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2022-08-17 第二小组张鑫学习笔记

2022-08-17 19:05:20 作者：互联网

实训三十九天

1.学习重点

1.子查询
2.案例
3.日期格式

2.学习心得

今天主要的学习内容就是练习数据库的查询操作，三十个查询的问题我都能独立完成了，感觉很有成就感，开头打好了，接下来要坚持住

3.学习内容

DQL查询语言

子查询

按照结果集的行列数不同，子查询可以分为以下几类

标量子查询：结果集只有一行一列
列子查询：结果集有一列多行
行子查询：结果集有一行多列
表子查询：结果集是多行多列

查询比小虎年龄大的所有学生

标量子查询

SELECT * FROM student WHERE age >( SELECT age FROM student WHERE NAME = '小虎' );

查询有一门学科分数大于90的学生信息

列子查询

SELECT * FROM student WHERE	id IN ( SELECT s_id FROM scores WHERE score > 90 );

查询男生且年龄最大的学生

行子查询

SELECT * FROM student WHERE (age,gender) =( SELECT MAX( age ),gender FROM student GROUP BY gender HAVING gender = '男' );

总结：

where 型子查询，如果是where列 =（内层sql）,则内层的sql返回的必须是单行单列，单个值

where 型子查询，如果是where（列1，列2） = （内层sql），内层的sql返回的必须是单列，可以是多行

取排名数学成绩前五的学生，正序排列

表子查询

SELECT
	* 
FROM
	(
	SELECT
		s.*,
		sc.score,
		c.NAME 科目 
	FROM
		student s
		LEFT JOIN scores sc ON s.id = sc.s_id
		LEFT JOIN course c ON c.id = sc.c_id 
	WHERE
		c.NAME = '数学' 
	ORDER BY
		score DESC 
		LIMIT 5 
	) t 
WHERE
	t.gender = '男';

经验分享
1.分析需求
2.拆步骤
3.分步写sql
4.整合拼装sql

查询每个老师的代课数

SELECT t.id,t.name,(SELECT COUNT(*) FROM course c WHERE c.id = t.id) as 代课的数量 FROM teacher t;

---

SELECT t.id,t.name,COUNT(*) '代课的数量'
FROM teacher t LEFT JOIN course c
ON c.t_id =t.id GROUP BY t.id,t.name

exists

SELECT
	* 
FROM
	teacher t 
WHERE
	EXISTS (
	SELECT
		* 
	FROM
		course c 
	WHERE
		c.t_id = t_id
		)

---

SELECT
	t.*,
	c.`name` 
FROM
	teacher t
	INNER JOIN course c ON t.id = c.t_id

总结：如果一个需求可以不用子查询，尽量不使用（sql可读性太低）

需求

-- 1.查询'01'号学生的姓名和各科成绩 
SELECT
	s.id sid,
	s.`name` sname,
	c.`name` cname,
	sc.score 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	s.id = 1;
	

---

-- 2.查询各个学科的平均成绩和最高成绩**
SELECT
	c.id,
	c.`name`,
	AVG( sc.score ),
	max( sc.score ) 
FROM
	course c
	LEFT JOIN scores sc ON c.id = sc.c_id 
GROUP BY
	c.id,
	c.`name`;
	

---

-- 3.查询每个同学的最高成绩和科目名称****（明天说，子查询）
SELECT
	t.id,
	t.NAME,
	c.id,
	c.NAME,
	r.score 
FROM
	(
	SELECT
		s.id,
		s.NAME,(
		SELECT
			max( score ) 
		FROM
			scores r 
		WHERE
			r.s_id = s.id 
		) score 
	FROM
		student s 
	) t
	LEFT JOIN scores r ON r.s_id = t.id 
	AND r.score = t.score
	LEFT JOIN course c ON r.c_id = c.id;
	

---

-- 4.查询所有姓张的同学的各科成绩**
SELECT
	s.id,
	s.`name`,
	c.`name` cname,
	sc.score 
FROM
SELECT
	s.id,
	s.`name`,
	c.`name` cname,
	sc.score 
FROM
	student s
	LEFT JOIN scores sc ON sc.s_id = s.id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	s.`name` LIKE '张%';
	

---

-- 5.查询每个课程的最高分的学生信息*****（明天说，子查询）
SELECT
	* 
FROM
	student s 
WHERE
	id IN (
	SELECT DISTINCT
		r.s_id 
	FROM
		(
		SELECT
			c.id,
			c.NAME,
			max( score ) score 
		FROM
			student s
			LEFT JOIN scores r ON r.s_id = s.id
			LEFT JOIN course c ON c.id = r.c_id 
		GROUP BY
			c.id,
			c.NAME 
		) t
		LEFT JOIN scores r ON r.c_id = t.id 
	AND t.score = r.score 
	)
	

---

-- 6.查询名字中含有'张'或'李'字的学生的信息和各科成绩
SELECT
	s.id,
	s.NAME sname,
	sc.score,
	c.NAME 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id 
WHERE
	s.NAME LIKE '%张%' 
	OR s.NAME LIKE '%李%';
	

---

-- 7.查询平均成绩及格的同学的信息。（子查询）
SELECT
	* 
FROM
	student 
WHERE
	id IN (
	SELECT
		sc.s_id 
	FROM
		scores sc 
	GROUP BY
		sc.s_id 
	HAVING
	avg( sc.score ) >= 70 
	)
	

---

-- 8.将学生按照总分数进行排名。（从高到低）
SELECT
	s.id,
	s.NAME,
	sum( sc.score ) score 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id 
GROUP BY
	s.id,
	s.NAME 
ORDER BY
	score DESC,
	s.id ASC;
	

---

-- 9.查询数学成绩的最高分、最低分、平均分。
SELECT
	c.NAME,
	max( sc.score ),
	min( sc.score ),
	avg( sc.score ) 
FROM
	course c
	LEFT JOIN scores sc ON c.id = sc.c_id 
WHERE
	c.NAME = '数学';
	

---

-- 10.将各科目按照平均分排序。
SELECT
	c.id,
	c.NAME,
	avg( sc.score ) score 
FROM
	course c
	LEFT JOIN scores sc ON c.id = sc.c_id 
GROUP BY
	c.id,
	c.NAME 
ORDER BY
	score DESC;
	

---
-- 11.查询老师的信息和他所带的科目的平均分
SELECT
	t.*,
	c.`name`,
	avg( sc.score ) 平均分 
FROM
	teacher t
	LEFT JOIN course c ON t.id = c.t_id
	LEFT JOIN scores sc ON sc.c_id = c.id 
GROUP BY
	t.id,
	t.`name`
---

-- 12.查询被"Tom"和"Jerry"教的课程的最高分和最低分
SELECT
	t.NAME,
	c.NAME,
	max( sc.score ),
	min( sc.score ) 
FROM
	teacher t
	LEFT JOIN course c ON t.id = c.t_id
	LEFT JOIN scores sc ON sc.c_id = c.id 
GROUP BY
	t.id,
	t.NAME 
HAVING
	t.NAME IN (
	'Tom',
	'Jerry')
---

-- 13.查询每个学生的最好成绩的科目名称（子查询）
SELECT
 t.NAME 学生姓名,
 c.NAME 科目名称,
 r.score 成绩
FROM
 (
 SELECT
  s.id,
  s.NAME,(
  SELECT
   max( score ) 
  FROM
   scores r 
  WHERE
   r.s_id = s.id 
  ) score 
 FROM
  student s 
 ) t
 LEFT JOIN scores r ON r.s_id = t.id 
 AND r.score = t.score
 LEFT JOIN course c ON r.c_id = c.id;
---

-- 14.查询所有学生的课程及分数
SELECT
	s.`name` 姓名,
	c.`name` 课程,
	sc.score 分数 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id
---

-- 15.查询课程编号为1且课程成绩在60分以上的学生的学号和姓名（子查询）
SELECT
s.id 学号 ,
s.`name` 姓名
FROM
(
SELECT
sc.s_id
FROM
scores sc
WHERE sc.c_id = 1 and sc.score >=60
) t
LEFT JOIN student s ON  t.s_id=s.id
---

-- 16. 查询平均成绩大于等于85的所有学生学号、姓名和平均成绩
SELECT
	s.id 学号,
	s.`name` 姓名,
	AVG( sc.score ) 平均成绩
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id 
GROUP BY
	sc.s_id 
HAVING
	AVG( sc.score ) >= 70
---

-- 17.查询有不及格课程的学生信息
SELECT
	s.id 学号,
	s.`name` 姓名,
	c.`name` 不及格学科,
	sc.score 分数 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id 
WHERE
	sc.score <= 60
---

-- 18.查询每门课程有成绩的学生人数
SELECT sc.c_id 课程号,c.`name` 课程名 ,COUNT(*) '有成绩的数量'
FROM scores sc 
LEFT JOIN course c
ON sc.c_id = c.id GROUP BY sc.c_id
---

-- 19.查询每门课程的平均成绩，结果按照平均成绩降序排列，如果平均成绩相同，再按照课程编号升序排列
SELECT
	c.`name` 课程名,
	AVG( sc.score ) 
FROM
	scores sc
	LEFT JOIN course c ON sc.c_id = c.id 
GROUP BY
	sc.c_id 
ORDER BY
	AVG( sc.score ) DESC,
	c.id
---

-- 20.查询平均成绩大于60分的同学的学生编号和学生姓名和平均成绩
SELECT
s.id 学号,
s.`name` 姓名,
AVG(sc.score)
FROM student s
LEFT JOIN scores sc ON s.id = sc.s_id
GROUP BY sc.s_id
HAVING AVG(sc.score) >60
---
-- 21.查询有且仅有一门课程成绩在80~90之间的学生信息
SELECT
	s.id 学号,
	s.`name` 姓名,
	c.`name` 科目,
	sc.score 成绩 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id 
WHERE
	sc.score >= 80 
GROUP BY
	sc.s_id 
HAVING
	COUNT( score ) =1
---

-- 22.查询只有三门课程的学生的学号和姓名
SELECT
	s.id 学号,
	s.`name` 姓名
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
GROUP BY
	sc.s_id 
HAVING
	COUNT(sc.s_id) = 3
 
---

-- 23.查询有不及格课程的课程信息
SELECT
 c.`name` 不及格学科
FROM
 course c
 LEFT JOIN scores sc ON sc.c_id = c.id 
 WHERE sc.score <= 60
 GROUP BY c.`name`

---

-- 24.查询至少选择四门课程的学生信息
SELECT
	s.id 学号,
	s.`name` 姓名
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
GROUP BY
	sc.s_id 
HAVING
	COUNT(sc.s_id) >=4

---

-- 25.查询没有学全所有课程的同学的信息
SELECT
	s.id 学号,
	s.`name` 姓名
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
GROUP BY
	sc.s_id 
HAVING
	COUNT(sc.s_id) <5
---
-- 26.查询选全所有课程的同学的信息
SELECT
	s.id,
	s.NAME,
	count(*) number 
FROM
	student s
	LEFT JOIN scores r ON s.id = r.s_id 
GROUP BY
	s.id,
	s.NAME 
HAVING
	number = ( SELECT count(*) FROM course );

---

-- 27.查询各学生都选了多少门课
SELECT
	s.id 学号,
	s.`name` 姓名,
	COUNT(*) 选课数 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id 
GROUP BY
	sc.s_id
---

-- 28.查询课程名称为"java"，且分数低于60分的学生姓名和分数
SELECT
	s.id 学号,
	s.`name` 姓名,
	sc.score java分数 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id 
WHERE
	c.`name` = 'java' 
	AND sc.score < 60
---

-- 29.查询学过"Tom"老师授课的同学的信息
SELECT
	s.id 学号,
	s.`name` 姓名 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON sc.c_id = c.id
	LEFT JOIN teacher t ON c.id = t.id 
WHERE
	t.`name` = 'Tom'
---

-- 30.查询没学过"Tony"老师授课的学生信息
SELECT
	* 
FROM
	student 
WHERE
	id NOT IN (
	SELECT DISTINCT
		s.id 
	FROM
		student s
		LEFT JOIN scores r ON r.s_id = s.id
		LEFT JOIN course c ON c.id = r.c_id
		LEFT JOIN teacher t ON t.id = c.t_id 
	WHERE
	t.NAME = 'Tom' 
	)

日期格式

格式	描述
%a	缩写的星期名
%b	缩写月名
%c	月，数值
%D	带有英文前缀的月中的天
%d	月的天，数值（00-31）
%e	月的天，数值（0-31）
%f	微秒
%H	小时（00-23）
%h	小时（01-12）
%I	小时（01-12）
%i	分钟，数值（00-59）
%j	年的天（001-366）
%k	小时（0-23）
%l	小时（0-12）
%M	月名
%m	月，数值（00-12）
%p	AM或PM
%r	时间，12-小时（hh:mm:ss AMh或PM）
%S	秒 (00-59)
%s	秒 (0-59)
%T	时间，24-小时（hh:mm:ss）
%U	周（00-53）星期日是第一天
%u	周（00-53）星期一是第一天
%W	星期名
%Y	年，2022
%y	年，22

标签：JOIN,17,张鑫,08,score,sc,id,SELECT,LEFT
来源： https://www.cnblogs.com/zxscj/p/16596402.html

2022-08-17 第二小组 张鑫 学习笔记