Home php教程 php手册 如何实现日期比较,暨实现显示5天内,显示10天内的记录

如何实现日期比较,暨实现显示5天内,显示10天内的记录

Jun 21, 2016 am 09:06 AM
days mysql select

比较|显示

主要的论点是集中在SELECT something FROM table
WHERE TO_DAYS(NOW()) - TO_DAYS(date_col)

MYSQL的TO_DAYS(DATE)函数是这样说明的:
传回DATE到公元0年的总日数,我测试了一下
mysql>select to_days(now(0));
+--------------------------+
| TO_DAYS(NOW()) |
+--------------------------+
| 730839 |
+--------------------------+

出来的是当前时间距离公元0年的总日数,接着我试着用上面的语句测试;

mysql>select TO_DAYS(NOW()) - TO_DAYS(date_col) 出现结果:
ERROR 1054:Unknown column 'date_col' in 'field first'

此路不通了,我就试着直接把5代到date_col里去

mysql>select to_days(now()) - to_days(5);
出现结果:
+---------------------------+
|to_days(now()) - to_days(5)|
+---------------------------+
| NULL |
+---------------------------+

啊?不会吧?这样也不行啊?
我接着试命令
mysql>select 。。。。

突然猛的想到,嘿嘿,to_days(now())出来的是整数,我直接跟整数运算就行了,何必再to_days(date)呢?马上试验

mysql>select to_days(now()) - 5;
+--------------------------+
| to_days(now()) -5 |
+--------------------------+
| 730834 |
+--------------------------+


OK,万岁,终于得到了我想要的结果,呵呵 下面就是在PHP代码中用SELECT 查询了

我存数据库向来的习惯就是DATEANDTIME用NOW()直接赋值,显示的时候不用格式化,直接取出来就能用,

下面是我的一个库的部分结构
CREATE TABLE infomess (
infoid int(11) NOT NULL auto_increment,
topic varchar(255) NOT NULL,
……
email varchar(50),
dateandtime datetime DEFAULT '0000-00-00 00:00:00' NOT NULL,
PRIMARY KEY (infoid)
);


这里的DATEANDTIME是标准的日期格式,然后我要查询5天内的记录,下面是SQL查询语句
$sql="select * from infomess where to_days(dateandtime) >= (to_days(now()) - 5) order by infoid desc limit $offset,$psize";

就要一个where to_days(dateandtime) >= (to_days(now()) - 5)就够了 后面的是另外的,这里的5可以设为一个变量

where to_days(dateandtime) >= (to_days(now()) - $limitdays)

然后$limitdays可以用GET方式传递(多数是有GET方式传递)

在你的PHP后面跟上?limitdays=5就行了 显示10天内也一样,$limitdasy改成10就行了

以上是利用MYSQL函数得到这样的结果,以上的结果都经过测试,因为时间匆忙,如果代码有什么问题,请跟帖提出,谢谢


还有朋友说利用UNIX戳记来得到这样的结果,请问哪位写过这样的代码,贴点出来,供大家参考比较,也可以测试判断一下PHP函数还是MYSQL函数实现的效率高

 



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