Generate pattern using SQL

I have a table with a single column with numbers 1-100. Want to generate an output where each number N is repeated N times (so 5 should appear 5 times):

1
2
2
3
3
3
4
4
4
4
and so on

For mysql versions prior to 8.0,

create table _set(n int);

insert into _set(n) values (1), (2), (3), (4), (5), (6); -- this goes on

select s2.* from _set as s
  join _set as s2 on s.n <= s2.n;

Temporary tables can't be opened twice documented issue.

For mysql versions 8.0, try this,

with _set as (select 1 as n 
            union all
            select n+1 as n from _set
            where n < 100
            )
select s2.* from _set s
    join _set s2 on s.n <= s2.n

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If you have a table with numbers, then use it in a self join:

select n.n
from numbers n join
     numbers n2
     on n2.n <= n.n
order by n.n;

Here is a db<>fiddle example.

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For MySql 5.x

Create a tally table with positive numbers, including 0.

Then self-join that number table.

select n1.n as num
from nums as n1
join nums as n2 on n2.n < n1.n
where n1.n between 1 and 100
order by n1.n;

An example of creating such a number table:

CREATE TABLE digits (n int primary key not null);

insert into digits (n) values 
(0),(1),(2),(3),(4),(5),(6),(7),(8),(9);

CREATE TABLE nums (n int primary key not null);

-- fill with numbers 0 to 999
INSERT INTO nums (n)
SELECT (d3.n*100 + d2.n*10 + d1.n) as num
FROM digits d1
CROSS JOIN digits d2
CROSS JOIN digits d3;

In MySql 8 a recursive CTE can be used instead

WITH RECURSIVE NUMS AS 
(
  SELECT 0 as n

  UNION ALL

  SELECT n+1
  FROM NUMS
  WHERE n < 100
)
SELECT n1.n
FROM NUMS n1
JOIN NUMS n2 ON n2.n < n1.n
ORDER BY n1.n;

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Comments
  • Which MySQL version?
  • I recommend using some sort of scripting language to manage this
  • @jarlh any version will be fine!
  • @jarlh is probably asking because, depending from the mysql version some features can or cannot be available
  • MySQL 5.7 is probably too old for recursive cte's. (Version 8 needed?)