Reputation: 847
In a column I have "1;2;3;6-9"
I need make this string in a array like this (1,2,3,6,7,8,9)
select range from my_table
return
| range |
|-----------|
| 1;2;3;6-9 |
I need run
select id from my_another_table where id in("1;2;3;6-9")
| id |
|----|
| 1 |
| 2 |
| 3 |
| 6 |
| 7 |
| 8 |
| 9 |
Upvotes: 2
Views: 35805
Reputation: 67
SELECT *
FROM [dbo].[Table_1]
where Id in (select [value] from openjson((select conf.[Value] FROM Table_2 conf WHERE conf.[Key]='Range'), '$'))
Where
Table_2 "data must be JSon" range (Nvarchar) =>'[1,2,3,4]'
Table_1 ID Int => 1,2,4,5
Upvotes: 0
Reputation: 2798
Just in case that you are still interested, you can create a function.
create or replace function parse_ranges(value TEXT) returns setof int as
$func$
DECLARE
range RECORD;
ra int;
rb int;
BEGIN
FOR range IN SELECT * FROM regexp_split_to_table(value, ';') as r(r) LOOP
IF range.r LIKE '%-%' THEN
ra = split_part(range.r, '-', 1)::int;
rb = split_part(range.r, '-', 2)::int;
RETURN QUERY SELECT * FROM generate_series(ra, rb, 1);
ELSE
RETURN NEXT range.r::int;
END IF;
END LOOP;
END;
$func$
LANGUAGE plpgsql;
Then you can use it:
select id from my_another_table where id in(SELECT parse_ranges('1;2;3;6-9'))
Upvotes: 1
Reputation: 1271151
This is a lousy structure for data. But you can do this with generate_series()
and string functions:
select generate_series(v2.lo, v2.hi, 1)
from (values ('1;2;3;6-9')) v(str) cross join lateral
regexp_split_to_table(v.str, ';') as r(range) cross join lateral
(values (case when range not like '%-%' then range::int else split_part(range, '-', 1)::int end,
case when range not like '%-%' then range::int else split_part(range, '-', 2)::int end
)
) v2(lo, hi);
Upvotes: 3