建立网站线上营销,上海企业所得税怎么征收,做pc网站如何实时预览,免费网站建设信息前言#xff1a; 记录一下sql学习#xff0c;仅供参考基本都对了#xff0c;不排除有些我做的太快做错了。里面sql不存在任何sql优化操作#xff0c;只以完成最后输出结果为目的#xff0c;包含我做题过程和思路最后一行才是结果。 1.过程: 1.1.插入数据 /* SQLyog Ul…前言 记录一下sql学习仅供参考基本都对了不排除有些我做的太快做错了。里面sql不存在任何sql优化操作只以完成最后输出结果为目的包含我做题过程和思路最后一行才是结果。 1.过程: 1.1.插入数据 /* SQLyog Ultimate v13.1.1 (64 bit) MySQL - 8.0.32 : Database - test ********************************************************************* *//*!40101 SET NAMES utf8 */;/*!40101 SET SQL_MODE*/;/*!40014 SET OLD_UNIQUE_CHECKSUNIQUE_CHECKS, UNIQUE_CHECKS0 */; /*!40014 SET OLD_FOREIGN_KEY_CHECKSFOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS0 */; /*!40101 SET OLD_SQL_MODESQL_MODE, SQL_MODENO_AUTO_VALUE_ON_ZERO */; /*!40111 SET OLD_SQL_NOTESSQL_NOTES, SQL_NOTES0 */; CREATE DATABASE /*!32312 IF NOT EXISTS*/test /*!40100 DEFAULT CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci */ /*!80016 DEFAULT ENCRYPTIONN */;USE test;/*Table structure for table course */DROP TABLE IF EXISTS course;CREATE TABLE course (c_id varchar(20) NOT NULL,c_name varchar(20) NOT NULL DEFAULT ,t_id varchar(20) NOT NULL,PRIMARY KEY (c_id) ) ENGINEInnoDB DEFAULT CHARSETutf8mb4 COLLATEutf8mb4_0900_ai_ci;/*Data for the table course */insert into course(c_id,c_name,t_id) values (01,语文,02), (02,数学,01), (03,英语,03), (04,化学,04), (05,物理,05), (06,生物,06);/*Table structure for table dept */DROP TABLE IF EXISTS dept;CREATE TABLE dept (id int NOT NULL AUTO_INCREMENT COMMENT ID,name varchar(50) NOT NULL COMMENT 部门名称,PRIMARY KEY (id) ) ENGINEInnoDB AUTO_INCREMENT7 DEFAULT CHARSETutf8mb4 COLLATEutf8mb4_0900_ai_ci COMMENT部门表;/*Data for the table dept */insert into dept(id,name) values (1,研发部), (2,市场部), (3,财务部), (4,销售部), (5,总经办), (6,人事部);/*Table structure for table score */DROP TABLE IF EXISTS score;CREATE TABLE score (s_id varchar(20) NOT NULL,c_id varchar(20) NOT NULL DEFAULT ,s_score int DEFAULT NULL,PRIMARY KEY (s_id,c_id) ) ENGINEInnoDB DEFAULT CHARSETutf8mb4 COLLATEutf8mb4_0900_ai_ci;/*Data for the table score */insert into score(s_id,c_id,s_score) values (01,01,80), (01,02,90), (01,03,99), (02,01,70), (02,02,60), (02,03,80), (03,01,80), (03,02,80), (03,03,80), (04,01,50), (04,02,30), (04,03,20), (05,01,76), (05,02,87), (06,01,31), (06,03,34), (07,02,89), (07,03,98);/*Table structure for table student */DROP TABLE IF EXISTS student;CREATE TABLE student (s_id varchar(20) NOT NULL,s_name varchar(20) NOT NULL DEFAULT ,s_brith varchar(20) NOT NULL DEFAULT ,s_sex varchar(10) NOT NULL DEFAULT ,PRIMARY KEY (s_id) ) ENGINEInnoDB DEFAULT CHARSETutf8mb4 COLLATEutf8mb4_0900_ai_ci;/*Data for the table student */insert into student(s_id,s_name,s_brith,s_sex) values (01,赵雷,1990-01-01,男), (02,钱电,1990-12-21,男), (03,孙风,1990-05-20,男), (04,李云,1990-08-06,男), (05,周梅,1991-12-01,女), (06,吴兰,1992-03-01,女), (07,郑竹,1989-07-01,女), (08,王菊,1990-01-20,女), (09,lpq,2002-**-**,女), (10,lpq,2002-**-**,女);/*Table