The Atlassian Community Forums are currently in read-only mode. We will be relaunching on a new platform on September 22 (read more here). We apologize for the extended downtime. For concerns or questions, please email communitymanagers@atlassian.com. See you on the other side, on the new Atlassian Community Forums! :)

×

Forums

Articles
Create
cancel
Showing results for 
Search instead for 
Did you mean: 

Find all Storys to an Epic by using a subquery on Epic Link

Numb007
Contributor
September 10, 2020

Hi together, 

in my organization we are working with Epics and Stories. Unfortunately we don't use the parent-child-connection between Epics and Stories, so in the query no function like epicsOf is working. So there is a field on Story called Epic Link and in this we are maintaining the corresponding Epic. 

 

Now I want to build a query to get all stories which are connected to an EPIC, where this Epic has a specific label or a specific fix version. 

Is this possible? 

Everytime if I try to create a subquery for the field Epic-Link this is not possible. DO you have any ideas? 

Thanks for you help in advance. 

 

Kind regards,

Numb007

2 answers

1 accepted

Comments for this post are closed

Community moderators have prevented the ability to post new answers.

0 votes
Answer accepted
Numb007
Contributor
September 17, 2020

I have found it issueFunction in epicsOf is working. 

Wendy Sorensen
April 6, 2021

Can you give an example of this?  Is epicsOf available in Cloud Jira

0 votes
Haddon Fisher
Rising Star
Rising Star
Rising Stars are recognized for providing high-quality answers to other users. Rising Stars receive a certificate of achievement and are on the path to becoming Community Champions.
September 11, 2020

You said you are not using the 'parent-child' relationship between epics and stories; how are you establishing which epic a story is a part of?

Assuming there is some kind of relationship between the two, you could use some of the JQL functions added by the 'Scriptrunner' plugin to do this.