```sql
use teaching2021;
select id,idcard,if(substring(idcard,17,1)%2=1,'男'，'女')as'性别'
from identity
```)
题目要求根据身份证号来辨别性别
substring(idcard,17,1)是字符串切割，切割结果也是字符串，但是，%2是数值运算，类型不匹配。你可以改为
select id,idcard,if(substring(idcard,17,1) in('1','3','5','7','9'),'男'，'女') as '性别'
from `teaching2021`.identity
里面的"use teaching2021;"也不需要了