Is there a way to create just using select statements a table that contains in a column the range of the repeating values like in this example?
Example:
from the input table:
id value:
1 25
2 25
3 24
4 25
5 25
6 25
7 28
8 28
9 11
should result:
range value
1-2 25
3-3 24
4-6 25
7-8 28
9-9 11
Note: The id values from the input table are always in order and the difference between 2 values ordered by id is always equal to 1
You want to find sequences of consecutive values. Here is one approach, using window functions:
select min(id), max(id), value
from (select id, value,
row_number() over (order by id) as rownum,
row_number() over (partition by value order by id) as seqnum
from t
) t
group by (rownum - seqnum), value;
This takes into account that a value might appear in different places among the rows. The idea is simple. rownum is a sequential number. seqnum is a sequential number that increments for a given value. The difference between these is constant for values that are in a row.
Let me add, if you actually want the expression as "1-2", we need more information. Assuming the id is a character string, one of the following would work, depending on the database:
select min(id)+'-'+max(id), . . .
select concat(min(id), '-', max(id)), . . .
select min(id)||'-'||max(id), . . .
If the id is an integer (as I suspect), then you need to replace the ids in the above expressions with cast(id as varchar(32)), except in Oracle, where you would use cast(id as varchar2(32)).
SELECT CONCAT(MIN(id), '-', MAX(id)) AS id_range, value
FROM input_table
GROUP BY value
Maybe this:
SELECT MIN(ID), MAX(ID), VALUE FROM TABLE GROUP BY VALUE
