mysql----导入导出、练习题

一、导入导出

1、Navicat软件使用

2、命令

导出现有数据库数据：

mysqldump -u用户名 -p密码数据库名称 >导出文件路径 # 结构+数据
mysqldump -u用户名 -p密码 -d 数据库名称 >导出文件路径 # 结构

导入现有数据库数据：

mysqldump -uroot -p密码数据库名称 < 文件路径

二、练习题

1、查询平均成绩大于60分的同学的学号和平均成绩；

思路：根据学生分组，使用avg获取平均值，通过having对avg进行筛选

SELECT student_id, AVG(num) FROM score GROUP BY student_id HAVING AVG(num) > 60

思考题：显示学生名字

SELECT
score.student_id,
student.sname,
AVG(score.num)
FROM
score
LEFT JOIN student on score.student_id=student.sid
GROUP BY
student_id
HAVING
AVG(num) > 60

2、查询所有同学的学号、姓名、选课数、总成绩；

SELECT
    score.student_id,
    student.sname,
    COUNT(score.course_id),
    SUM(score.num)
FROM
    score
LEFT JOIN student ON score.student_id = student.sid
GROUP BY
    student_id

3、查询姓“李”的老师的个数；

SELECT
	COUNT(tname)
FROM
	teacher
WHERE
	tname LIKE '李%'

4、查询没学过“李平”老师课的同学的学号、姓名；

5、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

SELECT
	b.student_id,
  student.sname,
	COUNT(b.course_id) AS course_num
FROM
	(
		SELECT
			student_id,
			course_id
		FROM
			score
		WHERE
			course_id = 1
		OR course_id = 2
	) AS b
LEFT JOIN student on b.student_id=student.sid
GROUP BY
	b.student_id
HAVING
  COUNT(b.course_id)>1

6、查询学过“李平”老师所教的所有课的同学的学号、姓名；

知识点：in

SELECT
	student_id as choice_liping
FROM
	score
WHERE
	course_id IN (
		SELECT
			course.cid
		FROM
			course
		LEFT JOIN teacher ON course.teacher_id = teacher.tid
		WHERE
			teacher.tname = '李平老师'
	)GROUP BY student_id

7、查询有课程成绩小于60分的同学的学号、姓名；

SELECT
	student.sname,
	a.student_id
FROM
	(
		SELECT
			student_id,
			course_id,
			num
		FROM
			score
		WHERE
			num < 60
	) AS a
LEFT JOIN student ON a.student_id = student.sid
GROUP BY
	a.student_id

　或（DISTINCT：自动删除重复的）：

SELECT
	student.sname,
	a.student_id

FROM
	(
		SELECT DISTINCT
			student_id
		FROM
			score
		WHERE
			num < 60
	) AS a
LEFT JOIN student ON a.student_id = student.sid

8、查询没有学全所有课的同学的学号、姓名；

SELECT
	student_id,
	COUNT(student_id)
FROM
	score
GROUP BY
	student_id
HAVING
   COUNT(student_id) 	< (
		SELECT
			COUNT(course.cid)
		FROM
			course
 )

9、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；

SELECT
	score.student_id,
	student.sname
FROM
	score
LEFT JOIN student ON score.student_id = student.sid
WHERE
	score.student_id != 1
AND score.course_id IN (
	SELECT
		course_id
	FROM
		score
	WHERE
		student_id = 1
)
GROUP BY
	score.student_id

10、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名

思路：1、课程数量与2号学生相同

2、课程与2号同学有交集

3、交集的课程数量与2号学生的相同

SELECT
	student_id,
	student.sname
FROM
	score
LEFT JOIN student ON student_id = student.sid
WHERE
	student_id IN (
		SELECT
			student_id
		FROM
			score
		WHERE
			student_id != 2
		GROUP BY
			student_id
		HAVING
			COUNT(course_id) = (
				SELECT
					COUNT(course_id)
				FROM
					score
				WHERE
					student_id = 2
			)
	)
AND course_id IN (
	SELECT
		course_id
	FROM
		score
	WHERE
		student_id = 2
)
GROUP BY
	student_id
HAVING
	COUNT(course_id) = (
		SELECT
			COUNT(course_id)
		FROM
			score
		WHERE
			student_id = 2
	)

