In SQL, how can you “group by” in ranges?

Question

Suppose I have a table with a numeric column (lets call it \"score\").

I\'d like to generate a table of counts, that shows how many times scores appeared in each ran

Answer 1

This will allow you to not have to specify ranges, and should be SQL server agnostic. Math FTW!

SELECT CONCAT(range,'-',range+9), COUNT(range)
FROM (
  SELECT 
    score - (score % 10) as range
  FROM scores
)

Answer 2

select t.blah as [score range], count(*) as [number of occurences]
from (
  select case 
    when score between  0 and  9 then ' 0-9 '
    when score between 10 and 19 then '10-19'
    when score between 20 and 29 then '20-29'
    ...
    else '90-99' end as blah
  from scores) t
group by t.blah

Make sure you use a word other than 'range' if you are in MySQL, or you will get an error for running the above example.

Answer 3

An alternative approach would involve storing the ranges in a table, instead of embedding them in the query. You would end up with a table, call it Ranges, that looks like this:

LowerLimit   UpperLimit   Range 
0              9          '0-9'
10            19          '10-19'
20            29          '20-29'
30            39          '30-39'

And a query that looks like this:

Select
   Range as [Score Range],
   Count(*) as [Number of Occurences]
from
   Ranges r inner join Scores s on s.Score between r.LowerLimit and r.UpperLimit
group by Range

This does mean setting up a table, but it would be easy to maintain when the desired ranges change. No code changes necessary!

Answer 4

James Curran's answer was the most concise in my opinion, but the output wasn't correct. For SQL Server the simplest statement is as follows:

SELECT 
    [score range] = CAST((Score/10)*10 AS VARCHAR) + ' - ' + CAST((Score/10)*10+9 AS VARCHAR), 
    [number of occurrences] = COUNT(*)
FROM #Scores
GROUP BY Score/10
ORDER BY Score/10

This assumes a #Scores temporary table I used to test it, I just populated 100 rows with random number between 0 and 99.

Answer 5

I would do this a little differently so that it scales without having to define every case:

select t.range as [score range], count(*) as [number of occurences]
from (
  select FLOOR(score/10) as range
  from scores) t
group by t.range

Not tested, but you get the idea...

Answer 6

In postgres (where || is the string concatenation operator):

select (score/10)*10 || '-' || (score/10)*10+9 as scorerange, count(*)
from scores
group by score/10
order by 1

gives:

 scorerange | count 
------------+-------
 0-9        |    11
 10-19      |    14
 20-29      |     3
 30-39      |     2