从ARM模板创建Azure Web应用程序插槽而无需复制原始Web应用程序配置

问题描述：

我正在尝试通过ARM模板创建Web应用程序插槽.

I am trying to create web app slots through ARM template.

我能够创建它们，但是看起来默认行为是将它们创建为当前Web应用程序状态的副本.这导致我的广告位继承了应用设置，连接字符串，虚拟目录等.

I was able to create those but it looks like the default behavior is to create the them as a copy of the current web app state. This result in my slot inheriting app settings, connection strings, virtual directories, ....

这里有一个复制示例，演示了行为 https://github.com/ggirard07/ARMSlotWebConfig

Here a reproduction sample which demonstrate the behavior https://github.com/ggirard07/ARMSlotWebConfig.

我想让我的插槽干净整洁，这是azure门户的默认行为.门户网站可以通过指定创建插槽时发布的"configSource": "",值来允许用户选择行为.

I want my slot clean and fresh instead, which is the azure portal default behavior. The portal is able to allow a user to select the behavior by specifying the "configSource": "", value it posts when creating the slot.

反正有从ARM模板内部实现相同目标的功能吗?

Is there anyway to achieve the same from inside an ARM template?

答

要防止从生产应用程序复制设置，只需在插槽properties中添加一个空的siteConfig对象.例如

To prevent the copying of settings from the production app, just add an empty siteConfig object in the slot properties. e.g.

    {
      "apiVersion": "2015-08-01",
      "type": "slots",
      "name": "maintenance",
      "location": "[resourceGroup().location]",
      "dependsOn": [
        "[resourceId('Microsoft.Web/Sites/', variables('webSiteName'))]"
      ],
      "properties": {
        "siteConfig": { }
      }
    }

我发送了 PR 来说明您的存储库.

I sent a PR to illustrate on your repo.

从ARM模板创建Azure Web应用程序插槽而无需复制原始Web应用程序配置

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