假设一个类似于这个问题的场景.我想得到以下结果:
score range | number of occurrences
-------------------------------------
0-9 | 11
10-19 | 14
20-29 | 3
... | ...
Run Code Online (Sandbox Code Playgroud)
我可以使用选定的答案作为解决方案:
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
Run Code Online (Sandbox Code Playgroud)
即使在该范围内没有结果,我怎样才能确保显示30-39的分数范围?
vye*_*rov 24
试试这个查询(也在SQL Fiddle上):
WITH ranges AS (
SELECT (ten*10)::text||'-'||(ten*10+9)::text AS range,
ten*10 AS r_min, ten*10+9 AS r_max
FROM generate_series(0,9) AS t(ten))
SELECT r.range, count(s.*)
FROM ranges r
LEFT JOIN scores s ON s.score BETWEEN r.r_min AND r.r_max
GROUP BY r.range
ORDER BY r.range;
Run Code Online (Sandbox Code Playgroud)
编辑:
您可以通过更改参数来轻松调整范围generate_series().可以使用以下构造来确保ranges始终覆盖您的分数:
SELECT (ten*10)::text||'-'||(ten*10+9)::text AS range,
ten*10 AS r_min, ten*10+9 AS r_max
FROM generate_series(0,(SELECT max(score)/10 FROM scores)) AS t(ten))
Run Code Online (Sandbox Code Playgroud)
对于rangesCTE.