sql查找模式并计算它出现的次数

Question

数据提供

22
22
22
22
22
36
54
40
22
22
22
22
36
22
22
54
22
22

这是表格中的列。使用sql查询我们需要找出诸如的模式 22 36 54 40是第一个模式，然后22 36是第二个，22 54是第三个模式。

Answer 1

您应该使用LEAD获取下一行的值以查看它是否为22，并使用它来删除列中的所有额外22。有这样的效果：

declare @t table (id int identity(1,1) not null, n int)
insert into @t 
select 22 union all 
select 22 union all 
select 22 union all 
select 22 union all 
select 22 union all 
select 36 union all 
select 54 union all 
select 40 union all 
select 22 union all 
select 22 union all 
select 22 union all 
select 22 union all 
select 36 union all 
select 22 union all 
select 22 union all 
select 54 union all 
select 22 union all 
select 22

select id,n from (select id,n ,lead(n) over (order by id) 
as lead_val from @t ) t where n<>22 or lead_val<>22

输出：

5   22
6   36
7   54
8   40
12  22
13  36
15  22
16  54

Answer 2

PostgreSQL

假设：

• 有一列确定元素的顺序
• 所有模式均以22
开头

select      array_to_string(array_agg(val order by i),',')  as pattern
           ,min  (i)                                        as from_i
           ,max  (i)                                        as to_i
           ,count(*)                                        as pattern_length           

from       (select  i,val
                  ,count(case when val = 22 then 1 end) over
                   (
                        order by i
                        rows unbounded preceding
                    ) as pattern_id

            from    mytable
            ) t

 group by   pattern_id

 having     count(*)>1
;

+-------------+--------+------+----------------+
|   pattern   | from_i | to_i | pattern_length |
+-------------+--------+------+----------------+
| 22,36,54,40 |      5 |    8 |              4 |
| 22,36       |     12 |   13 |              2 |
| 22,54       |     15 |   16 |              2 |
+-------------+--------+------+----------------+