我在这里是因为我有类似的问题，但我发现简短的答案是错误的，一个连续的“情况下”是太多的工作，看到任何重复在我的代码伤害我的眼睛。这就是解

SELECT --MIN(score), MAX(score),
    [score range] = CAST(ROUND(score-5,-1)AS VARCHAR) + ' - ' + CAST((ROUND(score-5,-1)+10)AS VARCHAR),
    [number of occurrences] = COUNT(*)
FROM order
GROUP BY  CAST(ROUND(score-5,-1)AS VARCHAR) + ' - ' + CAST((ROUND(score-5,-1)+10)AS VARCHAR)
ORDER BY MIN(score)

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)

也许你问的是如何让这样的事情继续下去……

当然，您将为查询调用全表扫描，如果包含需要统计(聚合)的分数的表很大，您可能想要一个性能更好的解决方案，您可以创建一个辅助表并使用规则，例如关于插入—您可能会研究它。

不过，并不是所有的RDBMS引擎都有规则!

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

这将允许您不必指定范围，并且应该与SQL server无关。数学增值!

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

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