Php 从日差小于某个数字的表格中选择
我试图获取当前日期和到期日期之间的日差小于4天的所有SQL条目 我的第一个方法是:Php 从日差小于某个数字的表格中选择,php,sql,Php,Sql,我试图获取当前日期和到期日期之间的日差小于4天的所有SQL条目 我的第一个方法是: $sql_i_requested = "SELECT *, (To_days(date_return)-TO_DAYS(NOW())) as daydif FROM ".$tbl_name." WHERE (status!='completed' AND status!='canceled') AND owner_id=".$owner_id." AND daydif < 4 ORDER BY da
$sql_i_requested = "SELECT *, (To_days(date_return)-TO_DAYS(NOW())) as daydif FROM ".$tbl_name."
WHERE (status!='completed' AND status!='canceled')
AND owner_id=".$owner_id."
AND daydif < 4
ORDER BY date_created DESC";
这两种方法都不起作用,所以如何从“日期返回”和now()
之间的日期差小于4天的表中进行选择
编辑:
改变
AND daydif < 4
和daydif<4
到
和(到天(日期返回)-到天(现在())<4
现在它开始工作了。无论如何,也许你们可以建议其他解决方案。试试:
SELECT DATEDIFF(date_return, NOW()) AS dayDiff;
及
具有dayDiff<4的
试试这个:
select *
from table
where DATEDIFF(day, present, future) < 4
选择*
从桌子上
其中DATEDIFF(日、现在、未来)<4
使用DATEDIFF
WHERE DATEDIFF(date_return, now()) < 4
WHERE DATEDIFF(date\u return,now())<4
在性能方面,什么更好(更快)DATEDIFF(date\u return,now())
或(To\u days(date\u return)-To\u days(now())
?
having dayDiff < 4
select *
from table
where DATEDIFF(day, present, future) < 4
WHERE DATEDIFF(date_return, now()) < 4