Issues assigned / resolved by user

I need to query JIRA to get a report of issues assigned versus resolved by user in a timeframe (say per month). This resembles the gadget Created versus Resolved in Jira.

Any hints how to query?

2 answers

1 accepted

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Ah! 

You should try this (MS/SQL):

SELECT ji.assignee, 
       Count(*)                                                       'tickets', 
       (SELECT Count(*) 
        FROM   jiraissue ja 
        WHERE  ja.assignee = ji.assignee 
               AND ja.resolutiondate IS NULL 
               AND ja.created >= CONVERT(DATETIME, '01.11.2014', 104) 
               AND ja.created < CONVERT(DATETIME, '05.11.2014', 104)) 'open', 
       (SELECT Count(*) 
        FROM   jiraissue ja 
        WHERE  ja.assignee = ji.assignee 
               AND ja.resolutiondate IS NOT NULL 
               AND ja.created >= CONVERT(DATETIME, '01.11.2014', 104) 
               AND ja.created < CONVERT(DATETIME, '05.11.2014', 104)) 'resolved',
       CONVERT(DATETIME, '01.11.2014', 104) 'from (inclusive)',
       CONVERT(DATETIME, '05.11.2014', 104) 'to (exclusive)'
FROM   jiraissue ji 
WHERE  ji.assignee IS NOT NULL 
       AND created >= CONVERT(DATETIME, '01.11.2014', 104) 
       AND created < CONVERT(DATETIME, '05.11.2014', 104) 
GROUP  BY ji.assignee 
HAVING Count(*) > 0;

Be sure to replace all from and to dates to your needs.

Cheers, Mario

This widget could not be displayed.

Hi Adolfo, 

I'm not sure if I correctly understand, but are you searching for something like that:

resolved > startOfMonth(-3) AND resolved < startOfMonth()

-3 means -3 month

Cheers, 
Mario 

Not quite, I need to group by user.

If you want that behaviour for each single user logged in you could use "currentUser()" in your query. If you need to overview all users Im not sure how to do it. However this would be the "by logged in user"-Query: resolved > startOfMonth(-3) AND resolved < startOfMonth() AND assignee = currentUser() AND reporter = currentUser() ORDER BY resolved DESC

Thanks, but what I actually need is a SQL query.

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