Question

考虑一个表

employee id    report_year  report_quarter  sequencenumber quarter1_wage quarter2_wage 
101                2015       1              1              1000           0
101                2015       1              2              2000           0
102                2016       2              1              3000           0
102                2016       2              2               0            4000

查询的结果必须是

      Total wages

         6000

与2015年一样，最高序列号为2的员工ID 101在第1季度以下为2000，因为报告_quarter表示1需要添加取此值并添加到

2016年最高序号为2的员工ID 102有4000（quarter2_wage）报告_quarter表示它是2

Answer 1

SELECT SUM(quarter1_wage )+SUM(quarter2_wage) AS total
,MAX(sequencenumber)
,report_year
FROM table 
GROUP BY report_year

Answer 2

与Ties和Order by一起使用，只获得EmployeeId的最高序列号，然后得到派生表的总和

 Select SUM(quarter1_wage )+SUM(quarter2_wage) AS total from (
    Select Top 1 with Ties * from table 
       ORDER BY row_number() over (partition by employee_id order by sequencenumber desc)) as DerivedTable

根据最近更新的序号计算所有员工的总工资

2 个答案: