mysql查询按照in里面的数据排序

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

SET FOREIGN_KEY_CHECKS=0; -- Table structure for t_user -- ---------------------------- DROP TABLE IF EXISTS t_user; CREATE TABLE t_user ( userId bigint(20) NOT NULL, fullName varchar(64) NOT NULL, userType varchar(16) NOT NULL, addedTime datetime NO

在项目当中用到Sphinx的时候,很多人遇到了这样的问题:使用mysql+Sphinx检索出了相关度的ID后,如何按照指定ID在Mysql中进行排序呢?这里是我在项目中的解决方法: 1 SELECT * FROM documents WHERE id IN (5,3,6,1) ORDER BY FIELD(id,5,3,6,1); 或者: 1 SELECT * FROM documents WHERE id IN (5,3,6,1) ORDER BY substring_index((id,5,

今天的数据 SELECT * FROM `statements` WHERE DATE_FORMAT( FROM_UNIXTIME( `happen_time` ) , '%Y-%m-%d' ) = DATE_FORMAT( NOW( ) , '%Y-%m-%d' ) 昨天的数据 SELECT * FROM statements where `happen_time`>UNIX_TIMESTAMP(DATE_SUB(CURDATE(), INTERVAL 1 DAY))

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

方法一: SELECT sc.*,c.`Cname` ,COUNT(sc.C) FROM sc LEFT JOIN sc aON sc.C = a.C AND sc.score >= a.scoreLEFT JOIN course cON sc.C = c.CWHERE sc.C = c.CGROUP BYsc.C,sc.S,sc.scoreHAVING COUNT(sc.C)>=4ORDER BY a.C,a.score DESC 解析: 先copy 一个sc 表中的数据,与原sc 进行对比

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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

1.用PHP代码循环把要插入的数据存入文件 //随机字符串 function getRandChar($length){ $str = null; $strPol = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz"; $max = strlen($strPol)-1; for($i=0;$i<$length;$i++){ $str.=$strPol[rand(0,$max)];//rand($min