为多个文件创建一个 msbuild

Question

我写了一个用于编译多个文件的文件，但它不起作用。

<项目 xmlns="http://schemas.microsoft.com/developer/msbuild/2003" DefaultTargets="Compile"> <物品组> <FilesToCompile Include="hola.cs"/> <FilesToCompile Include="hola2.cs"/> </项目组> <属性组><br> <程序集名称>Proyecto</程序集名称> <OutputPath>Bin\</OutputPath><br> <优化>假</优化> </物业组> <目标名称="Compile"> <MakeDir 目录="$(OutputDir)"/> <Csc Sources="@(FilesToCompile)" OutputAssembly="$(OutputPath)$(AssemblyName)" Optimize="$(Optimize)" TargetType="exe" /> </目标> </项目>

它告诉我这个错误。 enter image description here

英文：

error CS0017: Program 'output file name' has more than one entry point defined. Compile with /main to specify the type that contains the entry point.

Answer 1

我自己解决了

<Project DefaultTargets = "Compile" xmlns="http://schemas.microsoft.com/developer/msbuild/2003" >
<PropertyGroup>
<NombreClase1>hola1</NombreClase1>
<NombreClase2>hola2</NombreClase2>
</PropertyGroup>
<ItemGroup>
<Clase1 Include = "hola.cs"/>               
<Clase2 Include = "hola2.cs"/>              
</ItemGroup>
<Target Name = "Compile">            
<CSC Sources = "@(Clase1)" OutputAssembly = "$(NombreClase1).exe">            
<Output TaskParameter = "OutputAssembly" ItemName = "Ejecutable1" />                        
</CSC>        
<Message Text="Archivo compilado @(Ejecutable1)"/>          
<CSC Sources = "@(Clase1)" OutputAssembly = "$(NombreClase2).exe">
<Output TaskParameter = "OutputAssembly" ItemName = "Ejecutable2" />                    
</CSC>
<Message Text="Archivo compilado @(Ejecutable2)"/>      
</Target>                       
</Project>

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