11、删除学习“叶平”老师课的score表记录；

DELETE
FROM
	score
WHERE
	course_id IN (
		SELECT
			cid
		FROM
			course
		WHERE
			teacher_id IN (
				SELECT
					tid
				FROM
					teacher
				WHERE
					tname = '李平老师'
			)
	)

12、向SC表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；

INSERT INTO score(student_id,course_id,num) 
SELECT
	student_id,
  2,
  (SELECT AVG( num) FROM score WHERE course_id = 2)
FrOM
	score
WHERE
	student_id NOT IN (
		SELECT
			student_id
		FROM
			score
		WHERE
			course_id = 2
		GROUP BY
			student_id
	)
GROUP BY
	student_id

13、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示：学生ID,语文,数学,英语,有效课程数,有效平均分

SELECT 
 student_id as '学号',
 (SELECT num FROM score as innerdb LEFT JOIN course on course.cid=innerdb.course_id WHERE course.cname='生物' and  innerdb.student_id=oterdb.student_id)as '生物',
 (SELECT num FROM score as innerdb LEFT JOIN course on course.cid=innerdb.course_id WHERE course.cname='物理' and  innerdb.student_id=oterdb.student_id)as '物理',
 (SELECT num FROM score as innerdb LEFT JOIN course on course.cid=innerdb.course_id WHERE course.cname='体育' and  innerdb.student_id=oterdb.student_id)as '体育',
 COUNT(course_id)as '选修课程数',
 AVG(num)as '平均分'
FROM
	score AS oterdb
GROUP BY
	student_id
ORDER BY
	AVG(num) DESC

14、按各科平均成绩从低到高和及格率的百分数从高到低顺序；

SELECT
	course_id,
	COUNT(course_id),
	AVG(num),
	SUM(CASE WHEN score.num > 60 THEN 1 ELSE 0 end)/COUNT(course_id)*100 as persent
FROM
	score
GROUP BY
	course_id
ORDER BY persent DESC

15、课程平均分从高到低显示（显示任课老师）

SELECT
	cid,
	AVG(

		IF (
			ISNULL(score.num),
			0,
			score.num
		)
	) AS avg,
	teacher.tname
FROM
	course
LEFT JOIN score ON score.course_id = cid
LEFT JOIN teacher ON teacher.tid = course.teacher_id
GROUP BY
	cid
ORDER BY
	avg DESC

16、查询每个科目前三科成绩

SELECT *FROM (
SELECT course_id,
(SELECT num FROM score as s1 WHERE s1.course_id = s2.course_id ORDER BY num desc LIMIT 0,1) as first_num,
(SELECT num FROM score as s1 WHERE s1.course_id = s2.course_id ORDER BY num desc LIMIT 1,1) as second_num,
(SELECT num FROM score as s1 WHERE s1.course_id = s2.course_id ORDER BY num desc LIMIT 1,1) as third_num
FROM score as s2)as t GROUP BY course_id

三、知识点

1、where与having的区别：

where 子句的作用是在对查询结果进行分组前，将不符合where条件的行去掉，也就是在分组之前过滤数据，条件中不能包含聚和函数，使用where条件限制特定的行。

having 子句的作用是筛选满足条件的组，即在分组之后过滤数据，条件中经常包含聚合函数，使用having 条件过滤特定的组，也可以使用多个分组标准进行分组。

例子：

#where和having都可以使用的场景：
    select price,name from goods where price > 100
    select price,name from goods having price > 100

#只可以用where，不可以用having的情况
    select name from goods where price> 100
    select name from goods having price> 100 //报错！！！因为select没有筛选出price 字段，having不能用,而where是对表进行检索price。100

#只可以用having，不可以用where情况
    select id, avg(price) as agprice from goods group by id having agprice > 100
    select id, avg(price) as agprice from goods where agprice>100 group by id //报错！！因为from goods这表里面没有agprice这个字段

2、in

3、DISTINCT

4、inset into tb1(xx,xx) select x1,x2 from tb2;

5、三元操作

case when .. then... else...

6、if语句

IF(expr1,expr2,expr3)，如果expr1的值为true，则返回expr2的值，如果expr1的值为false，则返回expr3的值。

7、三张表之间互相有联系，可将中间联系的那张表作为主表，再用连续 left join 剩余的两张表

posted @ 2019-01-15 15:21 yujin123456 阅读(348) 评论(0) 编辑收藏举报

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mysql----导入导出、练习题

一、导入导出

二、练习题

三、知识点

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