Question

我正在尝试使用Visual Studio 2013自动部署Web Api 2项目。我已经创建了一个名为“Test”的发布配置文件，如下所示

<?xml version="1.0" encoding="utf-8"?>
    <!--
    This file is used by the publish/package process of your Web project. You can customize the behavior of this process
    by editing this MSBuild file. In order to learn more about this please visit http://go.microsoft.com/fwlink/?LinkID=208121. 
    -->
    <Project ToolsVersion="4.0" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
    <PropertyGroup>
        <WebPublishMethod>FileSystem</WebPublishMethod>
        <LastUsedBuildConfiguration>Release</LastUsedBuildConfiguration>
        <LastUsedPlatform>x86</LastUsedPlatform>
        <SiteUrlToLaunchAfterPublish />
        <LaunchSiteAfterPublish>True</LaunchSiteAfterPublish>
        <ExcludeApp_Data>False</ExcludeApp_Data>
        <publishUrl>C:\DbServiceDeploy</publishUrl>
        <DeleteExistingFiles>True</DeleteExistingFiles>
    </PropertyGroup>
    </Project>

即使它有行＆lt; LastUsedBuildConfiguration＆gt; Release＆lt; / LastUsedBuildConfiguration＆gt;似乎Visual Studio正在发布我的调试版本。我正在调用msbuild

    "C:\Program Files (x86)\MSBuild\12.0\Bin\MSBuild.exe" C:\somefolder\Myproj.csproj /p:
DeployOnBuild=true /p:PublishProfile=Test

Answer 1

这篇优秀的博客文章让我找到了答案http://sedodream.com/2012/10/27/MSBuildHowToSetTheConfigurationProperty.aspx。我将总结链接死亡的情况：当一个构建被启动时，MsBuild会对属性进行一次评估，并将该值用于构建的其余部分。由于Configuration属性正在解析为Debug，因此MsBuild使用它作为要发布的配置。

TLDR;

在命令行上传递配置，添加

/ P：配置=释放

到命令行调用

Visual Studio发布配置文件发布错误的构建配置

1 个答案: