Mysql SQL中靠近和的语法不正确

我需要修复以下查询，但我看不到错误。你能告诉我怎么了吗

select *
from(
     SELECT a.fisrt_name,a.last_name,a.job_id
     FROM employees A
     WHERE A.salary >(Select avg(salary) from employees)) and a.salary < select (max(salary) from employees)
) b
where b.job_id=SY_ANA;

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你需要在员工之后立即删除这个小家伙

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where子句条件字符串应被引用

 select *
    from(
         SELECT a.fisrt_name,a.last_name,a.job_id
         FROM employees a
         WHERE a.salary >(Select avg(salary) from employees)
             and a.salary < (select max(salary) from employees)
    ) b
    where b.job_id='SY_ANA';

我去掉了额外的括号

select b.*
from(
     SELECT a.fisrt_name,a.last_name,a.job_id
     FROM employees A
     WHERE A.salary >(Select avg(salary) from employees) and a.salary < (select max(salary) from employees)
) b
where b.job_id="SY_ANA";

内部查询中缺少括号，where条件中缺少引号

    select *
from(
     SELECT a.fisrt_name,a.last_name,a.job_id
     FROM employees A
     WHERE A.salary >(Select avg(salary) from employees)) and a.salary < (select max(salary) from employees)
) b
where b.job_id='SY_ANA';

从员工中选择maxsalary是从员工中选择maxsalary

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在where

中也缺少qoutes请尝试在a.salary>Select avgsalary中的>之后添加空格使用avg函数时您需要添加group by子句您的查询只剩下括号-这是不平衡的。First_name-您可能希望尽快解决此问题，而不是稍后解决。请参见：不要强迫人们找出差异。解释并强调你改变了什么以及为什么。

    select *
from(
     SELECT a.fisrt_name,a.last_name,a.job_id
     FROM employees A
     WHERE A.salary >(Select avg(salary) from employees)) and a.salary < (select max(salary) from employees)
) b
where b.job_id='SY_ANA';