Mysql中求出生日

Mysql中求出生日【待完善】

1.需求

给出一个出生表，求出某人的最近生日。

2.背景

• 创建表employee

CREATE TABLE `employee` (
  `emp_no` int(11) NOT NULL,
  `birth_date` date NOT NULL,
  `first_name` varchar(14) NOT NULL,
  `last_name` varchar(16) NOT NULL,
  `gender` enum('M','F') NOT NULL,
  `hire_date` date NOT NULL,
  PRIMARY KEY (`emp_no`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1

• • 
    
• 插入表数据【此处省略】
• 查看其中表的数据
  • 
    

mysql> select * from employee;
+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date  |
+--------+------------+------------+-----------+--------+------------+
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10003 | 1959-12-03 | Parto      | Bamford   | M      | 1986-08-28 |
|  10004 | 1954-05-01 | Chirstian  | Koblick   | M      | 1986-12-01 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |
|  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10007 | 1957-05-23 | Tzvetan    | Zielinski | F      | 1989-02-10 |
|  10008 | 1958-02-19 | Saniya     | Kalloufi  | M      | 1994-09-15 |
|  10009 | 1952-04-19 | Sumant     | Peac      | F      | 1985-02-18 |
|  10010 | 1963-06-01 | Duangkaew  | Piveteau  | F      | 1989-08-24 |
|  10011 | 1972-02-29 | Jiang      | David     | M      | 1990-02-20 |
+--------+------------+------------+-----------+--------+------------+
11 rows in set (0.00 sec)

3.代码

select 
name
,birth_date
,if(now() < cur,cur,next) as result
from
(
    select 
        name
        ,birth_date     
        ,DATE_ADD(cur,INTERVAL if(day(birth_date)=29 && day(cur) = 28 ,1,0) day) as cur
        ,DATE_ADD(next,INTERVAL if(day(birth_date)=29 && day(cur) = 28 ,1,0) day) as next
    FROM
    (
        select 
        name
        ,birth_date
        ,date_add(birth_date,interval diff year) as cur
        ,date_add(birth_date,interval diff+1 year) as next
        ,date(now())
        from 
        (
                select 
                concat(first_name,' ',last_name) as name
                ,birth_date
                ,(year(now())-year(birth_date)) as diff
                from employee
            )as a
    )as b
)as c

4.执行结果

+--------------------+------------+------------+
| name               | birth_date | result     |
+--------------------+------------+------------+
| Georgi Facello     | 1953-09-02 | 2019-09-02 |
| Bezalel Simmel     | 1964-06-02 | 2019-06-02 |
| Parto Bamford      | 1959-12-03 | 2018-12-03 |
| Chirstian Koblick  | 1954-05-01 | 2019-05-01 |
| Kyoichi Maliniak   | 1955-01-21 | 2019-01-21 |
| Anneke Preusig     | 1953-04-20 | 2019-04-20 |
| Tzvetan Zielinski  | 1957-05-23 | 2019-05-23 |
| Saniya Kalloufi    | 1958-02-19 | 2019-02-19 |
| Sumant Peac        | 1952-04-19 | 2019-04-19 |
| Duangkaew Piveteau | 1963-06-01 | 2019-06-01 |
| Jiang David        | 1972-02-29 | 2019-03-01 |
+--------------------+------------+------------+
11 rows in set (0.04 sec)

5.分析

上述代码的精妙之处在于：

• • 
    
• 没有自己实现一个闰年判断，而是借助系统的日期功能，顺带往后推了一段时间，这样就减少了很多麻烦。归根结底，这里还是用到了一个“构造变量”的思想，重新生成一列，然后重新生成的这列有很大的用处！
• 当然，这里也使用到了一部分的函数，比如​​if()​​，​​date_add()​​，​​day()​​，​​concat()​​,​​year()​​,​​now()​​等。下面我逐一讲解这些函数的使用。
  • if()函数
• 
  
  • 
    

mysql> select if(2>1,2,1);
+-------------+
| if(2>1,2,1) |
+-------------+
|           2 |
+-------------+
1 row in set (0.00 sec)

之前，我一直在考虑将日期部分的年取出来，然后我使用​​substring()​​函数，然后再强转成int型，最后做一个减法，就知道当前年到出生年之间的时间间隔了。但是这一切所有的步骤均可以通过一个函数实现： =>year函数：将一个日期取出年部分；同样，day()函数是取出一个日期中的日，month()函数是取出月部分。

• year()函数

mysql> select year(current_timestamp);
+-------------------------+
| year(current_timestamp) |
+-------------------------+
|                    2018 |
+-------------------------+
1 row in set (0.00 sec)

• month()函数

mysql> select month(current_timestamp);
+--------------------------+
| month(current_timestamp) |
+--------------------------+
|                        9 |
+--------------------------+
1 row in set (0.00 sec)

• day()函数

mysql> select day(current_timestamp);
+------------------------+
| day(current_timestamp) |
+------------------------+
|                      9 |
+------------------------+
1 row in set (0.00 sec)

因为最近的生日要么是今年，要么是明年，所以再根据diff+1生成明年的日期【在生成日期时，就考虑到了闰年的问题，根据法律的规定，如果闰年的生日在2/29，那么在平年的生日就是3/1】针对这个规定，就得到了如下的SQL语句。

select 
concat(first_name,' ',last_name) as name
,(year(now())-year(birth_date)) as diff
from employee;

+--------------------+------+
| name               | diff |
+--------------------+------+
| Georgi Facello     |   65 |
| Bezalel Simmel     |   54 |
| Parto Bamford      |   59 |
| Chirstian Koblick  |   64 |
| Kyoichi Maliniak   |   63 |
| Anneke Preusig     |   65 |
| Tzvetan Zielinski  |   61 |
| Saniya Kalloufi    |   60 |
| Sumant Peac        |   66 |
| Duangkaew Piveteau |   55 |
| Jiang David        |   46 |
+--------------------+------+
11 rows in set (0.00 sec)