Forums

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

JQL query for multiple epics

Steven Choiniere October 22, 2020

Helllo, 

 

I want to pull a timespent report for multiple epics (over 70+), and having trouble with the JQL query. 

 

Any tips on how to efficiently group many epics together?

2 answers

0 votes
final_openshift Certi October 22, 2020

Hi Steve,

You can achieve storing a subquery using the  JQL Search Extension plugin , once you create a subquery it can be saved in the JQL Subquery Section and then can be used from there in other complex queries , please refer to the documents it contains examples and other necessary information related to the usage of plugin you can use linkedBy in (Epic Name)

0 votes
Stephen Garber
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 Leaders.
October 22, 2020

What do these epics have in common?

 

JQL works well if there is a single attribute that can be found all the tickets that you are looking for. You may want to add the same label or component to each of these tickets to ensure you can easily query them.

Another tip that may help is to query based on update or create date. Something like this will show the issues created in the last 30 days.

issuetype = Epic AND created > startOfDay(-30d)

Steven Choiniere October 22, 2020

I have grouped all epics into one filter called "My Epic Filter". Do you know the JQL query to have all linked issues within the epics in my filter?

Stephen Garber
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 Leaders.
October 22, 2020

The only way that I know how to do this is to use the scriptrunner linkedIssuesOf function. It works like this:

 

issuefunction in linkedIssuesOf(SUBQUERY, "is Epic of")

You could also use this to do the same thing:

issueFunction in issuesInEpics(SUBQUERY)

 

The reverse of this function is:

issueFunction in epicsOf(SUBQUERY)

Suggest an answer

Log in or Sign up to answer
DEPLOYMENT TYPE
CLOUD
PRODUCT PLAN
STANDARD
PERMISSIONS LEVEL
Product Admin
TAGS
AUG Leaders

Atlassian Community Events