Amazon web services terraform生成一个文件，压缩并上传到s3

我想动态生成一个文件，压缩它并使用terraform将其上传到s3。以下是我到目前为止的情况

#Setup a local variable with docker container information
locals {
  file_content = jsonencode({
    "AWSEBDockerrunVersion" : "1",
    "Image" : {
      "Name" : "${var.aws_account_id}.dkr.ecr.${var.aws_region}.amazonaws.com/vin-${var.application}-api:${var.environment}",
      "Update" : "true"
    },
    "Ports" : {
      "ContainerPort" : 80,
      "HostPort" : 80
    }
  })
}

#Generate the Dockerrun.aws.json with the container information and environment tag
resource "local_file" "docker_container_info" {
  content  = local.file_content
  filename = "./${var.service_name}-${var.environment}-Dockerrun.aws.json"
}

#Zip the file
data "archive_file" "source" {
  type        = "zip"
  source_dir  = "./${var.service_name}-${var.environment}-Dockerrun.aws.json"
  output_path = "./${var.service_name}-${var.environment}-Dockerrun.aws.json.zip"

  depends_on = [
    local_file.docker_container_info
  ]
}

/*
#Upload the zip file to s3 bucket under DockerRunFiles folder
resource "aws_s3_bucket_object" "file_upload" {
  bucket           = var.run_file_bucket
  key              = "${var.bucket_name}/${var.service_name}-${var.environment}-Dockerrun.aws.json.zip"
  source           = "${data.archive_file.source.output_path}"
}
*/


variable "aws_region" {
  default = "us-east-1"
}

variable "environment" {
  default = "lab"
}

variable "aws_environment" {
  default = "lab"
}

variable "service_name" {
  default = "service_name"
}

variable "application" {
  default = "application_name"
}

variable "aws_account_id" {
  default = "1234567890"
}

variable "bucket_name" {
  default = "my_bucket"
}

我看到

service\u name-lab-dockrun.aws.json

文件正确生成，其中包含json。但是我得到的错误是：错误存档目录：无法存档文件目录：./service\u name-lab-dockrun.aws.json。terraform

数据存档文件是否只能压缩文件夹，而不能压缩json文件？感谢你的帮助

版本详情：

Terraform v0.12.28
+ provider.archive v2.0.0
+ provider.local v2.0.0

您的
“/${var.service_name}-${var.environment}-dockrun.aws.json”
是一个文件，而不是目录。因此，您应该使用
source\u file
而不是
source\u dir
，如地形图所示

天哪！谢谢这是我的经理经常对我说的，换一种眼光可能会发现问题。。
#Zip the file
data "archive_file" "source" {
  type        = "zip"
  source_file  = "./${var.service_name}-${var.environment}-Dockerrun.aws.json"
  output_path = "./${var.service_name}-${var.environment}-Dockerrun.aws.json.zip"

  depends_on = [
    local_file.docker_container_info
  ]
}