ИЗБИРАНЕ на денормализирани колони в отделни записи?

само ако имате по-малко от 10 000 имейла.... приемливо ли е това?

select 
       if(t1.c > 1, concat(e.employeename, ' (', e.employeeid, ')'), e.employeename) as Employee,
       replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') EmailAddress 
from 
       (select employeename, count(*) as c from Employees group by employeename) as t1, 
       (select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2,
       (SELECT @row := @row + 1 as row FROM 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x,
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x2, 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x3, 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x4, 
       (SELECT @row:=0) as ff) as n,
       Employees e
where 
      e.employeename = t1.employeename and
      e.employeeid = t2.employeeid and
      n.row <= t2.emails
order by e.employeeid;

РЕДАКТИРАНЕ:

С генерирани по-малко безполезни числа:

select 
       if(t1.c > 1, concat(e.EmployeeName, ' (', e.EmployeeID, ')'), e.EmployeeName) as Employee,
       replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') as EmailAddress 
from 
       (select EmployeeName, count(*) as c from Employees group by EmployeeName) as t1, 
       (select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2,
       (select `1` as row from (select 1 union all select 2 union all select 3 union all select 4) x) as n,
       Employees e
where 
      e.EmployeeName = t1.EmployeeName and
      e.EmployeeID = t2.EmployeeID and
      n.row <= t2.emails
order by e.EmployeeID;

И какво научихме? Лошият дизайн на базата данни води до ужасни запитвания. И можете да правите неща с SQL, които вероятно се поддържат само защото хората правят лош дизайн на база данни... :)