structure for table teacher */DROP TABLE IF EXISTS teacher;CREATE TABLE teacher (t_id varchar(20) NOT NULL,t_name varchar(20) NOT NULL DEFAULT ,PRIMARY KEY (t_id) ) ENGINEInnoDB DEFAULT CHARSETutf8mb4 COLLATEutf8mb4_0900_ai_ci;/*Data for the table teacher */insert into teacher(t_id,t_name) values (01,张三), (02,李四), (03,王五);/*!40101 SET SQL_MODEOLD_SQL_MODE */; /*!40014 SET FOREIGN_KEY_CHECKSOLD_FOREIGN_KEY_CHECKS */; /*!40014 SET UNIQUE_CHECKSOLD_UNIQUE_CHECKS */; /*!40111 SET SQL_NOTESOLD_SQL_NOTES */;1.2.解题过程  USE test;//查询01课程比02课程成绩高的学生 SELECT * FROM score s WHERE s.c_id01; SELECT * FROM score s WHERE s.c_id02; SELECT * FROM student st LEFT JOIN (SELECT * FROM score s WHERE s.c_id01) AS s1 ON s1.s_idst.s_id  LEFT JOIN  (SELECT * FROM score s WHERE s.c_id02) AS s2 ON st.s_ids2.s_id WHERE s1.s_scores2.s_score; //查询01课程比02课程成绩低的学生信息及课程分数 SELECT * FROM student st LEFT JOIN (SELECT * FROM score s WHERE s.c_id01) AS s1 ON s1.s_idst.s_id  LEFT JOIN  (SELECT * FROM score s WHERE s.c_id02) AS s2 ON st.s_ids2.s_id WHERE s1.s_scores2.s_score; //查询平均分数大于等于60的同学的学生编号和学生姓名和平均成绩 SELECT s.s_id,AVG(s_score) AS _avg FROM score s GROUP BY s.s_id HAVING AVG(s.s_score)60; SELECT st.*,s1._avg FROM student st JOIN (SELECT s.s_id,AVG(s_score) AS _avg FROM score s GROUP BY s.s_id HAVING AVG(s.s_score)60) AS s1 ON st.s_ids1.s_id; //查询平均成绩小于60的同学学生编号和学生姓名和平均成绩包括有成绩和无成绩的难度*** SELECT s.s_id,AVG(s_score) AS _avg FROM score s GROUP BY s.s_id HAVING AVG(s.s_score)60; SELECT * FROM student st LEFT JOIN (SELECT s.s_id,AVG(s_score) AS _avg FROM score s GROUP BY s.s_id HAVING AVG(s.s_score)) AS s1 ON s1.s_idst.s_id WHERE s1._avg60 OR s1._avg IS NULL;//查询所有同学的学生编号、学生姓名、选课总数、所有课程总成绩 SELECT st.*,st2._sum,st2._count FROM student st LEFT JOIN (SELECT SUM(s.s_score) AS _sum,s.s_id,COUNT(*) AS _count FROM score s GROUP BY s.s_id)AS st2 ON st2.s_idst.s_id; //查询李性老师的数量 SELECT COUNT(*) FROM teacher t WHERE t.t_name LIKE %李% //查询学过张三老师课程学生 SELECT st.* FROM score s JOIN course c ON s.c_idc.c_id JOIN student st ON st.s_ids.s_id WHERE c.t_id(SELECT t.t_id FROM teacher AS t WHERE t.t_name张三); //查询没学过张三老师课程学生 //1.先查询学过的 SELECT st.s_id FROM student st JOIN score s ON st.s_ids.s_id JOIN course c ON c.c_ids.c_id JOIN teacher t ON t.t_idc.t_id WHERE c.t_id IN (SELECT t.t_id FROM teacher AS t WHERE t.t_name张三); //取反 SELECT * FROM student st WHERE st.s_id NOT IN(SELECT st.s_id FROM student st JOIN score s ON st.s_ids.s_id JOIN course c ON c.c_ids.c_id JOIN teacher t ON t.t_idc.t_id WHERE c.t_id IN (SELECT t.t_id FROM teacher AS t WHERE t.t_name张三)) 剩下的就不一个个贴了。 //查询学过01和02课程学生信息 SELECT * FROM score s WHERE s.c_id01; SELECT * FROM score s WHERE s.c_id02; SELECT st.* FROM (SELECT * FROM score s WHERE s.c_id01) st1 JOIN (SELECT * FROM score s WHERE s.c_id02) st2 ON st1.s_idst2.s_id JOIN student st ON st.s_id st1.s_id;//查询学过01但是没学过02的 有点难度*** SELECT st.* FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id01 AND s.s_id NOT IN (SELECT s.s_id FROM score s WHERE s.c_id02)//查询没有学会全部课程的 SELECT s.s_id,COUNT(*) AS _count FROM score s GROUP BY s.s_id HAVING COUNT(*)!3; SELECT st.*,s1._count FROM student AS st LEFT JOIN (SELECT s.s_id,COUNT(*) AS _count FROM score s GROUP BY s.s_id) AS s1 ONst.s_ids1.s_id WHERE s1._count IS NULL OR s1._count!3;//查询和01号同学学习完全相同的其他同学的信息 SELECT s.c_id FROM score s WHERE s.s_id01;SELECT st.*,COUNT(*) FROM score s JOIN student st ON st.s_ids.s_idWHERE s.c_id IN (SELECT s.c_id FROM score s WHERE s.s_id01) AND s.s_id!01 GROUP BY s.s_id HAVING COUNT(*)3;//查询没学过张三老师讲授的任何一门课程的学生 SELECT c.c_id FROM teacher t JOIN course c ON c.t_idt.t_id WHERE t.t_name张三 SELECT s.s_id FROM score s WHERE s.c_id (SELECT c.c_id FROM teacher t JOIN course c ON c.t_idt.t_id WHERE t.t_name张三) GROUP BY s.s_id; SELECT * FROM student st WHERE st.s_id NOT IN (SELECT s.s_id FROM score s WHERE s.c_id (SELECT c.c_id FROM teacher t JOIN course c ON c.t_idt.t_id WHERE t.t_name张三) GROUP BY s.s_id);//查询两门及其以上不及格课程的同学的学号姓名及其平均成绩 SELECT s.s_id AS _count FROM score s WHERE s.s_score60 GROUP BY s_id; SELECT st.*,AVG(s.s_score) FROM student st JOIN score s ON st.s_ids.s_id WHERE st.s_id IN (SELECT s.s_id AS _count FROM score s WHERE s.s_score60 GROUP BY s_id) GROUP BY st.s_id;//检索01课程分数小于60按分数降序排列的学生信息 SELECT st.* FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id01 AND s.s_score60 ORDER BY s.s_score DESC;//按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩 SELECT st.s_id,s.s_score AS 语文 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id01; SELECT st.s_id,s.s_score AS 数学 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id02; SELECT st.s_id,s.s_score AS 英语 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id03; SELECT st.*,IFNULL(s1.语文,0) AS 语文,IFNULL(s2.数学,0) AS 数学,IFNULL(s3.英语,0) AS 英语, (IFNULL(s1.语文, 0) IFNULL(s2.数学, 0) IFNULL(s3.英语, 0)) / 3 AS 平均分 FROM student st LEFT JOIN (SELECT st.s_id,s.s_score AS 语文 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id01) AS s1 ON st.s_ids1.s_id LEFT JOIN (SELECT st.s_id,s.s_score AS 数学 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id02) AS s2 ON st.s_ids2.s_id LEFT JOIN (SELECT st.s_id,s.s_score AS 英语 FROM score s JOIN student st ON st.s_ids.s_id WHERE s.c_id03) AS s3 ON st.s_ids3.s_id//查询各科成绩最高分、最低分和平均分 //这题和标准答案不一样我随便写写这种用代码写更好sql性能太消耗了 SELECT s.c_id,MAX(s.s_score)AS _max,MIN(s.s_score) AS _min,AVG(s.s_score) AS _avg FROM score s GROUP BY s.c_id;//按各科成绩进行排序并显示排名 SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) FROM score s WHERE s.c_id01 SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) FROM score s WHERE s.c_id02 SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) FROM score s WHERE s.c_id03SELECT st.