SQL习题集锦

SQL习题

SQL如何找出同一个ID values中连续小于 60的记录。
比如标蓝色的，中间有一个大于60的就不算连续。2次或者以上就算连续；

table： product_id_value

index_id	product_id	date	id_values	time_count
1	01	2013/1/7	61	1
2	01	2013/1/8	57	2
3	01	2014/5/1	60	3
4	01	2014/5/2	58	4
5	01	2014/5/3	61	5
6	01	2014/5/4	46	6
7	02	2013/1/7	58	1
8	02	2013/1/8	54	2
9	02	2014/5/1	57	3
10	02	2014/5/2	69	4
11	02	2014/5/3	54	5
12	02	2014/5/4	77	6
13	02	2015/8/1	54	7
14	02	2015/8/2	58	8

上图中输出数据为

index	product_id	date	values	time_count
7	02	2013/1/7	58	1
8	02	2013/1/8	54	2
9	02	2014/5/1	57	3
13	02	2015/8/1	54	7
14	02	2015/8/2	58	8

点击查看

--方案一
--注意事项 
--1：自关联好像有字段相同的问题；
--2：需要union处理全部满足条件的数据
with table_a as 
(SELECT
*
from
(
(select *
from product_id_value) a 

left join 
 (select 
index_id as b_index_id,
product_id as b_product_id,
`date` as b_date_id,
id_values as b_id_values,
time_count as b_time_count
from product_id_value) b 
on a.product_id=b.b_product_id and a.time_count=b.b_time_count-1)t
where
id_values<60 and b_id_values<60)

select
index_id ,
product_id ,
`date` ,
id_values ,
time_count 
FROM
table_a 

union 

select
b_index_id,
b_product_id,
b_date_id,
b_id_values,
b_time_count
FROM
table_a

--方案二
--注意事项：
--1：rn2和rn3数值一样，校验函数可以用负值
--2：考虑排名第一位和最后一位的判断问题
SELECT
*
from
(select 
*,
lead(id_values,1,0) over(partition by product_id order by `date`) as rn1,
lag(id_values,1,0) over(partition by product_id order by `date`) as rn2,
lead(id_values,-1,0) over(partition by product_id order by `date`) as rn3
from product_id_value) t
where 
(id_values<60 and rn1<60 and rn1<>0 ) or (id_values<60 and  rn2<60 and rn2<>0)

方案三：
with t1 as (
select
*
#求出差值，因为id一定不会相同，所以使用最熟悉的rank就好
    index_id-rank() over(order by index_id) rk
from stadium
where people < 60
)
select
*
from t1
#where条件过滤出条数大于2的
where rk in (
select rk from t1 group by rk having count(1) >= 2);