date_sub()函数: DATE_SUB(date,INTERVAL expr type) 实例: SELECT NOW(),DATE_SUB(NOW(),INTERVAL HOUR) as the_time HOUR);

MySQL查询近一个月的数据 近一个月统计SQL select user_id, user_name, createtime from t_user where DATE_SUB(CURDATE(), INTERVAL 1 MONTH) <= date(createtime); 同理,近一个星期为: INTERVAL 7 DAY. @完

在mysql中查询不区分大小写重复的数据,往往会用到子查询,并在子查询中使用upper函数来将条件转化为大写.如: select * from staticcatalogue WHERE UPPER(Source) IN (SELECT UPPER(Source) FROM staticcatalogue GROUP BY UPPER(Source) having count(UPPER(Source))>1) ORDER BY upper(Source) DESC; 这条语句的执行效率是非常低

mysql 查询一个地点(经纬度) 附近N公里内的数据.(根据一个地点的经纬度查询这个地点方圆几公里内的数据)1.创建测试表 CREATE TABLE `location` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(50) NOT NULL, `longitude` decimal(13,10) NOT NULL, `latitude` decimal(13,10) NOT NULL, PRIMARY KEY

查询某个字段重复的数据 ; 查询股票重复的营业厅 ;

select * FROM biaoming WHERE TIMESTAMPDIFF(SECOND ,CREATE_TIME,now() ) > 3*60*60

问题描述 查询数据库表中最近7天的记录 select count(*),date(create_time) as date from task where datediff(now(),create_time)<=6 group by day(create_time); 但是发现某一天没有数据,结果中没有显示当天(2017-08-28)的数据 解决思路 思路一: 可以在自己的程序中做额外的补零处理 思路二: 构建一个最近七天的结果集,然后和查询的结果集合做left join(本文采用第二种方式)

如果你要严格要求是某一年的,那可以这样 查询一天: select * from table where to_days(column_time) = to_days(now()); select * from table where date(column_time) = curdate(); 查询一周: DAY) <= date(column_time); 查询一个月: select * from table where DATE_SUB(CURDATE(), INTERVAL 1 MONTH

select * from wap_content where week(created_at) = week(now) 如果你要严格要求是某一年的,那可以这样 查询一天: select * from table where to_days(column_time) = to_days(now());select * from table where date(column_time) = curdate(); 查询一周: select * from table where DATE_SUB(C

 date_format(date,'%H') = 8 and date_format(date,'%i') = 30   SELECT * FROM `t_pda_trucklog` WHERE DATE_FORMAT(operatetime,'%H') = 9 AND DATE_FORMAT(operatetime,'%i') = 42 AND DATE_FORMAT(operatetime,'%s') = 12       http://www.w3school.com.cn/sql/fu

SELECT ID,SERVICE FROM new_schedules_spider_full WHERE SERVICE = 'WSA2' and date_sub(curdate(), interval 7 day) <= date(CREATE_TIME);

语法:select 列名, length(列名) from 表名where length(列名) = ( select max(length(列名)) from 表名); 实例:select project_num, length(project_num) from project_infor_table where length(project_num) = ( select max(length(project_num)) from project_infor_table);

mysql查询当天的所有信息: select * from test where year(regdate)=year(now()) and month(regdate)=month(now()) and day(regdate)=day(now())这个有一些繁琐,还有简单的写法: select * from table where date(regdate) = curdate();另一种写法没测试过查询当天的记录 select * from hb_article_view where TO

mysql查询当天的所有信息: select * from test where year(regdate)=year(now()) and month(regdate)=month(now()) and day(regdate)=day(now())这个有一些繁琐,还有简单的写法: select * from table where date(regdate) = curdate();另一种写法没测试过查询当天的记录 select * from hb_article_view where TO

mysql查询当天的所有信息: SELECT * FROM 表名 WHERE year(时间字段名)=year(now()) and month(时间字段名) = month(now()) and day(时间字段名) = day(now()); 这个有一些繁琐,还有简单的写法: SELECT * FROM 表名 WHERE date(时间字段名) = curdate(); 另一种写法没测试过 查询当天的记录 SELECT * FROM 表名 WHERE TO_DAYS(时间字段名) = TO_

SELECT t1. HOUR HOUR, COUNT(t2. HOUR) count FROM ( SELECT DATE_FORMAT( HOUR), '%y-%m-%d %H' ) HOUR FROM ( SELECT @cdate := DATE_ADD( DATE_FORMAT(NOW(), '%y-%m-%d %H'), INTERVAL HOUR ) FROM city ) t0 LIMIT ) t1 LEFT JOIN ( SELECT DATE_FORMAT(visit_tim

需求:查询最近三个月的每一天的业绩总和 因为最近三个月每个月的天数是不一样,所以不能用这篇文章:Mysql查询最近30天的数据(每天的业绩总和数据) 介绍的用固定多少天去查数据.需要一个新方法. 一.生成每个月的每一天的时间序列 SELECT ADDDATE(y.first, x.d - 1) as d FROM (SELECT 1 AS d UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 U

以下代码中times为时间字段,类型为datetime 1.查询大于times十五分钟的数据 //大于号后面都是获取times十五分钟后的时间select*from table where now() >SUBDATE(times,interval -15 minute);select*from table where now() > SUBDATE(times,interval -900 second);select*from table where now() > date_add(

SELECT HOUR(e.time) as Hour,count(*) as Count FROM error_log e WHERE e.date = '2017-09-02' GROUP BY HOUR(e.time) ORDER BY Hour(e.time); 下面是查询结果截图 在另一篇文章里,我总结了查询每半小时统计一次的方法.Mysql 查询一天中每半小时记录的数量

MySQL中如何查询最近24小时. where visittime >= NOW() - interval 1 hour; 昨天. where visittime between CURDATE()-interval 1 day and CURDATE(); 最近7天. where visittime between CURDATE()-interval 7 day ; 上周. where week(visittime)= week(CURDATE()-interval 7 day ) and