*,s1.语文,s2.数学,s3.英语 FROM student st LEFT JOIN (SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) AS 语文 FROM score s WHERE s.c_id01) AS s1 ON s1.s_idst.s_id LEFT JOIN (SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) AS 数学 FROM score s WHERE s.c_id02) AS s2 ON s2.s_idst.s_id LEFT JOIN (SELECT s.s_id,RANK() OVER(ORDER BY s.s_score DESC) AS 英语 FROM score s WHERE s.c_id03) AS s3 ON s3.s_idst.s_id;//查询学生的总成绩并进行排名 SELECT SUM(s.s_score) AS _sum FROM score s GROUP BY s.s_id; SELECT st.*,IFNULL(s2._sum,0) FROM student st LEFT JOIN (SELECT SUM(s.s_score) AS _sum,s.s_id FROM score s GROUP BY s.s_id) s2 ON s2.s_idst.s_id ORDER BY _sum DESC;//查询不同老师所教不同课程平均分从高到低显示 SELECT t.t_name,s.c_id,AVG(s.s_score) AS _avg FROM score s JOIN course c ON c.c_ids.c_id JOIN teacher t ON t.t_idc.t_id GROUP BY s.c_id ORDER BY _avg DESC;//查询所有课程的成绩第2名到第3名的学生信息及该课程成绩 //错误示例 ORDER BY limit只能在语句最后面 但是写成子查询c SELECT * FROM score s WHERE s.c_id01 ORDER BY s.s_score DESC LIMIT 1,2 UNION ALL SELECT * FROM score s WHERE s.c_id02 ORDER BY s.s_score DESC LIMIT 1,2 UNION ALL SELECT * FROM score s WHERE s.c_id03 ORDER BY s.s_score DESC LIMIT 1,2SELECT * FROM (SELECT * FROM score s WHERE s.c_id01 ORDER BY s.s_score DESC LIMIT 1,2) AS s1 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id02 ORDER BY s.s_score DESC LIMIT 1,2) AS s2 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id03 ORDER BY s.s_score DESC LIMIT 1,2) AS s3//统计各科成绩各分数段人数课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 pass 程序更容易//查询学生平均成绩及其名次 SELECT st.*,RANK() OVER(ORDER BY AVG(s.s_score) DESC),AVG(s.s_score) FROM score s JOIN student st ON st.s_ids.s_id GROUP BY s.s_id;//查询各科成绩前三名的记录 SELECT * FROM (SELECT * FROM score s WHERE s.c_id01 ORDER BY s.s_score DESC LIMIT 0,3) AS s1 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id02 ORDER BY s.s_score DESC LIMIT 0,3) AS s2 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id03 ORDER BY s.s_score DESC LIMIT 0,3) AS s3//查询每门课程被选修的学生数 SELECT s.c_id,COUNT(*) FROM score s GROUP BY s.c_id;//查询出只有两门课程的全部学生的学号和姓名 SELECT st.*,COUNT(*) FROM score s JOIN student st ON st.s_ids.s_id GROUP BY s.s_id HAVING COUNT(*)2;//查询男生、女生人数 SELECT st.s_sex,COUNT(*) FROM student st GROUP BY st.s_sex;//查询名字中含有风字的学生信息 SELECT * FROM student st WHERE st.s_name LIKE %风%;//查询同名同性学生名单并统计同名人数 //手动加入了lpq测试数据这里开始不一样了 SELECT * FROM student st1 JOIN student st2 ON st1.s_id!st2.s_id AND st1.s_namest2.s_name;//查询1990年出生的学生名单 SELECT * FROM student st WHERE YEAR(st.s_brith)1990;//查询每门课程的平均成绩结果按平均成绩降序排列平均成绩相同时按课程编号升序排列 SELECT s.c_id,AVG(s.s_score) AS _avg FROM score s GROUP BY s.c_id ORDER BY _avg DESC,s.c_id DESC;//查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 SELECT st.*,AVG(s.s_score) AS _avg FROM score s JOIN student st ON st.s_ids.s_id GROUP BY st.s_id HAVING AVG(s.s_score)85;//查询课程名称为数学且分数低于60的学生姓名和分数 偷个懒知道数学是02 SELECT st.