📄 select sno,age=.sql
字号:
use test
select sno,age=
case
when datediff(year,[birth time],getdate())<=22 then '年龄较小'
when datediff(year,[birth time],getdate())<28 and datediff(year,[birth time],getdate())>22 then '年龄适中'
when datediff(year,[birth time],getdate())>=28 then '年龄偏大'
else '不详'
end
from XS
go
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