Question

我想要从ARM模板部署的应用程序服务中的公共IP地址，如下面的简化示例所示。我已经看到了从API网关，VM的VNET，App Service Environment等获取公共入站IP地址的示例，但是我没有发现任何指示如何从简单的App Service部署中获取公共入站IP地址的示例。我发现在浏览ARM API，以及如何将其转换为字符串并转换为JSON文件，而不是拜占庭。任何帮助将不胜感激。

{
    "$schema": "http://schema.management.azure.com/schemas/2015-01-01/deploymentTemplate.json#",
    "contentVersion": "1.0.0.0",
    "parameters": {
        "appServiceName": {
            "type": "string",
            "minLength": 1,
            "metadata": {
                "description": "Specifies the name of the Azure App Service"
            }
        },
        "appServicePlanName": {
            "type": "string",
            "minLength": 1
        }
    },
    "variables": {
    },
    "resources": [
        {
            "apiVersion": "2015-08-01",
            "name": "[parameters('appServiceName')]",
            "type": "Microsoft.Web/sites",
            "kind": "app",
            "location": "[resourceGroup().location]",
            "dependsOn": [],
            "properties": {
                "serverFarmId": "[resourceId('Microsoft.Web/serverfarms', parameters('appServicePlanName'))]",
                "clientAffinityEnabled": false
            },
            "resources": [],
        }
    ],
    "outputs": {
        "appServiceName": {
            "type": "string",
            "value": "[parameters('appServiceName')]"
        },
        "ipAddress": {
            "type": "string",
            "value": "whatingodsnamegoeshere"
        }
    }
}

Answer 1

您需要使用reference()函数并为其提供资源ID或名称（如果资源位于同一模板中则仅提供名称）：

reference(parameters('appServiceName'), '2016-03-01', 'Full').properties.inboundIpAddress

或使用resourceId()：

reference(resourceId('Microsoft.Web/serverfarms', parameters('appServicePlanName')), '2016-03-01', 'Full').properties.inboundIpAddress

如何输出通过ARM模板部署的App Service的IP地址

1 个答案: