假设我有这张桌子:
Price | OrderDate | OrderID
--------+-------------+-----
5.50000 | 2017-11-02 | 77319
5.30000 | 2017-11-02 | 77320
5.50000 | 2017-11-09 | 77463
5.50000 | 2017-11-16 | 77633
5.50000 | 2017-11-23 | 77839
5.25000 | 2017-11-23 | 77840
5.35000 | 2017-11-30 | 78012
5.50000 | 2017-12-07 | 78138
5.50000 | 2017-12-14 | 78283
我需要得到这个结果
Price | OrderDate | OrderID
--------+-------------+--------
5.50000 2017-11-02 77319
5.30000 2017-11-02 77320
5.50000 2017-11-09 77463
5.25000 2017-11-23 77840
5.35000 2017-11-30 78012
5.50000 2017-12-07 78138
以粗体显示的值,我需要分组并且仅获得1行。表的顺序应类似于结果。
我不知道如何执行此操作。
有什么想法吗?
谢谢!
答案 0 :(得分:2)
假设您可以使用lag()
:
with data as (
select *, lag(Price) over (order by OrderId) as lastPrice
from T
)
select *
from data
where coalesce(Price, -1) <> lastPrice;
否则,假设您可以使用cross apply
:
select t.*
from T t cross apply (
select max(OrderId) priorOrderId from T t2 where t2.OrderId < t.OrderId
) left outer join T t3 on t3.OrderId = t2.priorOrderId
where coalesce(t3.Price, -1) <> t.Price;
否则仍然可以重写:
with data as (
select *, (select max(OrderID from T t2 where t2.OrderId < t.OrderId) as priorOrderId
from T t
)
select d.*
from data d left outer join T t on t.OrderId = d.priorOrderId
where coalesce(t.Price, -1) <> d.Price;
答案 1 :(得分:1)
您可以使用lag
窗口函数获取上一行的价格,并忽略价格不变的行:
SELECT price, orderid, orderdate
FROM (SELECT price, orderid, orderdate,
LAG(price) OVER(ORDER BY orderid ASC) AS prev_price
FROM mytable)
WHERE price <> prev_price OR prev_price IS NULL
ORDER BY orderid ASC
答案 2 :(得分:1)
答案 3 :(得分:0)
这不是有效的方法,但可能易于理解,在现实生活中,您可能不会发现它效率低下。
SELECT
Price
,OrderDate
,OrderID
FROM TableName
WHERE Price <> ISNULL((SELECT TOP 1 Price FROM TableName T2 WHERE T2.OrderID < TableName.OrderID ORDER BY OrderID DESC), -1)
胆量在WHERE子句中。对于每条记录,它都基于[OrderID]查找前一条记录的价格。如果它们不同,则包括它们。但是,您必须包括ISNULL,因为第一条记录没有上一条记录。