计算一个分组和套装的mysql

Question

我试图通过计算他们的生日来获得每个人的年龄。现在我需要将它们分组，我已经实现了。现在我想要的是计算20组及以下的数量，依此类推等等。

这是我的查询

select case when datediff(now(), birthday) / 365.25 > 50 then '51 & over' when datediff(now(), birthday) / 365.25 > 30 then '31 - 50' when datediff(now(), birthday) / 365.25 > 19 then '20 - 30' else 'under 20' end as age_group from members

结果是

我试图通过下面的查询对其进行分组

select case when datediff(now(), birthday) / 365.25 > 50 then '51 & over' when datediff(now(), birthday) / 365.25 > 30 then '31 - 50' when datediff(now(), birthday) / 365.25 > 19 then '20 - 30' else 'under 20' end as age_group from members group by age_group

结果将是

但我需要的是这样的东西

** NOte：照片编辑。

我如何实现我需要将计数结果放在图表中。

Answer 1

将查询换行类似于：

Select age_group, count(*)
From (
  // your original query here
) t
Group by age_group

Answer 2

只需在上次查询中添加一个计数（*）：

   select case when datediff(now(), birthday) / 365.25 > 50 then '51 & over' when datediff(now(), birthday) / 365.25 > 30 then '31 - 50' when datediff(now(), birthday) / 365.25 > 19 then '20 - 30' else 'under 20' end as age_group, count(*) from members group by age_group

Answer 3

SELECT
concat(
    datediff(now(), birthday)/365.25 div 20 * 20 + 1,
    '-',
    (datediff(now(), birthday)/365.25 div 20 + 1) * 20) AS age,
count(*)
FROM members
GROUP BY datediff(now(), birthday)/365.25 DIV 20

此查询将以以下格式提供结果

+-------+----------+
| age   | count(*) |
+-------+----------+
| 1-20  |        2 |
| 21-40 |        1 |
| 41-60 |        1 |
+-------+----------+

基于以下表格定义和样本数据：

CREATE TABLE `stack` (
`id` int(6) NOT NULL AUTO_INCREMENT,
`birthdate` date NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

insert into members values (null,'1973-10-02');
insert into members values (null,'1977-04-07');
insert into members values (null,'2006-07-03');
insert into members values (null,'2011-10-11');