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A have a filter A which delivers a result of 33 issues. Filter B filters the same data basis but delivers 29 issues. How do I need to combine these two filters so that I can show the difference. In that case which 4 more issues are found in filter A?
Hi Markus,
In Jira JQL you can refer to a filter like: filter = "filter A"
Therefore you could do something like: filter = "filter A" and filter != "filter B"
I think this will give you the difference. Let me know if this works for you.
Hi Charlie, thx for the quick answer. I tried it out and it worked! Great thing!
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Hi Markus,
I'm glad I could help and that the solution is working because I did not have the time to try it out myself :-)
Please accept my answer to mark this question as answered to help out other users with the same question in the future.
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Suppose if I have two filters created as filter 1 with x, y,z fields and filter 2 with x,a,b,c fields. Then is there any jql query to combine filter 1 and 2 which provides us single output of all the fields present in both the filters.
Thanks in advance.. It would be of great help if could guide me here.
Thanks,
Himaja.
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Hi @himaja
Have you tried "filter A" and filter = "filter B"?
I'm not sure if this will provide the expected result. If it doesn't create a new question for this. You'll have more chance to get an answer.
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Thanks for your reply but I receive blank output i.e, no output,when I tired above solution.
The requirement is if filterA has x,y,z,... so on fields and filterB has x,a,b,c,.. so on fields then by combining filterA and filterB can I get output as x,a,b,c,y,z.. so on..
May be something as below:
prepare a report in jira using JQL query, where i need to fetch data from fliterA i.e, child(deploy issues) with fields like key, issuetype, environment, linked issues, status etc., and from fliter B i.e,parent(stories, defects) with fields like summary, category, status, key etc., here fields which are present in parent are not included in child. So only common field present in parent and child are linked issues(child) == key (parent). So can this be established without usage of plug-ins.
Please let me if we can connect on this..
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Hi @himaja
No I don't think it's possible.
But again: you asked this on an already resolved question so I suggest you create a new question so there's more chance for other community members to see it.
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Hi @Charlie Misonne good day hope you and your family is doing great.
If I may just ask, is it possible to search for issues that have similar Customer ID in Filter B from Filter A.
I would want the result to be from Filter A
Thank you in advance for responding. God bless. 😊
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