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
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)
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
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 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_
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