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使用msbuild生成wcf项目_Wcf_Msbuild - Fatal编程技术网

使用msbuild生成wcf项目

使用msbuild生成wcf项目,wcf,msbuild,Wcf,Msbuild,我正在尝试使用自定义MSBuild脚本构建WCf项目 在PropertyGroup中,我有 <DeployServicesRoot>$(MSBuildProjectDirectory)\$(Configuration)\Services</DeployServicesRoot> <IntermediateRoot>$(DeployServicesRoot)\obj</IntermediateRoot> <DevelopmentSymbo

我正在尝试使用自定义MSBuild脚本构建WCf项目

在PropertyGroup中,我有

 <DeployServicesRoot>$(MSBuildProjectDirectory)\$(Configuration)\Services</DeployServicesRoot>
 <IntermediateRoot>$(DeployServicesRoot)\obj</IntermediateRoot>
 <DevelopmentSymbols>DEVELOPMENT;DEBUG</DevelopmentSymbols>
$(MSBuildProjectDirectory)\$(配置)\Services
$(部署服务启动)\obj
发展;调试
在项目组中

<Solution Include="..\..\src\WCFProj\WCFProj.csproj" />
<DeployFiles Include="$(DeployServicesRoot)\**\*.*" />
<IntermediateFiles Include="$(IntermediateRoot)\**\*.*" />

和MSBuild任务:

<MSBuild Projects="@(Solution)" Properties="DefineConstants=$(DevelopmentSymbols);OutputPath=$(DeployServicesRoot)\bin\;BaseIntermediateOutputPath=$(IntermediateRoot)\Services\;DeployOnBuild=true;DeployTarget=Package"/>


问题是,部署的包位于obj子目录中,而不是bin中。bin中只有*.dll文件,没有WCF*.svc和转换后的web.config。是否有任何可用的附加参数可设置为输出包路径的MSBuild任务?我知道我可以通过复制任务来完成,但我很好奇还有其他(更简单的)解决方案吗?

在MSBuild属性中,我必须在DeployTarget之后添加_-PackageTempDir=$(DeployServicesRoot)”