我在从桌子上检索balance信息时遇到了麻烦。数据集看起来像这样：
| Name | Last Name | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John | Doe | $1600 | 2017-01-01 |
| John | Doe | $12 | 2017-01-02 |
| John | Doe | $1 | 2017-01-03 |
| John | Doe | $16 | 2017-01-04 |
| John | Doe | $16 | 2017-01-05 |
| John | Doe | $16 | 2017-01-06 |
任务是用Balance获取最新的Update Date，但如果相同的Balance在几天内相同，那么在这种情况下我们需要用这个Update Date得到第一个Balance，所以在这种情况下，我们需要以下结果：
| Name | Last Name | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John | Doe | $16 | 2017-01-04 |
我试图使用我的查询：
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
但它显然会回归：
| Name | Last Name | Balance | Update Date |
+---------------+---------------+---------+-------------+
| John | Doe | $16 | 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表，函数或类似内容。只是简单的选择，没什么特别的。
我很感激你的帮助。
您可以通过使用Tabibitosan查找包含最新update_date行的相同余额的行组（整个数据集顶部的行与最新余额之间的差异为0）然后按选择最早的update_date，如下所示：
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：
也许你可以尝试这个查询
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
在第一步中，我们获得之前的余额。在第二个，我们删除前一个余额等于当前余额的所有行。 BTW对不起我从智能手机上回答的布局。
我没有时间写出经过测试的解决方案，但分析函数lead()和lag()的目的是：
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