CODEWITHSUNDEEP
mysqlsqlgroup-bymysql-8.0
S. Berg
S. Berg

Reputation: 23

Select consecutive rows under a certain value in MySQL

Selecting consecutive rows with a column under a certain value

I have a table with the following data:

crashID     crash
-----------------------
1           189
2           144
3           8939        
4           748
5           988
6           102
7           392
8           482
9           185
10          101

I want to select the longest streak of consecutive rows that also have a crash value below a certain treshold. Let's say 500 for this example.

How do I go about doing this in a single MySQL query? (v8.0.1)

Desired output would be this:

crashID     crash
------------------
6           102
7           392
8           482
9           185
10          101

Upvotes: 2

Views: 1818

Answers (2)

Salman Arshad
Salman Arshad

Reputation: 272006

You can try to solve it using gaps and islands approach, assume every crash lte 500 is an island then find the largest island:

SET @threshold = 500;
WITH cte1 AS (
    SELECT
        crashID,
        CASE WHEN crash <= @threshold THEN 1 ELSE 0 END AS island,
        ROW_NUMBER() OVER (ORDER BY crashID) rn1,
        ROW_NUMBER() OVER (PARTITION BY CASE WHEN crash <= @threshold THEN 1 ELSE 0 END ORDER BY crashID) rn2
    FROM t
), cte2 AS (
    SELECT MIN(crashID) AS fid, MAX(crashID) AS tid
    FROM cte1
    WHERE island = 1
    GROUP BY rn1 - rn2
    ORDER BY COUNT(*) DESC
    LIMIT 1
)
SELECT *
FROM t
WHERE crashID BETWEEN (SELECT fid FROM cte2) AND (SELECT tid FROM cte2);

DB Fiddle

Upvotes: 2

Strawberry
Strawberry

Reputation: 33935

Here's one way, for older versions of MySQL... This solution assumes no ties for first place...

SELECT m.* 
  FROM my_table m
  JOIN 
     ( SELECT MIN(crash_id) range_start
            , MAX(crash_id) range_end
         FROM 
            ( SELECT x.*
                   , CASE WHEN FLOOR(crash/500) * 500 = 0 AND @prev = FLOOR(crash/500) * 500 THEN @i:=@i ELSE @i:=@i+1 END i
                   , @prev:=FLOOR(crash/500)*500 prev 
                FROM my_table x
                   , (SELECT @prev:=null,@i:=0) vars 
               ORDER 
                  BY crash_id
            ) a
        GROUP
           BY i
        ORDER
           BY COUNT(*) DESC LIMIT 1
     ) n
    ON m.crash_id BETWEEN n.range_start AND n.range_end;

Upvotes: 0

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