Is it possible to have the status of a workflow running in the same namespace and conditionally launching a new one based on the status of it?

[asant-leitha]

I need to launch a new workflow if and only if another workflow succeeded.

[wesleyscholl]

@asant-leitha

Yes, it's possible using onExit hook. Here is a basic functional example that checks the parent workflow status and submits the child workflow if successful. The parent workflow randomly fails to test the when: "{{workflow.status}} == Succeeded" condition.

Parent WorkflowTemplate

apiVersion: argoproj.io/v1alpha1
kind: WorkflowTemplate
metadata:
  name: parent-workflow
  namespace: argo
spec:
  templates:
    - name: main
      steps:
        - - name: parent-task
            template: parent-task
    - name: parent-task
      container:
        name: ""
        image: python:3.9
        command:
          - python
          - "-c"
        args:
          - |
            import random
            import sys

            # 50% chance to fail
            if random.random() < 0.5:
                print("Simulating failure: Random fail triggered!")
                sys.exit(1)  # Exit with failure
            else:
                print("Hello from Parent Workflow!")
    - name: exit-handler
      steps:
        - - name: launch-child
            templateRef:
              name: child-workflow
              template: main
            when: "{{workflow.status}} == Succeeded" # Child workflow will only run if parent workflow is successful
  entrypoint: main
  onExit: exit-handler

Child WorkflowTemplate

apiVersion: argoproj.io/v1alpha1
kind: WorkflowTemplate
metadata:
  name: child-workflow
  namespace: argo
spec:
  templates:
    - name: main
      container:
        name: ""
        image: docker/whalesay:latest
        command:
          - cowsay
        args:
          - Hello from Child Workflow!
  entrypoint: main