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;

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。

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;

对于PrestoSQL/Trino应用Ken https://stackoverflow.com/a/232463/429476的答案

select t.range, count(*) as "Number of Occurance", ROUND(AVG(fare_amount),2) as "Avg",
  ROUND(MAX(fare_amount),2) as "Max" ,ROUND(MIN(fare_amount),2) as "Min" 
from (
  select 
   case 
      when trip_distance between  0 and  9 then ' 0-9 '
      when trip_distance between 10 and 19 then '10-19'
      when trip_distance between 20 and 29 then '20-29'
      when trip_distance between 30 and 39 then '30-39'
      else '> 39' 
   end as range ,fare_amount 
  from nyc_in_parquet.tlc_yellow_trip_2022) t
  where fare_amount > 1 and fare_amount < 401092
group by t.range;

 range | Number of Occurance |  Avg   |  Max  | Min  
-------+---------------------+--------+-------+------
  0-9  |             2260865 |  10.28 | 720.0 | 1.11 
 30-39 |                1107 | 104.28 | 280.0 |  5.0 
 10-19 |              126136 |   43.8 | 413.5 |  2.0 
 > 39  |               42556 |  39.11 | 668.0 | 1.99 
 20-29 |               19133 |  58.62 | 250.0 |  2.5 

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

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…