检查键是否在地图中，然后使用 terraform 返回值

Question

我想：

1. 首先检查地图中是否存在键
2. 如果是，那么我想返回键的值
3. 如果没有，我想返回一张空地图。

  replication_configuration = (contains(keys(local.s3_bucket), "${local.bucket_prefix}-${var.buckets[count.index]}")
                               ? lookup(local.s3_bucket["${local.bucket_prefix}-${var.buckets[count.index]}"], "replication_configuration", {})
                               : {})

但是，我收到此错误

  44:   replication_configuration = (contains(keys(local.s3_bucket), "${local.bucket_prefix}-${var.buckets[count.index]}")
  45:                                ? lookup(local.s3_bucket["${local.bucket_prefix}-${var.buckets[count.index]}"], "replication_configuration", {})
  46:                                : {})
    |----------------
    | count.index is 7
    | local.bucket_prefix is "web-production"
    | local.s3_bucket is object with 1 attribute "web-production-data"
    | var.buckets is list of string with 17 elements

The true and false result expressions must have consistent types. The given
expressions are object and object, respectively.

我相信错误是因为条件的“if true”部分返回了一个不同类型的地图。

有没有更好的方法来解决这个问题？