将列值与 Oracle 中的先前记录进行比较答案

【问题标题】：Compare column value with previous record in Oracle将列值与 Oracle 中的先前记录进行比较
【发布时间】：2021-11-11 23:18:24
【问题描述】：

我的oracle表数据如下。

Org_ID	Product_ID	Order_Month	Amount
101	201	JAN-2021	2000
101	201	FEB-2021	2000
101	201	MAR-2021	2000
101	201	APR-2021	1500
101	201	MAY-2021	2000
101	202	JUN-2021	2000
101	202	JUL-2021	2000

我们需要比较之前的金额值，并找到金额不匹配的记录以及关于Product_ID的记录。

我的输出应该如下所示。尝试使用滞后但找不到解决方案。有人可以提供有关如何解决此问题的意见吗？

Org_ID	Product_ID	Order_Month	Amount
101	201	JAN-2021 to MAR-2021	2000
101	201	APR-2021 to APR-2021	1500
101	201	MAY-2021 to MAY-2021	1500
101	202	JUN-2021 to JUL-2021	1500

【问题讨论】：

MAY-2021 到 JUL-2021 的金额是 1500 而不是 2000 有什么原因吗？
在您的示例数据产品中，202 从未有过 1500 的数量，因此不清楚 gow 计算该值

标签： sql oracle

【解决方案1】：

你可以试试下面的

SELECT
    "Org_ID", 
    "Product_ID", 
    CONCAT(
         CONCAT(Order_Month_Group,' to '),
        TO_CHAR(MAX(actual_date),'MON-YYYY')
    ) as Order_Month, 
    "Amount"
FROM (
    SELECT
        t1.*,
        
        LAG(
             "Order_Month",
             CASE WHEN continued=0 THEN 0 ELSE seq_num-1 END
            ,"Order_Month") OVER (
                PARTITION BY "Org_ID","Product_ID","Amount"
                ORDER BY actual_date
        ) as Order_Month_Group
    FROM (
        SELECT 
            t.*,
            TO_DATE(t."Order_Month",'MON-YYYY') as actual_date,
            ROW_NUMBER() OVER (
                PARTITION BY t."Org_ID",t."Product_ID",t."Amount"
                ORDER BY TO_DATE("Order_Month",'MON-YYYY')
            ) as seq_num,
            CASE 
                WHEN t."Amount" = LAG(t."Amount",1,t."Amount") OVER (
                                      PARTITION BY t."Org_ID",t."Product_ID"
                                      ORDER BY TO_DATE("Order_Month",'MON-YYYY')
                                  ) THEN 1 
                ELSE 0 
            END as continued
        FROM 
            my_oracle_table t
    ) t1
) t2
GROUP BY "Org_ID", "Product_ID", Order_Month_Group, "Amount"
ORDER BY MIN(actual_date)

或

SELECT
    "Org_ID", 
    "Product_ID", 
    CONCAT(
         CONCAT(TO_CHAR(MIN(actual_date),'MON-YYYY'),' to '),
        TO_CHAR(MAX(actual_date),'MON-YYYY')
    ) as Order_Month, 
    "Amount"
FROM (
    SELECT
        t1.*,
        SUM(continued) OVER ( ORDER BY actual_date) as grp
    FROM (
        SELECT 
            t.*,
            TO_DATE("Order_Month",'MON-YYYY') as actual_date,
            CASE 
                WHEN t."Amount" = LAG(t."Amount",1,t."Amount") OVER (
                                      PARTITION BY t."Org_ID",t."Product_ID"
                                      ORDER BY TO_DATE("Order_Month",'MON-YYYY')
                                  ) THEN 0
                ELSE 1 
            END as continued
        FROM 
            my_oracle_table t
    ) t1
) t2
GROUP BY "Org_ID", "Product_ID", grp, "Amount"
ORDER BY MIN(actual_date)

View Demo on DB Fiddle

让我知道这是否适合你。

【讨论】：

【解决方案2】：

这是一种孤岛问题。在这种情况下，最简单的解决方案可能是行号的差异。以下假设order_month 实际上是一个字符串（而不是日期）：

select org_id, product_id, amount,
       min(order_month), max(order_month)
from (select t.*,
             row_number() over (partition by org_id, product_id order by to_date(order_month, 'MON-YYYY')) as seqnum,
             row_number() over (partition by org_id, product_id, amount order by to_date(order_month, 'MON-YYYY')) as seqnum_2
      from t
     ) t
group by org_id, product_id, amount, (seqnum - seqnm_2);

为什么这行得通有点难以解释。但是，如果您查看子查询的结果，您会发现这两个值之间的差异在相邻月份是如何保持不变的。

【讨论】：