Hello All,
I have user stories which have seperated with labels and I need to filter all test cases linked to the user stories which have labels "a" and "b". Could you help me?
Thanks in advance!
welcome to the community!
Unfortunately, this is trickier than one might think; as a hierarchical query, it would really require some kind of join or subquery, which isn't available in plain Jira/JQL.
A few directions forward:
If you want to run your search dynamically, without manually "stitching" two queries together, you'll need extra tooling:
Hope this helps,
Best,
Hannes
Just to expand on the last point, this is how this would look in the app that my team and I are working on, JXL for Jira. Put simply, you'd create a sheet with all issues that are potentially relevant to you, model your issue hierarchy (that's just a couple of clicks), and then use JXL filtering capabilities to narrow down to the issues that you care about:
Once you have your list of issues, you can work on these directly in JXL (much like you'd do in e.g. Excel or Google Sheets), trigger various operations in Jira, or export them for further processing.
Any questions just let me know!
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
I’m Charlotte, a support engineer at Appfire and I’m here to help you.
Unfortunately, using vanilla JQL, you’ll not be able to do it dynamically.
In the app where my team works, JQL Search Extensions for Jira, you can use this query to find all Test Cases linked to Stories that have labels “a” and “b”:
issue in linkedIssuesOfQuery("type = story AND labels = a AND labels = b") AND type = "Test Case"
Please contact our support if you have any other questions about this query.
We’ll be happy to help you!
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.