*,s.s_score FROM score s RIGHT JOIN student st ON st.s_id s.s_id WHERE s.s_score60 AND s.c_id02;//查询所有学生的课程及分数情况 SELECT s.s_score AS 语文,s.s_id FROM score s WHERE s.c_id01 SELECT s.s_score AS 数学,s.s_id FROM score s WHERE s.c_id02 SELECT s.s_score AS 英语,s.s_id FROM score s WHERE s.c_id03 SELECT st.*,IFNULL(s1.语文,0),IFNULL(s2.数学,0),IFNULL(s3.英语,0) FROM student st LEFT JOIN (SELECT s.s_score AS 语文,s.s_id FROM score s WHERE s.c_id01) AS s1 ON s1.s_idst.s_id LEFT JOIN (SELECT s.s_score AS 数学,s.s_id FROM score s WHERE s.c_id02)AS s2 ON st.s_ids2.s_id LEFT JOIN (SELECT s.s_score AS 英语,s.s_id FROM score s WHERE s.c_id03)AS s3 ON st.s_ids3.s_id //查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数 SELECT s.s_id FROM score s WHERE s.s_score70 GROUP BY s.s_id; SELECT st.* FROM score s RIGHT JOIN student st ON st.s_ids.s_id WHERE s.s_score70 GROUP BY st.s_id;//查询课程不及格的学生 SELECT st.* FROM score s RIGHT JOIN student st ON st.s_ids.s_id WHERE s.s_score60 GROUP BY s.s_id;//查询课程编号为01且课程成绩在80分以上的学生的学号和姓名 SELECT * FROM score s WHERE s.s_score80; SELECT st.* FROM score s JOIN student st ON s.s_idst.s_id WHERE s.s_score80 AND s.c_id01;//求每门课程的学生人数 SELECT s.c_id,COUNT(*) FROM score s GROUP BY s.c_id;//查询选修张三老师所授课程的学生中成绩最高的学生信息及其成绩 SELECT c.c_id FROM teacher t JOIN course c ON c.t_id t.t_id WHERE t.t_name张三 SELECT s.s_id,s.s_score FROM score s WHERE s.c_id (SELECT c.c_id FROM teacher t JOIN course c ON c.t_id t.t_id WHERE t.t_name张三) ORDER BY s.s_score DESC LIMIT 1 SELECT st.*,s1.s_score FROM student st JOIN (SELECT s.s_id,s.s_score FROM score s WHERE s.c_id (SELECT c.c_id FROM teacher t JOIN course c ON c.t_id t.t_id WHERE t.t_name张三) ORDER BY s.s_score DESC LIMIT 1) AS s1 ON st.s_ids1.s_id;//查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 *** SELECT s.s_score FROM score s GROUP BY s.s_score HAVING COUNT(*)1; SELECT * FROM score WHERE s_score IN (SELECT s.s_score FROM score s GROUP BY s.s_score HAVING COUNT(*)1)//查询每门课程成绩最好的前三名 SELECT c_id FROM score GROUP BY c_id; SELECT * FROM (SELECT * FROM score s WHERE s.c_id01 ORDER BY s.s_score DESC LIMIT 3) AS s1 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id02 ORDER BY s.s_score DESC LIMIT 3) AS s2 UNION ALL SELECT * FROM (SELECT * FROM score s WHERE s.c_id03 ORDER BY s.s_score DESC LIMIT 3)AS s3// 统计每门课程的学生选修人数超过5人的课程才统计 SELECT s.c_id,COUNT(*) FROM score s GROUP BY s.c_id HAVING COUNT(*)5;//检索至少选修两门课程的学生学号 SELECT s.s_id,COUNT(*) FROM score s GROUP BY s.s_id HAVING COUNT(*)2;//查询选修了全部课程的学生信息 SELECT s.s_id, FROM score s GROUP BY s.s_id HAVING COUNT(*)3; SELECT * FROM student AS st WHERE st.s_id IN (SELECT s.s_id FROM score s GROUP BY s.s_id HAVING COUNT(*)3)//查询各学生的年龄(周岁) 算个大概的 SELECT YEAR(NOW()) SELECT YEAR(NOW())-YEAR(st.s_brith) FROM student st;// 剩下4题没啥用不用了。 总结 1.前面比较难中间简单最后有几个也挺难的。最后几题跟日期有关的我没做这个用程序写更好。 2.仅供参考答案不一定对。