如何解决 terraform 错误“等待状态变为‘done: true’时超时（最后状态：‘done: false’，超时：10m0s）”？

Question

我正在尝试在 GCP 中使用 terraform 创建 Firestore 索引。下面是我的 Terraform 脚本：

resource "google_firestore_index" "job_config1_index" {
  project = var.projectId

  collection = var.job_config_firestore
  depends_on = [
    "google_firestore_index.job_config4_index"
  ]

  fields {
    field_path = "customer_id"
    order      = "ASCENDING"
  }

  fields {
    field_path = "job_type"
    order      = "ASCENDING"
  }

  fields {
    field_path = "start_date_time"
    order      = "ASCENDING"
  }

  fields {
    field_path = "__name__"
    order      = "ASCENDING"
  }
}

以下是日志：

Step #2: Error: Error waiting to create Index: Error waiting for Creating Index: timeout while waiting for state to become 'done: true' (last state: 'done: false', timeout: 10m0s)
Step #2: 
Step #2:   on firestore.tf line 298, in resource "google_firestore_index" "job_config1_index":
Step #2:  298: resource "google_firestore_index" "job_config1_index" {
Step #2: 
Step #2: 

我的其他 Firestore 索引创建得很好。如何增加每个索引的超时时间？

Answer 1

某些资源（包括资源）具有可选的可配置超时，用于使用块google_firestore_index创建、更新和/或删除：timeouts

resource "aws_db_instance" "example" {
  # ...

  timeouts {
    create = "60m"
    delete = "2h"
  }
}

因此，在您的情况下，您可以create向 Firestore 索引添加超时，如下所示：

resource "google_firestore_index" "job_config1_index" {
  project = var.projectId

  collection = var.job_config_firestore
  depends_on = [
    "google_firestore_index.job_config4_index"
  ]

  fields {
    field_path = "customer_id"
    order      = "ASCENDING"
  }

  fields {
    field_path = "job_type"
    order      = "ASCENDING"
  }

  fields {
    field_path = "start_date_time"
    order      = "ASCENDING"
  }

  fields {
    field_path = "__name__"
    order      = "ASCENDING"
  }

  timeouts {
    create = "60m"
  }
}