订单选择

Question

CustID Sequencing Total
2 1 10
4 2 20
3 2 30
4 4 40
5 8 50

我希望选择订单金额> gt = =订单金额的客户，如何实现？

Answer 1

对于2005,2008版本：

declare @t table(cust int, ord_s int, ord_a int);
insert @t values
(1,1,30), (1,3,40), (2,1,40), (2,6,30), (3,4,15),
(3,7,60),(3,8,50),(4,2,32),(4,5,2),(4,7,4);

with cte as
(
select row_number() over(partition by cust order by ord_s) as rn, *
from @t
)

select c1.cust,
        c1.ord_s,
        c1.ord_a,
        c2.ord_a as prv
from cte as c1 join cte as c2
         on c1.cust = c2.cust and  c1.rn = c2.rn + 1
where c1.ord_a > c2.ord_a;

听到的想法总是一样的，你应该找到以前的订单，如果你不能使用滞后，你在每个客户中列举订单并加入如上所述的相邻订单，如果你不能使用甚至row_number，你可以搜索最近的值当前这样的人不如：

select tt1.*
from 
(
select *,
       (select max(t2.ord_s) from @t t2 where t2.cust = t1.cust and t2.ord_s < t1.ord_s) as prv
from @t t1
) tt1 
join @t tt2
    on tt1.cust = tt2.cust and tt1.prv = tt2.ord_s
where tt1.ord_a > tt2.ord_a;

Answer 2

使用ANSI标准Array ( [0] => Some Stuff [1] => Some more stuff [2] => Some other stuff ) 功能，您可以选择值增加的所有场合：

lag()

编辑：

对于修订后的问题，您可以使用select t.* from (select t.*, lag(amount) over (partition by customer order by order_sequence) as prev_amount from t ) t where amount > prev_amount; 。但是，我并不是100％明确的确切逻辑：

first_value()

Answer 3

DECLARE @Temp table(cust int, ord_seq int, ord_amt int);
insert @Temp values
(1,1,30), (1,3,40), (2,1,40), (2,6,30), (3,4,15),
(3,7,60),(3,8,50),(4,2,32),(4,5,2),(4,7,4);

SELECT 
     Seq AS SeqNo
    ,cust AS CustomerID
    ,ord_seq AS OrderSequence
    ,ord_amt AS OrderAmount
FROM (
    SELECT *,CASE WHEN LAGord_amt IS NULL THEN 0
            ELSE 1
            END As  LAGord_amt1
        ,ROW_NUMBER() OVER (ORDER BY (SELECT 1)) AS Seq
        ,CASE 
            WHEN ord_amt > LAGord_amt
                THEN ord_amt
            ELSE 0
            END Lagval
    FROM (
        SELECT *,LAG(ord_amt) OVER (PARTITION BY cust ORDER BY ord_seq) As LAGord_amt
        FROM @Temp
        ) Dt
    ) Dt2
WHERE Dt2.LAGord_amt1 = 1
    AND Lagval <> 0

输出

    SeqNo   CustomerID  OrderSequence   OrderAmount
    ---------------------------------------------
     2          1           3               40
     6          3           7               60
     10         4           7               4