Invoke a lambda function (cross account) from event bridge

0

Is it possible to invoke a lambda function in account A directly from event bridge (custom event bus) in account B without integrating with an event bus in account A?

已提問 2 年前檢視次數 7384 次
2 個答案
7
已接受的答案

Hi there,

At this point sending events cross-account is only supported via Event Buses which means that it is not possible to directly set a target of Lambda-in-AccountA to an EventBridge Rule Rule-in-AccountB. The supported flow would be as below:

Event-Bus-AccountB => Rule-in-AccountB => Event-Bus-AccountA as one of the Target => Rule-in-AccountA to match the events coming from AccountB and with a Target of Lambda-in-AccountA.

To help you further understand, here are some useful resources Sending and Receiving events between AWS Accounts and Simplifying cross-account access with Amazon EventBridge Resource policies

Hope this helps, and please let me know if you have any further questions

AWS
支援工程師
已回答 2 年前
  • The flow Event-Bus-AccountB=>Rule-in-AccountB=>SQS-in-AccountB=>Lambda-in-AccountA looks to be supported now. As recently Lambda service announced the support of invocation from SQS Queue in a different AWS account. You can check the details from https://aws.amazon.com/about-aws/whats-new/2021/09/aws-lambda-lambda-function-amazon-sqs-queue/

  • In case it's not possible to use event bus in account A as a target. Any other options to invoke Lambda-in-AccountA when an event is published on Event-Bus-AccountB? Is Event-Bus-AccountB=>Rule-in-AccountB=>SQS-in-AccountB=>Lambda-in-AccountA a feasible alternative?

0

I don't think this is possible.

With EventBridge, it's possible to configure an event bus to receive events from different sources including sources from other AWS accounts. To invoke a lambda function - this would be done via Event Bus -> Rule -> Target (in this case, the target would be a lambda function). Currently, it seems that the Lambda function must be in the same account as the Rule.

If you're looking to do extra architecture, a couple solutions could be as follows:

jsonc
已回答 2 年前

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