mysql根据unix时间戳查询数据怎么使用?

假设我在数据库内插入的订单创建时间字段是用的10位数的unix时间戳,我想查询今天的所有订单,例如:

select count() where后面不会了

不知道如何操作了!

后面的条件该怎么写?

解决方案:

$todayzero = strtotime(date('Y-m-d',time()));
        //return  $todayzero;
        //今日访问量
        $today = pdo_fetchcolumn('select count(*) from '.tablename('wc_nongplus_count'). "where uniacid=:uniacid and (createtime>=$todayzero and createtime<=:createtime)",array(':uniacid'=>$uniacid,':createtime'=>time()));
        //总访问量
        $counts = pdo_fetchcolumn('select count(*) from '.tablename('wc_nongplus_count'). "where uniacid=:uniacid",array(':uniacid'=>$uniacid));
php

3个回答

time<当前时间戳 and time > 今天凌晨时间戳-1
查询其他日期的同理
time<日期结束次日凌晨时间戳 and time > 日期开始凌晨时间戳-1
建议增加一个冗余字段 day 20190320 这种格式的 这种格式的话进行查询压力要小特别多

qq_39733426
php—solitude 回复51digit: 解决了的话结下贴 不然之后回答的人会少 基本看结帖率特别低的很少有人回答
大约一年之前 回复
qdtravelsky
skywesdy 嗯嗯,谢谢你了,参考性很强
大约一年之前 回复

今天00:00
$todaystart = strtotime(date(‘Y-m-d’.’00:00:00′,time()));
今天24:00
$todayend = strtotime(date(‘Y-m-d’.’00:00:00′,time()+3600*24));
select count() where ctime > $todaystart and ctime < $todayend

qdtravelsky
skywesdy 谢谢您,思路基本和您这个差不多
大约一年之前 回复

select count(*) from orders where createdate>UNIX_TIMESTAMP(CAST(SYSDATE()AS DATE))

qq_39733426
php—solitude 在sql where条件内使用sql函数 是不会触发索引的吧
大约一年之前 回复
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