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…

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 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

如果在MySQL中，请确保使用'range'以外的单词，否则在运行上述示例时会出现错误。

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

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)

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。