Mysql 试图获得申请某一年龄段工作的男女比例

很抱歉，这个标题让人困惑，我想实现的是，通过下表获得一份工作的全部申请：

CREATE TABLE IF NOT EXISTS `applications` (
  `application_id` int(11) NOT NULL AUTO_INCREMENT,
  `application_user` varchar(100) NOT NULL,
  `application_date` datetime NOT NULL,
  `application_job` int(11) NOT NULL,
  `application_status` varchar(10) DEFAULT 'pending',
  `application_enabled` int(2) NOT NULL DEFAULT '1',
  `application_resume` int(11) NOT NULL,
  `application_description` text NOT NULL,
  PRIMARY KEY (`application_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

为了了解他们的年龄，我加入了

user\u personal\u information

进入

application\u user

，因为他们是申请工作的用户。我的问题是：

SELECT count(*) as total,
       user_gender as gender,
       TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE()) AS age,
       application_date
FROM applications
LEFT JOIN user_personal_information
    ON user_personal_information_user = application_user
WHERE application_job = ?

带有

user\u gender

的我的用户表，它可以等于

男性

/

女性

和

user\u生日

，在上述声明中，我将其转换为年龄

我正在尝试将年龄范围为的所有应用程序分组，例如：

16 - 21
22 - 30
31 - 45
45 - 64
65+

以及该年龄段的男性和女性百分比。要用于需要以下数据的数据图表：

 "dataProvider": [
 {
   "age": "85+",
   "male": 25, //
   "female": 25
 }, {
   "age": "80-54",
   "male": 25,//percentage
   "female": 25//percentage
}]

综上所述，25%的男性申请了85岁及以上，25%的女性申请。你明白要点了，所以这就是我如何让我的select语句工作的

这将创建如下图表：

---

因此，我想澄清一下，我想计算一下申请总数，并根据性别和年龄组计算出申请的百分比。如何使用上面的select语句实现这一点？

几个嵌套的GROUPBY子查询可以为您实现这一点。请注意，计算的百分比因性别而异：

select a.age, a.gender, a.cnt, 100*a.cnt/b.sm as percentage from
    (
        SELECT user_gender as gender,
               TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE()) AS age,
               count(*) as cnt,
        FROM applications
        LEFT JOIN user_personal_information
        ON user_personal_information_user = application_user
        WHERE application_job = ?
        GROUP BY user_gender, TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE())
    ) a,
    (
        SELECT TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE()) AS age,
               count(*) as sm
        FROM applications
        LEFT JOIN user_personal_information
        ON user_personal_information_user = application_user
        WHERE application_job = ?
        GROUP BY TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE())
    ) b
    where a.age = b.age;

如果要查找特定于总应用程序的百分比，则需要以下内容：

select a.age, 
       a.gender, 
       a.cnt, 
100*a.cnt/(
            select count(TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE()))
            from
            applications 
            LEFT JOIN user_personal_information
            ON user_personal_information_user = application_user
            WHERE application_job = ?
         ) as percentage 
from
    (
        SELECT user_gender as gender,
               TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE()) AS age,
               count(*) as cnt,
        FROM applications
        LEFT JOIN user_personal_information
        ON user_personal_information_user = application_user
        WHERE application_job = ?
        GROUP BY user_gender, TIMESTAMPDIFF(YEAR, user_birthdate, CURDATE())
    ) a;

---

这应该可以做到：

select sum(case when age between 16 and 21 and gender='male' then 1 end) as '[Male 16-21]', sum(case when age between 16 and 21 and gender='female' then 1 end) as 'Female [16-21]', sum(case when age between 22 and 30 and gender='male' then 1 end) as '[Male 22-30]', sum(case when age between 22 and 30 and gender='female' then 1 end) as '[Female 22-30]', sum(case when age between 31 and 45 and gender='male' then 1 end) as '[Male 31-45]', sum(case when age between 31 and 45 and gender='female' then 1 end) as '[Female 31-45]', sum(case when age between 46 and 64 and gender='male' then 1 end) as '[Male 46-64]', sum(case when age between 46 and 64 and gender='female' then 1 end) as '[Female 46-64]', sum(case when age > 64 and gender='male' then 1 end) as '[Male Over 64]', sum(case when age > 64 and gender='female' then 1 end) as '[Female Over 64]', sum(case when 1=1 and gender='female' then 1 end) as '[Male TOTAL]', sum(case when 1=1 and gender='male' then 1 end) as '[Female TOTAL]' FROM applications 挑选 总和（年龄在16至21岁之间且性别为“男性”的情况下，则1结束）为“[男性16-21]”， 总和（如果年龄在16至21岁之间，性别为“女性”，则1结束）为“女性[16-21]”， 总和（如果年龄在22-30岁之间，性别为“男性”，则以1结尾）为“[男性22-30]”， 总和（如果年龄在22-30岁之间，性别为“女性”，则1结束）为“[女性22-30]”， 总和（如果年龄在31-45岁之间，性别为“男性”，则以1结尾）为“[男性31-45]”， 总和（如果年龄在31-45岁之间，性别为“女性”，则以1结尾）为“[女性31-45]”， 总和（如年龄在46至64岁之间，性别为“男性”，则以1结尾）为“[男性46-64]”， 总和（如果年龄在46至64岁之间，性别为“女性”，则1结束）为“[女性46-64]”， 总和（当年龄>64岁且性别为“男性”时，则以1结尾）为“[男性超过64岁]”， 总和（当年龄>64岁且性别=女性时，则1结束）为“[女性超过64岁]”， 总和（当1=1且性别为女性时，则1结束）为“[男性总数]”， 总和（当1=1且性别=男性时，则为1结尾）为“[女性总数]” 来自应用程序

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抱歉，我也试图获取百分比并显示年龄范围，即（40-50）需要使用

大小写，当
语句对计算年龄进行限制时，将其限制为组（例如16-21等）&然后按此分组以及genderI选择嵌套两个group by，并查询其联接的损坏情况。我想它会工作的。我收到一个错误：
警告：PDOStatement:：execute（）：SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；检查与您的MariaDB服务器版本相对应的手册，以了解在第71行的C:\xampp\htdocs\job\application\model\AnalyticModel.php中第16行的“a”（选择TIMESTAMPDIFF（YEAR，user\u birthdate，CURDATE（））AS）附近使用的正确语法
好吧，
CURDATE（）
之后缺少close paranthes！