- Increase transparency
- Onboard new hires faster
- Access from mobile/offline
SELECT DISTINCT j.name, max(r.attempt) AS highest, j.active, j.jobID, r.lev FROM (journals j LEFT JOIN responses r ON r.jobID=j.jobID) WHERE j.contactID = 220 GROUP BY r.jobID
Join the community of 500,000 technology professionals and ask your questions.
Connect with top rated Experts
8 Experts available now in Live!