MySQL get difference between two smallest dates in different records grouped on

Question

I have a table with the following structure and example records:

I want to figure out how to get the difference between the two earliest created_at records for each name. It is a possibility the name only has one record, in which case any default value is okay, like 0, -1, or NULL. Running it on the above record I would anticipate a result like so:

nameA, 31
nameB, 59
nameC, -1

I can't think of any clean ways to do this purely in SQL. I thought about joining the table on itself, grouping on name, and selecting the min created_at from the first join, but I don't know how to then select the second minimum to bring it down to a single record.

Thanks

Answer 1

One option, using a CTE with ROW_NUMBER :

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY name ORDER BY created_at) rn
    FROM yourTable
)

SELECT
    name,
    DATEDIFF(MAX(created_at), MIN(created_at)) AS diff_in_days
FROM cte
WHERE rn <= 2
GROUP BY
    name
ORDER BY
    name;

Answer 2

If you mysql version didn't support window function you can try to use correlated-subquery make row number.

Query 1 :

SELECT t1.name,DATEDIFF(t1.max_dt,t1.min_dt) value
FROM (
  select name,
        max(create_date) max_dt,
        min(create_date) min_dt
  from (
        SELECT *,(SELECT COUNT(*) 
                  FROM T tt 
                  WHERE tt.name = t1.name and t1.create_date >= tt.create_date) rn
        FROM T t1
  )  t1
  where rn <= 2
  group by name
) t1 

Results :

|  name | value |
|-------|-------|
| NameA |    31 |
| NameB |    59 |
| NameC |     0 |

Answer 3

I would use lead() and `row_number():

select t.name,
       datediff(next_created_at, created_at) as diff_days
from (select t.*,
             lead(created_at) over (partition by name order by created_at) as next_created_at,
             row_number() over (partition by name order by created_at) as seqnum
      from t
     ) t
where seqnum = 1;