Question

我正在使用Terraform archive_file provider将多个文件打包为一个zip文件。当我这样定义归档文件时，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，我只希望其中的一部分。我注意到archive_file提供者具有source_file属性，因此我希望可以提供这些文件的列表并将其打包到归档中，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，将为local.source-files中定义的每个文件构建存档，因此我遇到了“最后一个获胜”的情况，其中构建的存档文件仅包含bar.txt。

我尝试过：

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}"
}

但毫不奇怪，它失败了：

data.archive_file.archive：source_file必须是单个值，而不是列表

是否有办法实现我的目标，即将文件列表传递给archive_file提供程序并将其打包到归档文件中？

Answer 1

----谢谢jamiet，我修改为您的评论----

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",
  ]
}

有没有一种方法可以为Terraform归档提供程序定义多个source_file？

1 个答案: