使用 Oracle SQL 中的 rank() 和子查询从更新日期检索最后余额

Question

我无法从 table 中检索 balance 信息。数据集如下所示：

| Name          | Last Name     | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John          | Doe           | 00   | 2017-01-01  |
| John          | Doe           |      | 2017-01-02  |
| John          | Doe           |       | 2017-01-03  |
| John          | Doe           |      | 2017-01-04  |
| John          | Doe           |      | 2017-01-05  |
| John          | Doe           |      | 2017-01-06  |

任务是用[=17=]得到最近的Balance，但是如果同一个Balance几天都一样，那么在那种情况下我们需要先得到Update Date 与此 Balance，所以在那种情况下，我们需要以下结果：

| Name          | Last Name     | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John          | Doe           |      | 2017-01-04  |

我尝试使用我的查询：

select 
    a.name,
    a.last_name,
    a.balance,
    a.update_date
from
    (select
        name,
        last_name,
        balance,
        update_date,
        rank () over (partition by name, last_name order by update_date desc) top
    from
        customer_balance) a
where
    a.top = 1

但显然 returns:

| Name          | Last Name     | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John          | Doe           |      | 2017-01-06  |

我不确定如何修改它以获得所需的结果。请注意，我的访问权限有限，因此不允许使用临时 table、函数或类似的东西。只是简单的选择，没有什么花哨的。

非常感谢你的帮助。

Answer 1

您可以使用查找具有相同余额的行组，其中包含最新的 update_date 行（整个数据集顶部的行与最新的余额将是 0) 然后一个分组来选择最早的 update_date，像这样：

WITH customer_balance AS (SELECT 'John' first_name, 'Doe' last_name, 1600 balance, to_date('01/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe' last_name, 12 balance, to_date('02/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe' last_name, 1 balance, to_date('03/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe' last_name, 16 balance, to_date('04/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe' last_name, 16 balance, to_date('05/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe' last_name, 16 balance, to_date('06/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 1600 balance, to_date('01/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 12 balance, to_date('02/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 1 balance, to_date('03/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 16 balance, to_date('04/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 15 balance, to_date('05/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 16 balance, to_date('06/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe2' last_name, 16 balance, to_date('07/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 1600 balance, to_date('01/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 12 balance, to_date('02/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 1 balance, to_date('03/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 16 balance, to_date('04/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 16 balance, to_date('05/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 16 balance, to_date('06/01/2017', 'dd/mm/yyyy') update_date FROM dual UNION ALL
                          SELECT 'John' first_name, 'Doe3' last_name, 17 balance, to_date('07/01/2017', 'dd/mm/yyyy') update_date FROM dual)
SELECT first_name,
       last_name,
       balance,
       min(update_date) update_date
FROM   (SELECT first_name,
               last_name,
               balance,
               update_date,
               row_number() OVER (PARTITION BY first_name, last_name ORDER BY update_date DESC) -- row number across the entire dataset (i.e. for each first_name and last_name)
                 - row_number() OVER (PARTITION BY first_name, last_name, balance ORDER BY update_date DESC) grp -- row number across each balance in the entire dataset.
        FROM   customer_balance)
WHERE  grp = 0
GROUP BY first_name,
         last_name,
         balance;

FIRST_NAME LAST_NAME    BALANCE UPDATE_DATE
---------- --------- ---------- -----------
John       Doe               16 04/01/2017
John       Doe2              16 06/01/2017
John       Doe3              17 07/01/2017

我提供了 3 个场景：

最新的行是相同的余额，但该余额没有在数据集（即您的原始数据集）中出现得更早
最新的行是相同的余额，但该余额在数据集中出现得更早
最新一行的余额与前一行不同。

Answer 2

也许你可以试试这个查询

WITH bal AS
  (SELECT 'John' first_name,
                 'Doe' last_name,
                       1600 balance,
                       to_date('20170101', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    12 balance,
                    to_date('20170102', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    1 balance,
                    to_date('20170103', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    16 balance,
                    to_date('20170104', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    16 balance,
                    to_date('20170105', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    16 balance,
                    to_date('20170106', 'YYYYMMDD') update_date
   FROM dual
   UNION ALL SELECT 'John',
                    'Doe',
                    328 balance,
                    to_date('20170107', 'YYYYMMDD') update_date
   FROM dual) -- The main query

SELECT *
FROM
  (SELECT bal.*,
          LAG(balance) OVER(PARTITION BY first_name, last_name
                            ORDER BY update_date)prev_balance
   FROM bal )
WHERE prev_balance IS NULL
  OR balance != prev_balance

第一步我们得到之前的余额。在第二个中，我们删除所有先前余额等于当前余额的行。顺便说一句，抱歉我用智能手机回答的布局。

Answer 3

我没有时间写出经过测试的解决方案，但是分析函数 lead() 和 lag() 是为此设计的：

select name, last_name, balance, update_date
  from (select name,
               last_name,
               balance,
               update_date,
               lead(balance) over (partition by first_name, last_name
                                      order by update_date) 
                 as next_balance
         where balance = :target_balance
         order by update_date
       )
 where balance <> next_balance
   and rownum = 1

使用 Oracle SQL 中的 rank() 和子查询从更新日期检索最后余额

Retrieve last balance from update date using rank() and subquery in Oracle SQL

sql

oracle

rank

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