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

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 

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;

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

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