MySQL 类似 SQL 的语句“%[^0-9]%”不起作用
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SQL like statement "%[^0-9]%" not working
提问by Jia-Luo
I have a table.
我有一张桌子。
USER TABLE Columns:
user_name (varchar)
age (int)
address (varchar)
telephone (varchar)
I want to seach for user whose address field contains digits.
我想搜索地址字段包含数字的用户。
I tried the following sql statement in MySQL Workbench:
我在 MySQL Workbench 中尝试了以下 sql 语句：
SELECT * FROM user WHERE address LIKE '%[^0-9]%';
However, it did not work. Can anyone help me out here?
但是，它不起作用。有人可以帮我从这里出去吗？
Thanks in advance.
提前致谢。
采纳答案by Brad Christie
回答by gkovacs90
You should remove the ^character from the brackets:
您应该^从括号中删除字符：
LIKE '%[0-9]%'
LIKE '%[0-9]%'
It doesn't want to be a regex, it is the [charlist]wildcard
它不想成为正则表达式，它是[charlist]通配符
回答by Jrf
select * from user where first_name REGEXP '^[0-9].*'
This gives you all records in your user table where the first name starts with numbers 0-9.
这将为您提供用户表中的所有记录，其中名字以数字 0-9 开头。
回答by Sushanth --
'%[^0-9]%'
Will search for the column that has %[^0-9]%as text
将搜索具有%[^0-9]%文本的列
It won't do a Regex on it
它不会对它做正则表达式
Try
尝试
'.[^0-9].' // That performs regex
