Max_Leb

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Jan 11th, 2023
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SQL 1.52 KB | None | 0 0
  1.  
  2. SELECT s.FullName, i.Title, i.CreatedAt AS date_made, s.Resolved, key_value FROM "Assignment" a      
  3. JOIN Specialists s ON s.ID = a.SpecialistID
  4. JOIN Incidents i ON a.IncidentID = i.ID
  5. JOIN (SELECT FullName, i.Title, i.CreatedAt AS date_made, s.Resolved,
  6. SUM(s.Resolved) OVER (PARTITION BY s.FullName ORDER BY i.Title) AS key_value    
  7. FROM "Assignment" a
  8. JOIN Specialists s ON s.ID = a.SpecialistID
  9. JOIN Incidents i ON a.IncidentID = i.ID) AS tmp
  10. ON s.FullName = tmp.FullName AND i.Title = tmp.Title AND key_value NOT IN (SELECT * FROM (SELECT key_value FROM (SELECT FullName, i.Title, i.CreatedAt AS date_made, s.Resolved,
  11. SUM(s.Resolved) OVER (PARTITION BY s.FullName ORDER BY i.Title) AS key_value    
  12. FROM "Assignment" a
  13. JOIN Specialists s ON s.ID = a.SpecialistID
  14. JOIN Incidents i ON a.IncidentID = i.ID) WHERE key_value != 0) GROUP BY key_value HAVING key_value <= (SELECT key_value FROM (SELECT FullName, i.Title, i.CreatedAt AS date_made, s.Resolved,
  15. SUM(s.Resolved) OVER (PARTITION BY s.FullName ORDER BY i.Title) AS key_value    
  16. FROM "Assignment" a
  17. JOIN Specialists s ON s.ID = a.SpecialistID
  18. JOIN Incidents i ON a.IncidentID = i.ID) WHERE key_value != 0 ORDER BY key_value ASC LIMIT 1) * (SELECT key_value FROM (SELECT FullName, i.Title, i.CreatedAt AS date_made, s.Resolved,
  19. SUM(s.Resolved) OVER (PARTITION BY s.FullName ORDER BY i.Title) AS key_value    
  20. FROM "Assignment" a
  21. JOIN Specialists s ON s.ID = a.SpecialistID
  22. JOIN Incidents i ON a.IncidentID = i.ID) WHERE key_value != 0 ORDER BY key_value ASC LIMIT 1));
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