如何确定MySQL中的人是否超过21岁

Question

我正在尝试在MySQL中创建一个视图。我希望它显示用户电子邮件，first-lastname，他们的DOB以及他们当前的年龄。我只想向21岁以上的用户展示。

这就是我所拥有的：

CREATE VIEW 'valid_Users' AS 
SELECT Persons.Email AS Email, Persons.FName + ' ' + Persons.LName AS Name,  Persons.DOB AS Birth_Date, DATEDIFF(Persons.DOB, NOW()) AS Current_Age'
FROM Persons
WHERE DATEDIFF(Persons.DOB, NOW()) > 21;

问题是Where子句返回一个负数，这显然不是一个正确的年龄。我如何编辑此代码以确定一个人的正确年龄？

我知道名字还不起作用 - 但是......这里是一些当前数据的图像。实际的表格是＆＃39;用户＆＃39;

Answer 1

尝试：

CREATE VIEW 'valid_Users' AS 
SELECT Persons.Email AS Email, Persons.FName + ' ' + Persons.LName AS Name, Persons.DOB AS Birth_Date, DATE_FORMAT(FROM_DAYS(DATEDIFF(NOW(), Persons.DOB)), "%Y")+0 AS Current_Age'
FROM Persons
WHERE DATE_FORMAT(FROM_DAYS(DATEDIFF(NOW(), Persons.DOB)), "%Y")+0 > 21;

OR：

CREATE VIEW 'valid_Users' AS 
SELECT Persons.Email AS Email, Persons.FName + ' ' + Persons.LName AS Name, Persons.DOB AS Birth_Date, TIMESTAMPDIFF(YEAR, Persons.DOB, NOW()) AS Current_Age'
FROM Persons
WHERE TIMESTAMPDIFF(YEAR, Persons.DOB, NOW()) > 21;

Answer 2

我认为你想要这样的东西：

CREATE VIEW valid_Users AS 
    SELECT p.Email, 
           CONCAT(p.FName, ' ', p.LName) AS Name,  
           p.DOB AS BirthDate,
           TIMESTAMPDIFF('year', p.DOB, CURDATE()) AS CurrentAge
    FROM Persons p
    WHERE p.DOB < CURDATE() - INTERVAL 21 YEARS;

Answer 3

试试切换参数：

SELECT DATEDIFF(NOW(), Persons.DOB) AS days; 

告诉我们你得到了什么？