Question

我在使用数据透视查询（SQL Server）时遇到了一些问题。任务非常简单：对于一个人，我必须收集一年中每个月的收入统计数据，但每个新月收入都基于previuos income加current month income

仅举例。让一个人每月有3k的工资（为简单起见它是常数），那么查询结果应该是这样的：

Year | Jan | Feb | ... | Dec
2016 | 3k  | 6k  | ... | 36k
2015 | 3k  | 6k  | ... | 36k
...

伪SQL查询是：

select * from (
    select 
        year(date) as year,
        month(date) as month
        salary,
    from income
    where personId = 'some id'
) as tmp
pivot (
    sum(salary),
    for month in ([1], [1..2], [1..3], ...)
) as pvt

问题是SQL中没有[1..2]表达式。使用标准SQL执行此类查询的方法是什么？

Answer 1

也许是这样的？（此OVER适用于2008 R2版本及之后的版本）

create table #income (
    personid int,
    salary int,
    [date] date
)

insert into #income 
(personid,salary,[date])
values
(1,3000,'2016-01-31'),
(1,3000,'2016-02-29'),
(1,3000,'2016-03-31'),
(1,3000,'2016-04-30'),
(1,3000,'2016-05-31');

select * from (
    select 
        year(date) as year,
        month(date) as month,
        SUM(salary) OVER ( PARTITION BY personid ORDER BY [date]) salary
    from income
    where personId = 1
) as tmp
pivot (
    sum(salary)
    for month in ([1], [2], [3],[4],[5])
) as pvt;

drop table #income;

具有复杂聚合的{s}数据透视查询

1 个答案: