08-06-2022 07:28 AM
Hi, follow the example to create one user. It's working however I want to create multiple users, I have tried many ways but still cannot get it work, please share some idea.
https://registry.terraform.io/providers/databricks/databricks/latest/docs/resources/user
data "databricks_group" "admins" {
display_name = "admins"
}
resource "databricks_user" "me" {
user_name = "me@example.com"
}
resource "databricks_group_member" "i-am-admin" {
group_id = data.databricks_group.admins.id
member_id = databricks_user.me.id
}
08-10-2022 05:27 PM
as @Cedric Law Hing Ping mentioned, you can use `count` or `foreach` in Terraform to do this.
Here's a simple example:
```
locals {
user_emails = toset([
"user1@example.com",
"user2@example.com",
"user3@example.com"
])
}
resource "databricks_user" "engineers" {
for_each = local.user_emails
user_name = each.value
}
data "databricks_group" "admins" {
display_name = "admins"
}
resource "databricks_group_member" "engineer-admins" {
for_each = databricks_user.engineers
group_id = data.databricks_group.admins.id
member_id = each.value.id
}
```
08-07-2022 11:52 PM
count is one of the meta-parameters available to all resources. We can specify the number of identical resources to create.
08-10-2022 07:11 AM
08-10-2022 05:27 PM
as @Cedric Law Hing Ping mentioned, you can use `count` or `foreach` in Terraform to do this.
Here's a simple example:
```
locals {
user_emails = toset([
"user1@example.com",
"user2@example.com",
"user3@example.com"
])
}
resource "databricks_user" "engineers" {
for_each = local.user_emails
user_name = each.value
}
data "databricks_group" "admins" {
display_name = "admins"
}
resource "databricks_group_member" "engineer-admins" {
for_each = databricks_user.engineers
group_id = data.databricks_group.admins.id
member_id = each.value.id
}
```
08-10-2022 10:10 PM
Thanks @Zach King this solution work for me.
01-22-2024 11:10 PM
What if I want to give User Name along with the email ID?
I used below code but its not helping(code is not failing, but not adding user name)
It seems this code line: "display_name = each.key" is not working. Pls suggest.
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