You're on your way to the next level! Join the Kudos program to earn points and save your progress.
Level 1: Seed
25 / 150 points
Next: Root
1 badge earned
Challenges come and go, but your rewards stay with you. Do more to earn more!
What goes around comes around! Share the love by gifting kudos to your peers.
Keep earning points to reach the top of the leaderboard. It resets every quarter so you always have a chance!
Join now to unlock these features and more
The Atlassian Community can help you and your team get more value out of Atlassian products and practices.
Hi,
I have issue type Story and two sub-tasks issue type Task and QA. Is there any way to find all Stories with closed Tasks but open QA issue?
Hi @Alena Samruk - Unfortunately, not natively. You may be able to accomplish this with a 3rd party app like scriptrunner
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Making sure I understand this a bit better. Your comment is a bit misleading...
two sub-tasks issue type Task and QA
Task is its own issue type that is at the same level hierarchically as Story. Assuming our "Task" is really "Sub-Task", you could do this with scriptrunner:
issueFunction in subtasksOf("(issueType = Sub-Task AND statusCategory = Done) AND (issueType = QA AND statusCategory != Done)")
Now, if Task/QA are really just standard issue types that are linked as children of the story you'll want to do something like this (Note - I'm using blocks, but you could go with something else):
issueFunction in linkedIssuesOf("(issueType = Task AND statusCategory = Done) AND (issueType = QA AND statusCategory != Done)", "blocks")
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.