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mysql中on,in,As,whErE如何用,意思是什么?

Where查询条件,on内外连接时候用,as作为别名,in查询某值是否在某条件里

as不是给表里的字段取别名,而是给查询的结果字段取别名。其目的是让查询的结果展现更符合人们观看习惯,在多张表查询的时候可以直接的区别多张表的同名的字段。 比如: 1、selec name as “姓名” ,sex as "性别" from user 2、select u.name as “...

select count(*) as amount from user order by UID asc 以降序的方式(asc)查询user表里字段为UID的值的数目(count(*)),返回的结果存在别名 amount 里.

SELECT t1.a, (SELECT group_concat(name) FROM t2 WHERE id IN ( 1,2,3 ) ) AS t2n1, (SELECT group_concat(name) FROM t2 WHERE FIND_IN_SET(id ,t1.a) ) AS t2n2 FROM t1 WHERE t1.id=1

mysql中,inner join和where的结合问题 1 2 select u.id,username,tx,postnum,replynum,nb,regtime,points,n.title,n.contents from userinfo as u inner join note as n on u.id=n.userid where n.id='$edit'

as不是给表里的字段取别名,而是给查询的结果字段取别名。其目的是让查询的结果展现更符合人们观看习惯,在多张表查询的时候可以直接的区别多张表的同名的字段。 比如: 1、selec name as “姓名” ,sex as "性别" from user 2、select u.name as “...

select *,aaa as a from table1 where id = 1; 我试过了好使。不要把*放到查询字段的最后,否则不好使。 还有啊,既然语句都写出来了,就先在mysql中执行一下啊,不就知道好不好用了吗? 希望可以帮到你

COUNT() 函数返回匹配指定条件的行数 COUNT(column_name) 函数返回指定列的值的数目(NULL 不计入) 我们拥有下列 "Orders" 表: O_Id OrderDate OrderPrice Customer 1 2008/12/29 1000 Bush 2 2008/11/23 1600 Carter 3 2008/10/05 700 Bush 4...

方法一:SELECT CAST('123' AS SIGNED); 方法二:SELECT CONVERT('123',SIGNED); 方法三:SELECT '123'+0;

SELECT d.*,pm.label, ( SELECT COUNT(*) from (select js,jifen from deal where uid=m.id and time>DATE_FORMAT(time,'$Y-$d-$m') and state!=1 and state!=6 GROUP BY stoptime ) ) as jys from deal as d,members as m,pricemsg as pm where...

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