根据SQL中的序列对值进行分组

Question

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

Answer 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 id s in the above expressions with cast(id as varchar(32)) , except in Oracle, where you would use cast(id as varchar2(32)) .

Answer 2

SELECT CONCAT(MIN(id), '-', MAX(id)) AS id_range, value
  FROM input_table
 GROUP BY value

Answer 3

也许这样：

SELECT MIN(ID), MAX(ID), VALUE FROM TABLE GROUP BY VALUE