解决异常值范围，如何计算来自同一输出的两个不同行？

如何解决解决异常值范围，如何计算来自同一输出的两个不同行？

我有以下查询：

SELECT
    age_quartile,MAX(age) AS quartile_break
    from 

    (SELECT
        full_name,age,NTILE(4) OVER (ORDER BY age) AS age_quartile
    FROM friends) AS quartiles
WHERE age_quartile IN (1,3)
GROUP BY age_quartile)

这给了我看起来像这样的输出：

age_quartile | quantile_break
    1                 31
    3                 35

期望的输出：

outlier range 
25 
41

其中 25 = 31-6 和 41 = 35 + 6

如何将我的查询添加到我最终想要的输出的位置？我的查询目前为我提供了哪些数字，我需要做一个额外的步骤来解决离群值范围。谢谢！

表格数据看起来像：

friends

full_name        | age
Ameila Lara        1
Evangeline Griffin 21
Kiara Atkinson     31
Isobel Nieslen     31
Genevuve Miles     32
Jane Jenkins       99
Marie Acevedo      null

解决方法

不要现在 ntile 是在这里使用的正确函数。但一种方法是在临时表中定义年龄四分位数并与年龄表连接并找到结果。只是一个尝试。可能有更好的方法。有兴趣查看其他答案。

示例查询：

with friends as 
(
  select 'user1' as full_name,31 as age union all
  select 'user2' as full_name,55 as age union all
  select 'user3' as full_name,75 as age 
),quartiles_age as 
(
  select 1 as quartile,0 as st_range,25 as end_range union all
  select 2 as quartile,26 as st_range,50 as end_range union all
  select 3 as quartile,51 as st_range,75 as end_range union all
  select 4 as quartile,76 as st_range,100 as end_range 
)


SELECT
  fr.full_name,fr.age,qrtl_age.quartile,qrtl_age.end_range - fr.age as diff_age
FROM 
  friends fr
  join quartiles_age qrtl_age on fr.age between qrtl_age.st_range and qrtl_age.end_range

小提琴网址：(https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=c48d209264e90d276cea6ae03f2a7af6)

,

您可以使用以下方法计算范围：

MAX(age) - MIN(age) AS age_range

这回答了您提出的问题。您的示例数据有一个任意的 6 用于计算，问题没有解释。