Question

我尝试创建一个查询，该查询从名为$ incorporation_date的文本字段中获取用户输入。

查询理念

$sql = "SELECT * FROM companies WHERE incorporation_date LIKE '%%%%/%%/%%" . $incorporation_date . "%%%%/%%/%%'";

我如何制作它以便您可以使用SQL来显示您输入的提交格式的值。

搜索范例2015-06-15

Answer 1

使用Date或Datetime并免除其本来会遇到的悲伤

drop table theGuy;

create table theGuy
(
    id int not null auto_increment primary key,
    fullName varchar(60) not null,
    birthDate date not null
);

insert theGuy (fullName,birthDate) values ('Kim Smithers','2002-3-1'),('John Doe','2014-4-5');

select * from theGuy where birthDate>='2000-1-1' and birthDate<='2007-12-31';

select * from theGuy where birthDate between '2000-1-1' and '2007-12-31';


select *,birthDate+interval 45 DAY as BirthCertAvail from theGuy;

select *,datediff(curdate(),birthDate) as DaysAlive from theGuy;

您可能会发现内置例程足够，例如间隔，而不必重写它们。 ;）

PHP MySQL日期格式搜索

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