Mysql中的嵌套子查询问题

--查询公司之中工资最低的雇员的完整信息
SELECT * 
FROM emp e
WHERE e.sal=(
  SELECT MIN(sal) 
  FROM emp);

-- 查询出基本工资比ALLEN低的全部雇员信息
SELECT * 
FROM emp e
WHERE e.sal<(
 SELECT sal 
 FROM emp 
 WHERE ename='ALLEN'
);

--查询基本工资高于公司平均薪金的全部雇员信息
SELECT *
FROM emp e
WHERE e.sal>(
SELECT AVG(sal)
FROM emp);

--查找出与ALLEN从事同一工作，并且基本工资高于雇员编号为7521的全部雇员信息,
SELECT *
FROM emp e
WHERE e.job=(
  SELECT job
  FROM emp 
  WHERE ename='ALLEN') 
  AND e.sal>(
  SELECT sal
  FROM emp 
  WHERE empno=7521);

SELECT *
FROM emp e
WHERE (e.job,e.sal) = (
  SELECT job,sal
  FROM emp 
  WHERE ename='SCOTT')
  AND ename<>'SCOTT';

  --查询与雇员7566从事同一工作且领导相同的全部雇员信息
SELECT *
FROM emp e
WHERE (e.job,e.mgr) =(
        SELECT job,mgr
        FROM emp
      WHERE empno=7566
);

--查询与ALLEN从事同一工作且在同一年雇佣的全部雇员信息（包括ALLEN）
SELECT *
FROM emp e
WHERE (e.job,to_char(e.hiredate,'yyyy'))=(
           SELECT job,to_char(hiredate,'YYYY')
            FROM emp
            WHERE ename='ALLEN'
);

--查询出与每个部门中最低工资相同的全部雇员信息
SELECT *
FROM emp e
WHERE e.sal IN(
     SELECT MIN(sal)
    FROM emp
    GROUP BY deptno
);

--查询出不与每个部门中最低工资相同的全部雇员信息
SELECT *
FROM emp e
WHERE e.sal NOT IN(
     SELECT MIN(sal)
    FROM emp
    GROUP BY deptno
);

第一种方式：能够使用HAVING字句实现

SELECT job,AVG(sal)
FROM emp
GROUP BY job
HAVING AVG(sal)>2000;

第二种方式：使用FROM子查询

SELECT job,avgsal
FROM (
  SELECT job,AVG(sal) AS avgsal
    FROM emp
      GROUP BY job) temp
WHERE temp.avgsal>2000;

SELECT e1.empno,e1.ename,e1.comm,e2.sal
FROM emp e1,emp e2
WHERE e1.comm IS NOT NULL AND e1.comm<e2.sal
AND e1.empno=e2.empno;

SELECT d.*,(
    SELECT COUNT(*)
    FROM employees
    WHERE employees.`department_id`=d.department_id
)

FROM
    departments d;

SELECT department_name
FROM
employees e
INNER JOIN
    departments d
ON
    e.`department_id`=d.`department_id`
WHERE
    e.`employee_id`=102;

SELECT (
SELECT department_name
FROM
    employees e
INNER JOIN
    departments d
ON
    e.`department_id`=d.`department_id`
WHERE
    e.`employee_id`=102
);
`department_id`=d.`department_id`
WHERE
    e.`employee_id`=102;

SELECT (
SELECT department_name
FROM
    employees e
INNER JOIN
    departments d
ON
    e.`department_id`=d.`department_id`
WHERE
    e.`employee_id`=102
);