terraform - 有没有办法为 Terraform 存档提供程序定义多个 source_file？

Question

我正在使用 Terraform archive_file provider将多个文件打包成一个 zip 文件。当我像这样定义存档时，它工作正常:

data "archive_file" "archive" {
  type        = "zip"
  output_path = "./${var.name}.zip"
  source_dir  = "${var.source_dir}"
}

但是我不希望存档包含 var.source_dir 中的所有文件，我只想要其中的一个子集。我注意到 archive_file 提供程序有一个 source_file属性，所以我希望我可以提供这些文件的列表并将它们打包到存档中，如下所示:

locals {
  source_files = ["${var.source_dir}/foo.txt", "${var.source_dir}/bar.txt"]
}

data "archive_file" "archive" {
  type        = "zip"
  output_path = "./${var.name}.zip"
  count       = "2"
  source_file = "${local.source_files[count.index]}"
}

但这不起作用，文件是为 local.source-files 中定义的每个文件构建的。因此，我有一个“最后一个获胜”的场景，其中构建的存档文件仅包含 bar.txt。

我试过这个:

locals {
  source_files = ["${var.source_dir}/main.py", "${var.source_dir}/requirements.txt"]
}

data "archive_file" "archive" {
  type        = "zip"
  output_path = "./${var.name}.zip"
  source_file = "${local.source_files}"
}

但不出所料，失败了:

data.archive_file.archive: source_file must be a single value, not a list

有没有办法实现我在这里之后的目标，即将文件列表传递给 archive_file 提供程序并将它们全部打包到存档文件中？

Answer 1

----谢谢jamiet，我修改为你的评论----

将文件复制到临时目录并存档

locals {
  source_files = ["${var.source_dir}/main.py", "${var.source_dir}/requirements.txt"]
}

data "template_file" "t_file" {
  count = "${length(local.source_files)}"

  template = "${file(element(local.source_files, count.index))}"
}

resource "local_file" "to_temp_dir" {
  count    = "${length(local.source_files)}"
  filename = "${path.module}/temp/${basename(element(local.source_files, count.index))}"
  content  = "${element(data.template_file.t_file.*.rendered, count.index)}"
}

data "archive_file" "archive" {
  type        = "zip"
  output_path = "${path.module}/${var.name}.zip"
  source_dir  = "${path.module}/temp"

  depends_on = [
    "local_file.to_temp_dir",
  ]
}

使用 archive_file 的来源

locals {
  source_files = ["${var.source_dir}/main.py", "${var.source_dir}/requirements.txt"]
}

data "template_file" "t_file" {
  count = "${length(local.source_files)}"

  template = "${file(element(local.source_files, count.index))}"
}


data "archive_file" "archive" {
  type        = "zip"
  output_path = "./${var.name}.zip"

  source {
    filename = "${basename(local.source_files[0])}"
    content  = "${data.template_file.t_file.0.rendered}"
  }

  source {
    filename = "${basename(local.source_files[1])}"
    content  = "${data.template_file.t_file.1.rendered}"
  }
}

创建 shell 脚本并使用外部数据资源调用它。

locals {
  source_files = ["${var.source_dir}/main.py", "${var.source_dir}/requirements.txt"]
}

data "template_file" "zip_sh" {
  template = <<EOF
#!/bin/bash
zip $* %1>/dev/null %2>/dev/null
echo '{"result":"success"}'
EOF
}

resource "local_file" "zip_sh" {
  filename = "${path.module}/zip.sh"
  content  = "${data.template_file.zip_sh.rendered}"
}

data "external" "zip_sh" {
  program = ["${local_file.zip_sh.filename}", "${var.name}", "${join(" ", local.source_files)}"]

  depends_on = [
    "data.template_file.zip_sh",
  ]
}