显示 3 个或更多连续行（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

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 id s 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 3

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 4

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 5

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 6

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 7

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 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