select * from student where name like '%[^ja,as,on]%'

来源：学生作业帮编辑：作业帮分类：综合作业时间：2024/07/18 12:16:08

select * from student where name like '%[^ja,as,on]%'
这是我自己建的一个表,查询的结果

oracle 的正则表达式
再问： ^A匹配A是什么意思啊？我才刚学数据库，很多都不懂
再答：如果A是字符串的第一个字符，^A 匹配 A 你仔细看看那个文档我感觉很好正则表达式你以前听过吗？ js里也有 java里面也有很多领域都有用到其实正则表达式就是为了描述字符串特征 ^A就是说以A开头的字符串这是其中的一个特征继续还可以追加更多特征这样描述出来的特征就可以匹配到相应的字符串了

select * from student where name like '%[^ja,as,on]%' select id,name from student where id in ( select id from sco select * from teacher where s_id in(select s_id from student select s.name,s.course,s.grade from student as s inner join SQL语句:SELECT * from stu where name like "%伟%";的作用是:_________ ms sql :select * from (select *from b) as xxx where 和 select $query = "SELECT * FROM student_info WHERE name = '".$login_ SELECT name FROM [BookStore].[dbo].[user] WHERE userid=?AND SQL语句理解 select * from B where (select count(1) as num from A select count(*) as total from ddcm_news where key=a if exists (select 1 from #tmp_dj_xsg212 WHERE shangplx LIKE select *from A where A.a like '%1%'||'%'