Question

我有一张这样的表：

+---------+----------+
| post_id | reply_to |
+---------+----------+
|       1 |        0 |
|       2 |        1 |
|       3 |        2 |
|       4 |        2 |
|       5 |        3 |
|       6 |        5 |
|       7 |        1 |
|       8 |        7 |
|       9 |        8 |
|      10 |        7 |
+---------+----------+

reply_to只是正在回复的帖子的ID（即post_id的2是对1 {1}的回复。

这是嵌入表单时的样子：

post_id

如何创建执行以下操作的单个查询：

查询1：获取回复1 2 3 5 6 4 7 8 9 10（post_id = 1）的所有帖子，并将结果数量限制为5
查询2 ：使用查询1 中检索到的值，获取子帖子（2 and 7和3 and 4）并限制数量每个帖子的结果为3
查询3：使用查询2 中检索到的值，获取子帖子（8 and 10和5）并限制每个帖子的结果为

所以最后，结果应该包含这些post_id：9。

以下是我创建的SQL小提琴： http://sqlfiddle.com/#!2/23edc/21

请帮忙！

Answer 1

这适用于SQL Server，我认为它是通用SQL，但我现在无法让SQL Fiddle工作。

create table test1 (post_id int, reply_to int);

insert into test1 (post_id, reply_to) values
(1,0),(2,1),(3,2),(4,2),(5,3),(6,5),(7,1),(8,7),(9,8),(10,7),
(11,2),(12,2),(13,2),(14,3); /* Added records to test conditions */

/* All replies to post_id=1 */
with q1 as (
    select post_id
    from test1
    where reply_to = 1
)
/* Top 3 replies to all results in q1 */
, q2 as (
    select
            q1.post_id as parent_post,
            t1.post_id as child_post,
            count(*) as row_num
    from test1 as t1
        inner join q1 on t1.reply_to = q1.post_id
        left outer join test1 as t2 on t1.reply_to = t2.reply_to 
                and t1.post_id >= t2.post_id
    group by q1.post_id, t1.post_id
    having count(*) <= 3
)
/* Get 0 or 1 grandchild posts */
, q3 as (
    select
            q2.parent_post,
            q2.child_post,
            t1.post_id as grandchild_post,
            count(*) as row_num
    from q2
        left outer join test1 as t1 on q2.child_post = t1.reply_to
        left outer join test1 as t2 on t1.reply_to = t2.reply_to 
                and t1.post_id >= t2.post_id
    group by q2.parent_post, q2.child_post, t1.post_id
    having count(*) = 1
)
/* Aggregate the different post ids */
select distinct parent_post as post_id from q3
union
select distinct child_post from q3
union
select distinct grandchild_post from q3 
        where grandchild_post is not null;

drop table test1;

使用先前查询中的值获得更多结果

1 个答案: