display 3 or more consecutive rows(Sql) - sql

Question

I have a table with below data
+------+------------+-----------+
| id | date1 | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
now what i am trying to do is to display 3 consecutive rows where people were >=100 like this
+------+------------+-----------+
| id | date1 | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
can anyone help me how to do this query using oracle database. I am able to display rows which are above 100 but not in a consecutive way
Table creation(reducing typing time for people who will be helping)
CREATE TABLE stadium
( id int
, date1 date, people int
);
Insert into stadium values (
1,TO_DATE('2017-01-01','YYYY-MM-DD'),10);
Insert into stadium values
(2,TO_DATE('2017-01-02','YYYY-MM-DD'),109);
Insert into stadium values(
3,TO_DATE('2017-01-03','YYYY-MM-DD'),150);
Insert into stadium values(
4,TO_DATE('2017-01-04','YYYY-MM-DD'),99);
Insert into stadium values(
5,TO_DATE('2017-01-05','YYYY-MM-DD'),145);
Insert into stadium values(
6,TO_DATE('2017-01-06','YYYY-MM-DD'),1455);
Insert into stadium values
(7,TO_DATE('2017-01-07','YYYY-MM-DD'),199);
Insert into stadium values(
8,TO_DATE('2017-01-08','YYYY-MM-DD'),188);
Thanks in advance for the help

Answer 1

Assuming you mean >= 100, there are a couple of ways. One method just uses lead() and lag(). But a simple method defines each group >= 100 by the number of values < 100 before it. Then it uses count(*) to find the size of the consecutive values:
select s.*
from (select s.*, count(*) over (partition by grp) as num100pl
from (select s.*,
sum(case when people < 100 then 1 else 0 end) over (order by date) as grp
from stadium s
) s
) s
where num100pl >= 3;
Here is a SQL Fiddle showing that the syntax works.

Answer 2

You can use the following sql script to get the desired output.
WITH partitioned AS (
SELECT *, id - ROW_NUMBER() OVER (ORDER BY id) AS grp
FROM stadium
WHERE people >= 100
),
counted AS (
SELECT *, COUNT(*) OVER (PARTITION BY grp) AS cnt
FROM partitioned
)
select id , visit_date,people
from counted
where cnt>=3

Answer 3

I'm assuming that both the id and date columns are sequential and correspond to each other (there will need to be additional ROW_NUMBER() if the ids are not sequential with the dates, and more complex logic included if the dates are not necessarily sequential).
SELECT
*
FROM
(
SELECT
*
,COUNT(date) OVER (PARTITION BY sequential_group_num) AS num_days_in_sequence
FROM
(
SELECT
*
,(id - ROW_NUMBER() OVER (ORDER BY date)) AS sequential_group_num
FROM
stadium
WHERE
people >= 100
) AS subquery1
) AS subquery2
WHERE
num_days_in_sequence >= 3
That produces the following output:
id date people sequential_group_num num_days_in_sequence
----------- ---------- ----------- -------------------- --------------------
5 2017-01-05 145 2 4
6 2017-01-06 1455 2 4
7 2017-01-07 199 2 4
8 2017-01-08 188 2 4

Answer 4

By using joins we can display the consecutive rows like this
SELECT id, date1, people FROM stadium a WHERE people >= 100
AND (SELECT people FROM stadium b WHERE b.id = a.id + 1) >= 100
AND (SELECT people FROM stadium c WHERE c.id = a.id + 2) >= 100
OR people >= 100
AND (SELECT people FROM stadium e WHERE e.id = a.id - 1) >= 100
AND (SELECT people FROM stadium f WHERE f.id = a.id + 1) >= 100
OR people >= 100
AND (SELECT people FROM stadium g WHERE g.id = a.id - 1) >= 100
AND (SELECT people FROM stadium h WHERE h.id = a.id - 2) >= 100
order by id;

Answer 5

select distinct
t1.*
from
stadium t1
join
stadium t2
join
stadium t3
where
t1.people >= 100
and t2.people >= 100
and t3.people >= 100
and
(
(t1.id + 1 = t2.id
and t2.id + 1 = t3.id)
or
(
t2.id + 1 = t1.id
and t1.id + 1 = t3.id
)
or
(
t2.id + 1 = t3.id
and t3.id + 1 = t1.id
)
)
order by
id;

Answer 6

SQL script:
SELECT DISTINCT SS.*
FROM STADIUM SS
INNER JOIN
(SELECT S1.ID
FROM STADIUM S1
WHERE 3 = (
SELECT COUNT(1)
FROM STADIUM S2
WHERE (S2.ID=S1.ID OR S2.ID=S1.ID+1 OR S2.ID=S1.ID+2)
AND S2.PEOPLE >= 100
)) AS SS2
ON SS.ID>=SS2.ID AND SS.ID<SS2.ID+3

Answer 7

select *
from(
select * , count(*) over (partition by grp) as total
from
(select * , Sum(case when people < 100 then 1 else 0 end) over (order by date) as grp
from stadium) T -- inner Query 1
where people >=100 )S--inner query 2
where total >=3 --outer query

Answer 8

I wrote the following solution for this similar leetcode problem:
with groupVisitsOver100 as (
select *,
sum(
case
when people < 100 then 1
else 0
end
) over (order by date1) as visitGroups
from stadium
),
filterUnder100 as (
select
*
from groupVisitsOver100
where people >= 100
),
countGroupsSize as (
select
*,
count(*) over (partition by visitGroups) as groupsSize
from filterUnder100
)
select id, date1, people from countGroupsSize where groupsSize >= 3 order by date1