Question

我有这张桌子：

+------+-------+---------+----------------------------+
| id   | date  | value   |  value2 | value3           |
+------+-------+---------+----------------------------+
| 1    | 5/28  | 100     | NULL    | NULL             |
| 1    | 5/27  | 200     | NULL    | NULL             |
| 1    | 5/26  | 300     | 10      | NULL             |
| 1    | 5/25  | NULL    | NULL    | NULL             |
| 1    | 5/24  | ...     | 20      | NULL             |
| 1    | 5/23  | ...     | ....    | 50               |
| 1    | 5/22  | ...     | ....    | 30               |
| 1    | 5/21  | ...     | ....    | NULL             |
| 1    | 5/20  | ...     | ....    | ...              |
| 1    | 5/19  | ...     | ....    | ...              |
| 1    | 5/18  | ...     | ....    | ...              |
| 1    | 5/17  | ...     | ....    | ...              |
+------+-------+--------------------------------------+

我需要查询每列的最新可能值。因此，如果在value3不为一天的那一天value2为NULL，它将获取先前的记录，其中value2不为null。我不确定这是否有意义，但这是我想要的结果表：

+------+-------+-------+--------+----------+
| id   | date  | value | value2 |  value3  |
+------+-------+-------+--------+----------+
| 1    | 5/28  | 100   | 10     | 50       |
+------+-------+---------------------------+

日期将始终为最新日期（在此示例中为5/28），对于其余日期，我需要获取每列的最新值，即使它与日期不匹配。

这甚至有可能吗？

Answer 1

这里是一个选择：

SQL> alter session set nls_date_format = 'dd.mm.yyyy';

Session altered.

SQL> with test (datum, value1, value2, value3) as
  2    (select date '2018-05-28', 100, null, null from dual union
  3     select date '2018-05-27', 200, null, null from dual union
  4     select date '2018-05-26', 300, 10, null   from dual union
  5     select date '2018-05-23', 500, null, 50   from dual
  6    ),
  7  inter as
  8    (select
  9      first_value(datum  ignore nulls) over (order by datum desc) datum,
 10      first_value(value1 ignore nulls) over (order by datum desc) value1,
 11      first_value(value2 ignore nulls) over (order by datum desc) value2,
 12      first_value(value3 ignore nulls) over (order by datum desc) value3
 13    from test
 14    order by 1 desc nulls last, 2 desc nulls last, 3 desc nulls last, 4 desc nulls last
 15   )
 16  select *
 17  from inter
 18  where rownum = 1;

DATUM          VALUE1     VALUE2     VALUE3
---------- ---------- ---------- ----------
28.05.2018        100         10         50

SQL>

Answer 2

last_value窗口函数也可以用于此。

示例：

create table test_table(id int, "date" date, value int, value2 int, value3 int);

insert into test_table (id, "date", value, value2, value3)
select 1, date '2018-05-28', 100, null, null from dual union
select 1, date '2018-05-27', 200, null, null from dual union
select 1, date '2018-05-26', 300, 10, null   from dual union
select 1, date '2018-05-23', 500, null, 50   from dual;

insert into test_table (id, "date", value, value2, value3)
select 2, "date" + 1, value+100, value2+10, value3+10 from test_table;

select
 id,
 max("date") as "date",
 max(lastValue1) as value,
 max(lastValue2) as value2,
 max(lastValue3) as value3
from
(
   select id, "date",
    last_value(value ignore nulls) over (partition by id order by "date" ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as lastValue1,
    last_value(value2 ignore nulls) over (partition by id order by "date" ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as lastValue2,
    last_value(value3 ignore nulls) over (partition by id order by "date" ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as lastValue3
   from test_table
) q
group by id;

ID | date      | VALUE | VALUE2 | VALUE3
-- | --------- | ----- | ------ | ------
 1 | 28-MAY-18 |   100 |     10 |     50
 2 | 29-MAY-18 |   200 |     20 |     60

db <>提琴here

使用ORACLE SQL查询不同行中的数据

2 个答案: