小徐帶你瞭解sql99總結以及相關練習題重點難點

內連接：

內連接特點

外連接

外連接特點

交叉連接

交叉連接特點:

	類似於笛卡爾乘積

子查詢 含義

分類：

示例

where或having後面1：標量子查詢

查詢最低工資的員工姓名和工資

SELECT last_name, salary FROM employees WHERE salary =( SELECT min( salary ) FROM employees );

查詢所有是領導的員工姓名 重點

SELECT
	last_name 
FROM
	employees 
WHERE
	manager_id IN ( SELECT employee_id FROM employees );

分頁查詢應用場景

語法：

子查詢經典案例 重點難點

案例一：查詢工資最低的員工信息last_name，salary

SELECT last_name, salary FROM employees WHERE salary =( SELECT MIN( salary ) FROM employees );

案例二：查詢平均工資最低的部門信息 重點難點

#第一種方式使用limit
SELECT * FROM departments WHERE department_id =( SELECT department_id FROM employees GROUP BY department_id ORDER BY avg( salary ) ASC LIMIT 1 );
#第二種方式
SELECT
	d.* 
FROM
	departments d 
WHERE
	d.department_id =(
	SELECT
		department_id 
	FROM
		employees 
	GROUP BY
		department_id 
	HAVING
		avg( salary )=(
		SELECT
			min( ag ) 
		FROM
		( SELECT AVG( salary ) ag, department_id FROM employees GROUP BY department_id ) ag_dep 
	));

案例三：查詢平均工資最低的部門信息和該部門的平均工資 重點難點

SELECT
	* 
FROM
	( SELECT AVG( salary ), department_id id FROM employees GROUP BY department_id ORDER BY AVG( salary ) ASC LIMIT 0, 1 ) avg_temp
	INNER JOIN departments ON departments.department_id = avg_temp.id;

案例四：查詢平均工資最高的job信息

SELECT * FROM jobs WHERE job_id =( SELECT job_id FROM employees GROUP BY job_id ORDER BY AVG( salary ) DESC LIMIT 1 );

案例五：查詢平均工資高於公司平均工資的部門有哪些？

SELECT AVG( salary ), department_id FROM employees GROUP BY department_id HAVING AVG( salary ) >( SELECT AVG( salary ) FROM employees );

案例六：查詢公司中所有manager的詳細信息

SELECT
	* 
FROM
	employees e 
WHERE
	e.employee_id IN ( SELECT DISTINCT manager_id FROM employees );

案例七：各個部門中，最高工資中最低的那個部門的最低工資多少

SELECT min( salary ), department_id FROM employees WHERE department_id =( SELECT department_id FROM employees GROUP BY department_id ORDER BY MAX( salary ) LIMIT 1 );

案例八：查詢平均工資最高的部門的manager的詳細信息，last_name，department_id，email，salary

SELECT
	last_name,
	d.department_id,
	email,
	salary 
FROM
	employees e
	INNER JOIN departments d ON e.employee_id = d.manager_id 
WHERE
	d.department_id =(
	SELECT
		department_id 
	FROM
		employees 
	GROUP BY
		department_id 
	ORDER BY
		AVG( salary ) DESC 
	LIMIT 1 
	);

練習題：

一：查詢每個專業的學生人數

SELECT
	majorid,
	COUNT(*) 
FROM
	student 
GROUP BY
	majorid;

二：查詢參加考試的學生中，每個學生的平均分，最高分

SELECT
	AVG( score ),
	MAX( score ),
	studentno 
FROM
	result 
GROUP BY
	studentno;

三：查詢姓張的每個學生的最低分大於60的學號，姓名

SELECT
	s.studentno,
	s.studentname,
	MIN( score ) 
FROM
	student s
	JOIN result r ON s.studentno = r.studentno 
WHERE
	s.studentname LIKE '%張%' 
GROUP BY
	s.studentno 
HAVING
	MIN( score )> 60;

四：查詢生日在1988-1-1後的學生姓名，專業名稱

SELECT
	studentname,
	majorname 
FROM
	student s
	JOIN major m ON s.majorid = m.majorid 
WHERE
	datediff ( borndate, '1988-1-1' )> 0;

五：查詢每個專業的男生人數和女生人數分別是多少

#方式一
SELECT COUNT(*) '個數',sex,majorid
FROM student GROUP BY sex,majorid;
#方式二
SELECT
	majorid,
	( SELECT COUNT(*) FROM student WHERE sex = '男' AND majorid = s.majorid ) '男',
	( SELECT COUNT(*) FROM student WHERE sex = '女' AND majorid = s.majorid ) '女' 
FROM
	student s 
GROUP BY
	majorid;

案例六：查詢專業和張翠山一樣的學生的最低分

SELECT MIN( score ) FROM result WHERE studentno IN ( SELECT studentno FROM student WHERE majorid =( SELECT majorid FROM student WHERE studentname = '張翠山' ));

案例七：查詢大於60分的學生的姓名，密碼，專業名

SELECT
	studentname,
	loginpwd,
	majorname 
FROM
	student s
	JOIN major m ON s.majorid = m.majorid
	JOIN result r ON s.studentno = r.studentno 
WHERE
	r.score > 60;

案例八：按郵箱位數分組，查詢每組的學生個數

SELECT
	COUNT(*),
	LENGTH( email ) 
FROM
	student 
GROUP BY
	LENGTH( email );

案例九：查詢學生名，專業名，分數

SELECT
	studentname,
	score,
	majorname 
FROM
	student s
	JOIN major m ON m.majorid = s.majorid
	JOIN result r ON s.studentno = r.studentno;

案例十：查詢哪個專業沒有學生，分別用左連接和右連接實現

#左連接
SELECT
	m.majorid,
	m.majorname,
	s.studentno 
FROM
	major m
	LEFT JOIN student s ON m.majorid = s.majorid 
WHERE
	s.studentno IS  NULL;
#右連接
	SELECT
	m.majorid,
	m.majorname,
	s.studentno 
FROM
	student s
	RIGHT  JOIN major m	ON m.majorid = s.majorid 
WHERE
	s.studentno IS  NULL;

案例十一：查詢沒有成績的學生人數

SELECT COUNT(*),score
FROM student s
LEFT JOIN result r
on r.studentno=s.studentno WHERE r.id is null;