declare @RangeWidth int

set @RangeWidth = 10

select
   Floor(Score/@RangeWidth) as LowerBound,
   Floor(Score/@RangeWidth)+@RangeWidth as UpperBound,
   Count(*)
From
   ScoreTable
group by
   Floor(Score/@RangeWidth)

我会用一种稍微不同的方式来做，这样它就不必定义每种情况:

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

没有测试过，但你懂的…

select cast(score/10 as varchar) + '-' + cast(score/10+9 as varchar), 
       count(*)
from scores
group by score/10

因为要排序的列(Range)是一个字符串，所以使用字符串/单词排序而不是数值排序。

只要字符串有0来填充数字长度，排序在语义上仍然是正确的:

SELECT t.range AS ScoreRange,
       COUNT(*) AS NumberOfOccurrences
  FROM (SELECT CASE
                    WHEN score BETWEEN 0 AND 9 THEN '00-09'
                    WHEN score BETWEEN 10 AND 19 THEN '10-19'
                    ELSE '20-99'
               END AS Range
          FROM Scores) t
 GROUP BY t.Range

如果范围是混合的，简单地填充一个额外的0:

SELECT t.range AS ScoreRange,
       COUNT(*) AS NumberOfOccurrences
  FROM (SELECT CASE
                    WHEN score BETWEEN 0 AND 9 THEN '000-009'
                    WHEN score BETWEEN 10 AND 19 THEN '010-019'
                    WHEN score BETWEEN 20 AND 99 THEN '020-099'
                    ELSE '100-999'
               END AS Range
          FROM Scores) t
 GROUP BY t.Range

我在这里看到的答案在SQL Server的语法中行不通。我会用:

select t.range 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 range
  from scores) t
group by t.range

编辑:见评论

在SQL Server 2000上，投票最多的答案都不正确。也许他们用的是另一个版本。

下面是在SQL Server 2000上这两个软件的正确版本。

select t.range 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'
    else '20-99' end as range
  from scores) t
group by t.range

or

select t.range as [score range], count(*) as [number of occurrences]
from (
      select user_id,
         case when score >= 0 and score< 10 then '0-9'
         when score >= 10 and score< 20 then '10-19'
         else '20-99' end as range
     from scores) t
group by t.range