构建一个azure云项目

构建一个azure云项目

本文介绍了构建一个azure云项目的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

您好,


我正在尝试使用Powershell构建(然后发布)Azure项目。


我是什么到目前为止已完成(以下文章:  https://docs.microsoft.com/en-us/azure/cloud-services/cloud-services-dotnet-continuous-delivery)


 function Get-ScriptDirectory 
{
Split-Path $ script:MyInvocation.MyCommand.Path
}

$ currentPath = Get-ScriptDirectory

[string] $ Path_MS_BUILD =" C:\Program Files(x86)\ MSBuild\14.0 \ Bin \ MSBUILD.exe" ;

$ Publish_Stagging_Profile =" Publish_Stagging"

& $ Path_MS_BUILD $ PublishVersion_PathToResources / t:Publish / p:PublishProfile = $ Publish_Stagging_Profile

在visual studio中我有一个名为  Publish_Stagging.azurePubxml的个人资料,其中包含23行,包括这些:

 

< AzureServiceConfiguration> Cloud< / AzureServiceConfiguration>

< AzureSolutionConfiguration> Release< / AzureSolutionConfiguration>

然而,当我运行我的Powershell并检查我的文件夹时,它会更新我项目的bin / Debug文件夹。


你知道吗?为什么?



谢谢

解决方案

Hello,

I'm trying to build (and then publish) an Azure project with Powershell.

What I've done so far (following this article : https://docs.microsoft.com/en-us/azure/cloud-services/cloud-services-dotnet-continuous-delivery)

function Get-ScriptDirectory
{
    Split-Path $script:MyInvocation.MyCommand.Path
}

$currentPath = Get-ScriptDirectory

[string]$Path_MS_BUILD="C:\Program Files (x86)\MSBuild\14.0\Bin\MSBUILD.exe"

$Publish_Stagging_Profile="Publish_Stagging"

& $Path_MS_BUILD $PublishVersion_PathToResources/t:Publish /p:PublishProfile=$Publish_Stagging_Profile

In visual studio I have a Profile called Publish_Stagging.azurePubxml which contain 23 lines including these ones :

<AzureServiceConfiguration>Cloud</AzureServiceConfiguration>

<AzureSolutionConfiguration>Release</AzureSolutionConfiguration>

However when I run my Powershell and check my folder it updates the bin/Debug folder of my project.

Do you know why ?

Thank you

解决方案


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08-16 05:43