你可以用 replace , substr 以及 instr 合称为：

SELECT replace( 'aaa@test.com',
         substr('aaa@test.com',instr('aaa@test.com','@'),length('aaa@test.com'))
                              ,'@other.net') as result_str;

 result_str
-------------
aaa@other.net

或者从你的table上( tab )有一列称为 email :

select replace(email,substr(email,instr(email,'@'),length(email)),'@other.net') result_str
  from tab;

result_str
-------------
aaa@other.net
bbb@other.net
ccc@other.net

rextester演示

使用 SUBSTRING_INDEX 以及 concat 功能

select concat(SUBSTRING_INDEX("aaa@test.com", "@", 1),'@other.net')

输出 aaa@other.net 那你的专栏呢 user_email 会的

select concat(SUBSTRING_INDEX(user_email, "@", 1),'@other.net')

select concat
        (substring
          ('bbb@test.com',1,char_length
            ('bbb@test.com')-
            char_length
              (substring_index
                ('bbb@test.com','.',-1))),'net') x;

| x            |
| ------------ |
| bbb@test.net |

db fiddle视图

我找到了以下解决方案，似乎可以做到这一点：

UPDATE table_name SET user_email = REPLACE(user_email, '@test.com', '@other.net');

使用替换功能
演示

select replace(name,substring(name,position('@' in name),length(name)-position('@' in name)+1),'@other.net')

select replace('aaa@test.com',substring('aaa@test.com',position('@' in 'aaa@test.com'),
length('aaa@test.com')-position('@' in 'aaa@test.com')+1),'@other.net')

输出：

val             n
aaa@test.com    aaa@other.net