通过Azure应用程序服务的Terraform脚本将变量传递给Docker Compose

Question

我怎么做

VAR=VAULE docker-compose up

使用Docker编写文件从Terraform脚本构建Azure应用程序服务

 site_config {
    app_command_line = ""
    linux_fx_version = "COMPOSE|${filebase64("../docker-compose.yaml")}"
  }

  depends_on = [ null_resource.docker_push ]

  name                = "myapps-apps-appservice"
  location            = "${azurerm_resource_group.mine.location}"
  resource_group_name = "${azurerm_resource_group.mine.name}"
  app_service_plan_id = "${azurerm_app_service_plan.asp.id}"

  site_config {
    app_command_line = ""
    linux_fx_version = "COMPOSE|${filebase64("../docker-compose.yaml")}"
  }

  app_settings = {
    "WEBSITES_ENABLE_APP_SERVICE_STORAGE" = "false",
    "DOCKER_REGISTRY_SERVER_USERNAME" = azurerm_container_registry.container_registry.admin_username,
    "DOCKER_REGISTRY_SERVER_URL" = azurerm_container_registry.container_registry.login_server,
    "DOCKER_REGISTRY_SERVER_PASSWORD" = azurerm_container_registry.container_registry.admin_password
  }
}

terraform apply -auto-approve \
    -var "app_version=$VERSION" \
    -var "client_id=$ARM_CLIENT_ID" \
    -var "client_secret=$ARM_CLIENT_SECRET" \
    -var "tenant_id=$ARM_TENANT_ID" \
    -var "subscription_id=$ARM_SUBSCRIPTION_ID"

export VERSION=1

docker-compose.yaml代码段网

    image: "myacr.azurecr.io/my-container:${VERSION}"
    build: 
      context: ./my-container
      dockerfile: Dockerfile-prod
    container_name: my-container

有什么想法吗？

Answer 1

我找到了一份和Sed在一起的工作。在部署脚本中，我只是使用sed替换docker-compose.yaml文件中的值，然后用令牌替换它们，这样版本控制就保持不变。似乎有点老套，但很管用：）

eval "sed -i 's/MY_VERSION/$VERSION/' ../docker-compose.yaml"

cat ../docker-compose.yaml

terraform init 
terraform apply -auto-approve \
    -var "app_version=$VERSION" \
    -var "client_id=$ARM_CLIENT_ID" \
    -var "client_secret=$ARM_CLIENT_SECRET" \
    -var "tenant_id=$ARM_TENANT_ID" \
    -var "subscription_id=$ARM_SUBSCRIPTION_ID"

eval "sed -i 's/$VERSION/MY_VERSION/' ../docker-compose.yaml"