Question

我想在我的网站上显示用户统计信息，返回年龄组的百分比，如：

-13 years : $percent %
13-15 years : $percent %
15-20 years : $percent %
23+ : $percent %

在我的mysql表中，我有一列birth_date返回datatime（yyyy-mm-dd）。

你有提示或想法吗？

Answer 1

select case when year(curdate() - birth_date) < 13 
            then '< 13 years'
            when year(curdate() - birth_date) between 13 and 14 
            then '13 - 14 years'
            when year(curdate() - birth_date) between 15 and 20 
            then '15 - 20 years'
            when year(curdate() - birth_date) >= 23 
            then '+23 years'
         end as `description`,
       (select count(*) from your_table) / count(*) 
from your_table
group by case when year(curdate() - birth_date) < 13 then 1
              when year(curdate() - birth_date) between 13 and 14 then 2
              when year(curdate() - birth_date) between 15 and 20 then 3
              when year(curdate() - birth_date) >= 23 then 4
         end

Answer 2

如果你想要纯sql：

SELECT COUNT(*)    
FROM [table_name]
WHERE birth_date < DATE_ADD(NOW(),INTERVAL 13 YEAR)


WHERE birth_date >= DATE_ADD(NOW(),INTERVAL 13 YEAR)
AND birth_date < DATE_ADD(NOW(),INTERVAL 15 YEAR)

--etc

否则我建议使用php来创建日期。

您可以将所有查询合并在一起并创建和sql视图，例如：

CREATE VIEW statistics AS
SELECT "0-13" as age ,COUNT(*) as total
FROM table_name
WHERE birth_date < DATE_ADD(NOW(),INTERVAL 13 YEAR)
UNION
SELECT "13-15" as age ,COUNT(*) as total
FROM table_name
WHERE birth_date >= DATE_ADD(NOW(),INTERVAL 13 YEAR)
    AND birth_date < DATE_ADD(NOW(),INTERVAL 15 YEAR)
UNION
SELECT "15-20" as age ,COUNT(*) as total
FROM table_name
WHERE birth_date >= DATE_ADD(NOW(),INTERVAL 15 YEAR)
    AND birth_date < DATE_ADD(NOW(),INTERVAL 20 YEAR)
UNION
SELECT "20-23" as age ,COUNT(*) as total
FROM table_name
WHERE birth_date >= DATE_ADD(NOW(),INTERVAL 20 YEAR)
    AND birth_date < DATE_ADD(NOW(),INTERVAL 23 YEAR)
UNION
SELECT "23+" as age ,COUNT(*) as total
FROM table_name
WHERE birth_date >= DATE_ADD(NOW(),INTERVAL 23 YEAR)

然后你可以查询：

SELECT * from statistics

Answer 3

Pure SQL：

SELECT
    `group`,
    COUNT(*) as `count`
FROM
`user`
INNER JOIN (
    SELECT 
        0 as `start`, 12 as `end`, '0-12' as `group`
    UNION ALL 
    SELECT
        13, 14, '13-14'
    UNION ALL
    SELECT
        15, 19, '15-19'
    UNION ALL
    SELECT
        20, 150, '20+'
) `sub`
    ON TIMESTAMPDIFF(YEAR, `birth_date`, NOW()) BETWEEN `start` AND `end`
GROUP BY `group` WITH ROLLUP;

其他任何内容都可以通过PHP计算。

Answer 4

将datetime作为时间戳返回，与now.offset（years = ??）进行比较，在数组中进行分类，然后计算记录。

使用PHP和MYSQL的年龄组统计信息

4 个答案: