SQL语句能计算增长率？

[qq_27797971]

数据是这样的
table1:
id num date
1 50 2018-3-31
2 40 2017-3-31
3 30 2016-3-31

能通过一条SQL语句计算 2018-3-31相对2017-3-31的增长率？

[IEEE_China]

同比增长率=（当年的指标值-去年同期的值）÷去年同期的值*100%

[IEEE_China]

;with cte(Id,num,date) as(
			select 1, 50.00,'2018-03-31' 
union all select 2,40.00,'2017-03-31'			
union all select 3,30.00,'2016-03-31'
)
select a.num as ANum,b.num as BNum,(a.num-b.num)/b.num as Rate  from cte  a, cte  b 
where a.date='2018-03-31' and b.date='2017-03-31'

[二月十六]

--测试数据
if not object_id(N'Tempdb..#table1') is null
	drop table #table1
Go
Create table #table1([id] int,[num] FLOAT,[date] Date)
Insert #table1
select 1,50,'2018-3-31' union all
select 2,40,'2017-3-31' union all
select 3,30,'2016-3-31'
Go
--测试数据结束
SELECT
    a.date,
    CASE
        WHEN b.num IS NOT NULL
            THEN (a.num - b.num) / b.num
        ELSE
            0
    END AS 增长率
FROM
    #table1     a
    LEFT JOIN
        #table1 b
            ON a.id = b.id - 1;

[qq_27797971]

看不太懂，我用的是mysql：
只有一个table，类似下面这种：

table_shouru:
日期 收入 支出
2018/3/31 369 119
2017/12/31 168 507
2017/9/30 124 389
2017/6/30 83 262

计算各个日期的增长率。

[Dear SQL(燊)]

sql server 2016+

--测试数据
if not object_id(N'Tempdb..#table1') is null
    drop table #table1
Go
Create table #table1([id] int,[num] FLOAT,[date] Date)
Insert #table1
select 1,50,'2018-3-31' union all
select 2,40,'2017-3-31' union all
select 3,30,'2016-3-31'
Go
--测试数据结束

select *,增长率=([num]-lead([num])over(order by [date] desc))/lead([num])over(order by [date] desc)*100
from #table1

id          num                    date       增长率
----------- ---------------------- ---------- ----------------------
1           50                     2018-03-31 25
2           40                     2017-03-31 33.3333333333333
3           30                     2016-03-31 NULL