How to find the same record in SQL
select * from table group by field having count (field)> 1
Table 1: job number, date, wages
Table 2: job number, name
you want displayed as a result: job number, name, year of wages in January, February, salaries, wages March until December wages
select e.[G_ID],e.[G_XM],
sum(case when month([A_RQ])=1 then [A_DRGZ] else 0 end) [一月],
sum(case when month([A_RQ])=2 then [A_DRGZ] else 0 end) [二月],
sum(case when month([A_RQ])=3 then [A_DRGZ] else 0 end) [三月],
sum(case when month([A_RQ])=4 then [A_DRGZ] else 0 end) [四月],
sum(case when month([A_RQ])=5 then [A_DRGZ] else 0 end) [五月],
sum(case when month([A_RQ])=6 then [A_DRGZ] else 0 end) [六月],
sum(case when month([A_RQ])=7 then [A_DRGZ] else 0 end) [七月],
sum(case when month([A_RQ])=8 then [A_DRGZ] else 0 end) [八月],
sum(case when month([A_RQ])=9 then [A_DRGZ] else 0 end) [九月],
sum(case when month([A_RQ])=10 then [A_DRGZ] else 0 end) [十月],
sum(case when month([A_RQ])=11 then [A_DRGZ] else 0 end) [十一月],
sum(case when month([A_RQ])=12 then [A_DRGZ] else 0 end) [十二月]
from WAGE w
left join WORKER e on e.[G_ID]=w.[A_GH] where (DATEDIFF(yy,A_RQ,GETDATE())=0)
group by e.[G_ID], e.[G_XM]
