假设您有

 < ProjectReference ...> ...< / ProjectReference> 
  

或

 < Reference ...> ...< / Reference> 
  

您要手动编辑.proj文件以包含

 < ProjectReference Condition ='$（Configuration）'！='DebugVB'...> ...< / ProjectReference> 
< Reference Condition ='$（Configuration）'=='DebugVB'...> ...< / Reference> 
  

I´m still trying to eleminate the need of a cobol compiler in a Project with cobol-Projects in it.

Is it possible to create following build behaviour:

If the Configuration is Debug then use ProjectReferences on ExCobol.cblproj if the Configuration is DebugVB then use FileReferences on ExCobol.dll

When Yes, How to achieve it?

I assume the use of tags in the project file will do the trick.

And does this really eliminate the need of a cobol compiler for the DebugVB Configuration?

解决方案

Regarding the conditional 'how', assuming you have either

<ProjectReference ...>...</ProjectReference>

or

<Reference ...>...</Reference>

you want to hand-edit the .proj file to include both thusly

<ProjectReference Condition="'$(Configuration)'!='DebugVB'" ...>...</ProjectReference>
<Reference Condition="'$(Configuration)'=='DebugVB'" ...>...</Reference>

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