Arm Template条件输出参数

joh*_*ley 7 azure azure-rm-template

我有一个有条件地创建资源的ARM模板:

    {
  "type": "Microsoft.Storage/storageAccounts",
  "sku": {
    "name": "Standard_GRS",
    "tier": "Standard"
  },
  "kind": "BlobStorage",
  "name": "[variables('storageAccounts_name')]",
  "condition": "[equals(parameters('is_Not_Development'), 'True')]",
  "apiVersion": "2017-06-01",
  "location": "[resourceGroup().location]",
  "scale": null,
  "properties": {
    "accessTier": "Hot"
  },
  "dependsOn": []
},
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在我的输出参数中,我有以下内容,如果未创建资源,则会导致错误:

    "storageAccountConnectionString": {
  "type": "string",
  "value": "[Concat('DefaultEndpointsProtocol=https;AccountName=',variables('StorageAccounts_name'),';AccountKey=',listKeys(resourceId('Microsoft.Storage/storageAccounts', variables('StorageAccounts_name')), providers('Microsoft.Storage', 'storageAccounts').apiVersions[0]).keys[0].value)]"
},
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我试过这个:

    "storageAccountConnectionString": {
  "type": "string",
  "condition": "[equals(parameters('is_Not_Development'), 'True')]",
  "value": "[Concat('DefaultEndpointsProtocol=https;AccountName=',variables('StorageAccounts_name'),';AccountKey=',listKeys(resourceId('Microsoft.Storage/storageAccounts', variables('StorageAccounts_name')), providers('Microsoft.Storage', 'storageAccounts').apiVersions[0]).keys[0].value)]"
},
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条款条款,但不承认.如何使输出参数有条件?

更新:

我尝试过以下方法:

    "storageAccountConnectionString": {
  "type": "string",
  "value": "[if(equals(parameters('is_Not_Development'),'False'),'null',Concat('DefaultEndpointsProtocol=https;AccountName=',variables('StorageAccounts_name'),';AccountKey=',listKeys(resourceId('Microsoft.Storage/storageAccounts', variables('StorageAccounts_name')), providers('Microsoft.Storage', 'storageAccounts').apiVersions[0]).keys[0].value))]"
},
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但它给了我相同的错误信息,它必须评估真假条件.

xab*_*kos 7

有一个技巧可以解决这个问题,我们成功地使用它.

让我们看一下,例如,如果已部署相应的资源,以下模板如何返回值.

"$schema": "http://schema.management.azure.com/schemas/2015-01-01/deploymentTemplate.json#",
  "contentVersion": "1.0.0.0",
  "parameters": {
    "appInsightsLocation": {
      "type": "string",
      "defaultValue": "",
      "allowedValues": [
        "",
        "northeurope",
        "westeurope"
      ]
    }
  },
  "variables": {
    "appInsightsName": "exampleAppInsights",
    "planName": "example-plan",
    "appInsightsEnabled": "[if(greater(length(parameters('appInsightsLocation')), 0), 'true', 'false')]",
    "appInsightsOrPlanResource": "[if(bool(variables('appInsightsEnabled')), concat('Microsoft.Insights/components/', variables('appInsightsName')), concat('Microsoft.Web/serverFarms/', variables('planName')))]",
    "appInsightsKeyOrPlanName": "[if(bool(variables('appInsightsEnabled')), 'InstrumentationKey', 'name')]"
  },
  "resources": [
    {
      "comments": "The example service plan",
      "apiVersion": "2015-08-01",
      "type": "Microsoft.Web/serverfarms",
      "location": "[resourceGroup().location]",
      "name": "[variables('planName')]",
      "sku": {
        "name": "B1",
        "capacity": 1
      },
      "properties": {
        "numberOfWorkers": 1,
        "name": "[variables('planName')]"
      }
    },
    {
      "comments": "The application insights instance",
      "apiVersion": "2014-04-01",
      "condition": "[bool(variables('appInsightsEnabled'))]",
      "type": "Microsoft.Insights/components",
      "location": "[parameters('appInsightsLocation')]",
      "name": "[variables('appInsightsName')]",
      "properties": {}
    }
  ],
  "outputs": {
    "appInsightsKey": {
      "value": "[if(bool(variables('appInsightsEnabled')), reference(variables('appInsightsOrPlanResource'))[variables('appInsightsKeyOrPlanName')], '')]",
      "type": "string"
    }
  }
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模板声明了两个资源.一个应用服务计划和一个Application Insights实例.仅当location参数不为空字符串时才部署AppInsights实例.因此,只有在创建实例时,才会返回此实例的检测项.

为实现这一目标,我们还需要一个始终存在的资源.在我们的例子中,这是服务计划.我们使用此资源在未部署AppInsights时获取引用.当然,这可能是任何天蓝色的资源.

诀窍发生在两个变量上appInsightsOrPlanResource,appInsightsKeyOrPlanName我们声明.当appInsightsLocation设置然后这两个变量最终引用其从输出返回的实例的关键.

appInsightsLocation另一方面,当未提供时,这两个变量包含对未使用但有效的服务计划的有效引用.我们需要这样做,因为if函数总是评估双方.但是,在这种情况下,输出会返回一个空字符串.