MySQL自连接以仅查找下一个修订版
实际上我在和Wordpress合作。我想创建一个自连接或类似的东西来查找一篇文章的修订版,以及同一篇文章的后续修订版MySQL自连接以仅查找下一个修订版,mysql,sql,self-join,Mysql,Sql,Self Join,实际上我在和Wordpress合作。我想创建一个自连接或类似的东西来查找一篇文章的修订版,以及同一篇文章的后续修订版 create table wp_posts (post_id int, revision_id int); INSERT INTO wp_posts(post_id, revision_id) VALUES (1, 1); INSERT INTO wp_posts(post_id, revision_id) VALUES (1, 2); INSERT INTO wp_posts(
create table wp_posts (post_id int, revision_id int);
INSERT INTO wp_posts(post_id, revision_id) VALUES (1, 1);
INSERT INTO wp_posts(post_id, revision_id) VALUES (1, 2);
INSERT INTO wp_posts(post_id, revision_id) VALUES (1, 3);
INSERT INTO wp_posts(post_id, revision_id) VALUES (2, 11);
INSERT INTO wp_posts(post_id, revision_id) VALUES (2, 12);
INSERT INTO wp_posts(post_id, revision_id) VALUES (2, 13);
SELECT a.post_id, a.revision_id "PreviousRevision", b.revision_id "FollowingRevision"
FROM `wp_posts` a
JOIN `wp_posts` b
ON a.post_id = b.post_id #the id of every revision of a post is different but the post_id is the same
WHERE a.revision_id < b.revision_id
AND a.revision_id != b.revision_id
上一个查询不起作用,因为它需要对a的每个记录进行所有修改,而不仅仅是下一个修改
这就是我得到的,我已经删除了我不想要的线条。我只需要父子行
如何只获取一个元素?在MySQL 8+中,您将使用窗口函数:
SELECT p.post_id, p.revision_id,
lag(p.revision_id) over (partition by p.post_id order by p.revision_id) as prev_revision_id,
lead(p.revision_id) over (partition by p.post_id order by p.revision_id) as next_revision_id
FROM `wp_posts` p;
在早期版本中,我将使用相关子查询:
select p.post_id, p.revision_id,
(select max(p2.revision_id)
from wp_posts p2
where p2.post_id = p.post_id and p2.revision_id < p.revision_id
) as prev_revision_id,
(select min(p2.revision_id)
from wp_posts p2
where p2.post_id = p.post_id and p2.revision_id > p.revision_id
) as next_revision_id
from wp_posts p;
对于此示例数据,您需要a.post\u id和b.revision\u id的每个组合的最大a.revision\u id:
我不会将@forpas中的查询与group by一起使用,因为我不喜欢查询的explain看起来像是临时+文件排序 在这种情况下,我通常会这样做:
SELECT
a.post_id
, a.revision_id "PrevRevision"
, b.revision_id "NextRevision"
FROM
`wp_posts` AS a
INNER JOIN `wp_posts` AS b ON (
b.post_id = a.post_id
AND b.revision_id > a.revision_id
)
LEFT JOIN `wp_posts` AS c ON (
c.post_id = a.post_id
AND c.revision_id > a.revision_id
AND c.revision_id < b.revision_id
)
WHERE
c.revision_id IS NULL
在某些数据集上,@Gordon Linoff建议的查询子查询会更快。谢谢,我无法理解为什么group by有效。。你能帮我吗?在这方面:你可以不分组就看到结果。正如您所看到的,对于a.post_id、b.revision_id的每个组合,您希望在结果中显示a.revision_id的最大值。这是GROUP BY和MAX所做的。非常聪明的解决方案我不能同时接受这两个答案,但这也是非常重要的great@Revious . . . 使用窗口函数的版本显然性能更好。我希望相关子查询也比聚合解决方案快。
| post_id | PreviousRevision | FollowingRevision |
| ------- | ---------------- | ----------------- |
| 1 | 1 | 2 |
| 1 | 2 | 3 |
| 2 | 11 | 12 |
| 2 | 12 | 13 |
SELECT
a.post_id
, a.revision_id "PrevRevision"
, b.revision_id "NextRevision"
FROM
`wp_posts` AS a
INNER JOIN `wp_posts` AS b ON (
b.post_id = a.post_id
AND b.revision_id > a.revision_id
)
LEFT JOIN `wp_posts` AS c ON (
c.post_id = a.post_id
AND c.revision_id > a.revision_id
AND c.revision_id < b.revision_id
)
WHERE
c.revision_id IS NULL
id select_type table partitions type possible_keys key key_len ref rows filtered Extra
1 SIMPLE a index IX_wp_post_idx IX_wp_post_idx 10 6 100.00 Using where; Using index
1 SIMPLE b ref IX_wp_post_idx IX_wp_post_idx 5 test.a.post_id 4 33.33 Using where; Using index
1 SIMPLE c ref IX_wp_post_idx IX_wp_post_idx 5 test.a.post_id 4 16.67 Using where; Using index