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php爱好者> php文档>sql学习6

sql学习6

时间:2010-09-25  来源:为你抒写

 use zhongfei

select * from t_employee select fsubcompany,fdepartment from t_employee group by fsubcompany,fdepartment   select fage,count(*) count from t_employee  group by fage   select fsubcompany,fage,count(*) count from t_employee group by fsubcompany,fage   select fsubcompany,sum(fsalary) salary from t_employee group by fsubcompany   select fdepartment,sum(fsalary) salary from t_employee group by fdepartment     select fage,count(*) count from t_employee group by fage having count(*)>1     select fage,count(*) count from t_employee group by fage having count(*)>1 or count(*)<3     select fage,count(*) count from t_employee group by fage having count(*) in (1,3)   select row_number() over (order by fsalary),fnumber,fname,fsalary,fage from t_employee   select * from ( select row_number() over(order by fsalary desc) as rownum,fnumber,fname,fsalary,fage from t_employee ) as a where a.rownum between 3 and 5 where a.rownum>=3 and a.rownum<=5     select * from t_employee order by fsalary desc select top 3 * from t_employee order by  fsalary desc   select distinct fdepartment from t_employee   select 'nanwang',3222,fname,fage,fsubcompany from t_employee   select fnumber,fname,fage * fsalary from t_employee   select * from t_employee   where fsalary/(fage-15)>500     select '工号为'+fnumber +'的员工姓名为'+fname from t_employee where fname is not null   update t_employee set fage=fage+1
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