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

在我看来，James Curran的回答是最简洁的，但输出并不正确。对于SQL Server，最简单的语句如下:

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

这假设了一个我用来测试它的#Scores临时表，我只是用0到99之间的随机数填充了100行。

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!

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

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)

Try

SELECT (str(range) + "-" + str(range + 9) ) AS [Score range], COUNT(score) AS [number of occurances]
FROM (SELECT  score,  int(score / 10 ) * 10  AS range  FROM scoredata )  
GROUP BY range;