Question

我们说我们有下一个数据

id |    date    | price
------------------------
1  | 10-09-2016 |   200    
2  | 11-09-2016 |   190    
3  | 12-09-2016 |   210    
4  | 13-09-2016 |   220    
5  | 14-09-2016 |   200    
6  | 15-09-2016 |   200    
7  | 16-09-2016 |   230    
8  | 17-09-2016 |   240

我们必须先按日期订购，然后按价格订购，但是如果价格必须按顺序排列。如果当前价格低于先前价格，我们应该省略此行，结果将是：

id |    date    | price
------------------------
1  | 10-09-2016 |   200
3  | 12-09-2016 |   210
4  | 13-09-2016 |   220
7  | 16-09-2016 |   230
8  | 17-09-2016 |   240

没有加入是否可能？

Answer 1

使用LAG窗口功能

SELECT *
FROM   (SELECT *,
               Lag(price)OVER( ORDER BY date) AS prev_price
        FROM   Yourtable) a
WHERE  price > prev_price
        OR prev_price IS NULL -- to get the first record

Answer 2

如果“previous”应该表示输出中的上一行，则跟踪运行的最大值。在子查询中带有window function的Postgres解决方案：

SELECT id, date, price
FROM  (
   SELECT *, price >= max(price) OVER (ORDER BY date, price) AS ok
   FROM   tbl
   ) sub
WHERE  ok;

Answer 3

如果Postgres：

select id, date, price
from
(select
    t.*,
    price - lag(price, 1, price) over (order by id) diff
from
    your_table) t
where diff > 0;

如果是MySQL：

select id, date, price from
(
    select t.*, 
        price - @lastprice diff,
        @lastprice := price
    from
    (select *
    from your_table
    order by id) t
    cross join (select @lastprice := 0) t2
) t where t.diff > 0;

SQL列顺序为两列，省略第二列是否符合订单

3 个答案: