我有两个表,第二个表通过多个条目(1:n
)连接到第一个表。
第一个表具有id_something
(另一个表,但目前不重要)和每天date_day
的条目。
+-----------+
| Table ONE |
+----+------+-------+----------+----------+
| id | id_something | date_day | created |
+----+--------------+----------+----------+
| 1 | 666 | 2019-1-1 | 2019-1-1 |
| 2 | 666 | 2019-1-1 | 2019-7-7 |
| 3 | 123 | 2019-1-1 | 2019-1-1 |
+----+--------------+----------+----------+
第二个表与此id
连接,并包含键值对。
+-----------+
| Table TWO |
+--------+--+--+-----+
| id_one | foo | bar |
+--------+-----+-----+
| 1 | 1 | 20 |
| 1 | 2 | 21 |
| 2 | 1 | 30 |
| 2 | 2 | 31 |
| 2 | 3 | 32 |
| 3 | 1 | 10 |
+--------+-----+-----+
我想查询所有可能的连接,很简单,这是一个JOIN:
SELECT *
FROM one
LEFT JOIN two
ON two.id_one = one.id;
+----+--------------+----------+----------+--------+-----+-----+
| id | id_something | date_day | created | id_one | foo | bar |
+----+--------------+----------+----------+--------+-----+-----+
| 1 | 666 | 2019-1-1 | 2019-1-1 | 1 | 1 | 20 |
| 1 | 666 | 2019-1-1 | 2019-1-1 | 1 | 2 | 21 |
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 1 | 30 |
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 2 | 31 |
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 3 | 32 |
| 3 | 123 | 2019-1-1 | 2019-1-1 | 3 | 1 | 10 |
+----+--------------+----------+----------+--------+-----+-----+
现在,如您所见,我还有一个created
字段。 id_something
与date_day
可能是相同的-但我只想与第二张表具有最新的(created DESC
)对。
因此,在这种情况下,查询应返回:
+----+--------------+----------+----------+--------+-----+-----+
| id | id_something | date_day | created | id_one | foo | bar |
+----+--------------+----------+----------+--------+-----+-----+
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 1 | 30 |
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 2 | 31 |
| 2 | 666 | 2019-1-1 | 2019-7-7 | 2 | 3 | 32 |
| 3 | 123 | 2019-1-1 | 2019-1-1 | 3 | 1 | 10 |
+----+--------------+----------+----------+--------+-----+-----+
但是我无法使其工作...我试图使用DISTINCT甚至子查询和case构造,但是它要么不起作用,要么非常。分组依据也不会返回每个加入的对,而对于表一中的每个id
仅返回一行。
要实现理想的结果,我缺少什么?
(如果不使用特定于Oracle的synthax,那将是一个奖励。)
答案 0 :(得分:1)
您可以使用分析功能:
SELECT *
FROM (SELECT o.* ROW_NUMBER() OVER (PARTITION BY id ORDER BY created DESC) as seqnum
FROM one o
) o LEFT JOIN
two t
ON t.id_one = o.id
WHERE o.seqnum = 1;
答案 1 :(得分:1)
使用row_number()
获取最新的id_something,然后使用加入
select a.*,b.* from
(
select *,
row_number()over(partition by id_something order by created desc) rn from one
) a join two b ON b.id_one = a.id;
where rn=1