Question

我在terraform中有这个代码：

data "archive_file" "lambdazip" {

type        = "zip"
output_path = "lambda_launcher.zip"

source_dir = "lambda/etc"
source_dir = "lambda/node_modules"

source {
  content  = "${data.template_file.config_json.rendered}"
  filename = "config.json"
 }
}

我terraform plan时出现以下错误：

* data.archive_file.lambdazip: "source": conflicts with source_dir 
("lambda/node_modules")
* data.archive_file.lambdazip: "source_content_filename": conflicts 
with source_dir ("lambda/node_modules")
* data.archive_file.lambdazip: "source_dir": conflicts with 
source_content_filename ("/home/user1/experiments/grascenote-
poc/init.tpl")

我正在使用terraform版本v0.9.11

Answer 1

@Ram是正确的。您不能在同一window.history.ignoreUrlVariables()块中使用source_dir和source。

archive_file

Terraform运行

data "template_file" "config_json" {
  template = "${file("${path.module}/config_json.tpl")}"
  vars = {
    override = "my value"
  }
}

# create the template file config_json separately from the archive_file block
resource "local_file" "config" {
  content  = "${data.template_file.config_json.rendered}"
  filename = "${path.module}/lambda/etc/config.json"
}

# now you can grab the entire lambda source directory or specific subdirectories
data "archive_file" "lambdazip" {
  type        = "zip"
  output_path = "lambda_launcher.zip"

  source_dir = "lambda/"
  #source_dir = "lambda/etc"
  #source_dir = "lambda/node_modules"
}

压缩文件内容

$ terraform init
$ terraform apply
data.template_file.config_json: Refreshing state...
data.archive_file.lambdazip: Refreshing state...

An execution plan has been generated and is shown below.
Resource actions are indicated with the following symbols:
  + create

Terraform will perform the following actions:

  + local_file.config
      id:       <computed>
      content:  "{\"test\": \"my value\"}\n"
      filename: "/Users/user/lambda/config.json"


Plan: 1 to add, 0 to change, 0 to destroy.

Do you want to perform these actions?
  Terraform will perform the actions described above.
  Only 'yes' will be accepted to approve.

  Enter a value: yes

local_file.config: Creating...
  content:  "" => "{\"test\": \"my value\"}\n"
  filename: "" => "/Users/user/lambda/config.json"
local_file.config: Creation complete after 0s (ID: 05894e86414856969d915db57e21008563dfcc38)

Apply complete! Resources: 1 added, 0 changed, 0 destroyed.

Answer 2

对于node.js lambda函数，

您需要将“ resource.local_file”与“ depends_on”一起使用。并将“渲染文件”与“目录”分开。

首先，将静态目录（等，node_modules）放入“ lambda”文件夹中而不渲染文件。

第二，将渲染文件放入其他任何路径。

data "template_file" "config_json" {
  template = "${file("${path.module}/config_json.tpl")}"
  vars = {
    foo = "bar"
  }
}

resource "local_file" "config_json" {
  content  = "${data.template_file.config_json.rendered}"
  filename = "${path.module}/lambda/config.json"
}

data "archive_file" "lambda_zip" {
  type        = "zip"
  output_path = "${path.module}/lambda_function.zip"
  source_dir = "${path.module}/lambda"

  # It is important to this process.
  depends_on = [
    "local_file.config_json" 
  ]
}

resource "aws_lambda_function" "lambda" {
  filename         = "${path.module}/lambda_function.zip"
  function_name    = "lambda_function"
  role             = "${aws_iam_role.lambda.arn}"
  handler          = "index.handler"
  runtime          = "nodejs10.x"
}

resource "aws_iam_role" "lambda" {
...

如何在Terraform中创建存档文件？

2 个答案: