3大核心功能揭秘:暗黑破坏神2存档编辑器如何重塑你的游戏体验
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1.学⽣表 Student(SId,Sname,Sage,Ssex)
SId 学⽣编号
Sname 学⽣姓名
Sage 出⽣年⽉
Ssex 学⽣性别
2.课程表 Course(CId,Cname,TId)
CId 课程编号
Cname 课程名称
TId 教师编号
3.教师表 Teacher(TId,Tname)
TId 教师编号
Tname 教师姓名
4.成绩表 SC(SId,CId,score)
SId 学⽣编号
CId 课程编号
score 分数
学⽣表 Student
create table Student( SId varchar(10), Sname varchar(10), Sage datetime, Ssex varchar(10) );
课程表 Course
create table Course( CId varchar(10), Cname nvarchar(10), TId varchar(10) );
教师表 Teacher
create table Teacher( TId varchar(10), Tname varchar(10) );
成绩表 SC
create table SC( SId varchar(10), CId varchar(10), score decimal(18,1) );
学⽣表 Student
-- 学生表 Student insert into Student values('01' , '赵雷' , '1990-01-01' , '男'); insert into Student values('02' , '钱电' , '1990-12-21' , '男'); insert into Student values('03' , '孙⻛' , '1990-12-20' , '男'); insert into Student values('04' , '李云' , '1990-12-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '⼥'); insert into Student values('06' , '吴兰' , '1992-01-01' , '⼥'); insert into Student values('07' , '郑⽵' , '1989-01-01' , '⼥'); insert into Student values('09' , '张三' , '2017-12-20' , '⼥'); insert into Student values('10' , '李四' , '2017-12-25' , '⼥'); insert into Student values('11' , '李四' , '2012-06-06' , '⼥'); insert into Student values('12' , '赵六' , '2013-06-13' , '⼥'); insert into Student values('13' , '孙七' , '2014-06-01' , '⼥');课程表 Course
-- 科⽬表 Course insert into Course values('01' , '语⽂' , '02'); insert into Course values('02' , '数学' , '01'); insert into Course values('03' , '英语' , '03');教师表 Teacher
-- 教师表 Teacher insert into Teacher values('01' , '张三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五');成绩表 SC
-- 成绩表 SC insert into SC values('01' , '01' , 80); insert into SC values('01' , '02' , 90); insert into SC values('01' , '03' , 99); insert into SC values('02' , '01' , 70); insert into SC values('02' , '02' , 60); insert into SC values('02' , '03' , 80); insert into SC values('03' , '01' , 80); insert into SC values('03' , '02' , 80); insert into SC values('03' , '03' , 80); insert into SC values('04' , '01' , 50); insert into SC values('04' , '02' , 30); insert into SC values('04' , '03' , 20); insert into SC values('05' , '01' , 76); insert into SC values('05' , '02' , 87); insert into SC values('06' , '01' , 31); insert into SC values('06' , '03' , 34); insert into SC values('07' , '02' , 89); insert into SC values('07' , '03' , 98);sql
-- 步骤:先分别查询01和02课程成绩,再关联比较 SELECT s.*, sc1.score AS '01课程分数', sc2.score AS '02课程分数' FROM Student s JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01' JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02' WHERE sc1.score > sc2.score;sql
-- 方法1:关联查询 SELECT s.SId, s.Sname, sc1.score AS '01课程', sc2.score AS '02课程' FROM Student s JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01' JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02'; -- 方法2:分组筛选 SELECT SId FROM SC WHERE CId IN ('01', '02') GROUP BY SId HAVING COUNT(DISTINCT CId) = 2;sql
SELECT s.SId, s.Sname, sc1.score AS '01课程', sc2.score AS '02课程' FROM Student s JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01' LEFT JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02';sql
SELECT s.*, sc2.score AS '02课程分数' FROM Student s JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02' WHERE s.SId NOT IN (SELECT SId FROM SC WHERE CId = '01');sql
SELECT s.SId, s.Sname, ROUND(AVG(sc.score), 1) AS 平均成绩 FROM Student s JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname HAVING AVG(sc.score) >= 60;sql
-- 方法1:DISTINCT去重 SELECT DISTINCT s.* FROM Student s JOIN SC sc ON s.SId = sc.SId; -- 方法2:IN子查询 SELECT * FROM Student WHERE SId IN (SELECT DISTINCT SId FROM SC);sql
SELECT s.SId, s.Sname, COUNT(sc.CId) AS 选课总数, SUM(sc.score) AS 总成绩 FROM Student s LEFT JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname;sql
SELECT COUNT(*) AS 李姓老师数量 FROM Teacher WHERE Tname LIKE '李%';sql
-- 步骤:先找到张三老师的课程,再关联成绩表和学生表 SELECT DISTINCT s.* FROM Student s JOIN SC sc ON s.SId = sc.SId JOIN Course c ON sc.CId = c.CId JOIN Teacher t ON c.TId = t.TId WHERE t.Tname = '张三';sql
-- 步骤:先统计总课程数,再筛选选课数不足的学生 SELECT s.* FROM Student s LEFT JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname HAVING COUNT(DISTINCT sc.CId) < (SELECT COUNT(*) FROM Course);sql
SELECT DISTINCT s.* FROM Student s JOIN SC sc ON s.SId = sc.SId WHERE sc.CId IN (SELECT CId FROM SC WHERE SId = '01') AND s.SId != '01'; -- 排除01号学生自身sql
-- 步骤:1. 获取01号课程列表 2. 筛选选课数相同且课程完全匹配的学生 SELECT s.* FROM Student s JOIN SC sc ON s.SId = sc.SId WHERE s.SId != '01' GROUP BY s.SId, s.Sname HAVING COUNT(DISTINCT sc.CId) = (SELECT COUNT(*) FROM SC WHERE SId = '01') AND GROUP_CONCAT(sc.CId ORDER BY sc.CId) = (SELECT GROUP_CONCAT(CId ORDER BY CId) FROM SC WHERE SId = '01');sql
-- 步骤:先找到张三老师的课程,再筛选未选这些课程的学生 SELECT s.Sname FROM Student s WHERE s.SId NOT IN ( SELECT DISTINCT sc.SId FROM SC sc JOIN Course c ON sc.CId = c.CId JOIN Teacher t ON c.TId = t.TId WHERE t.Tname = '张三' );sql
SELECT s.SId, s.Sname, ROUND(AVG(sc_all.score), 1) AS 平均成绩 FROM Student s JOIN SC sc_fail ON s.SId = sc_fail.SId AND sc_fail.score < 60 JOIN SC sc_all ON s.SId = sc_all.SId GROUP BY s.SId, s.Sname HAVING COUNT(DISTINCT sc_fail.CId) >= 2;sql
SELECT s.*, sc.score FROM Student s JOIN SC sc ON s.SId = sc.SId WHERE sc.CId = '01' AND sc.score < 60 ORDER BY sc.score DESC;sql
-- 步骤:先计算每个学生的平均成绩,再关联所有课程成绩 SELECT s.SId, s.Sname, MAX(CASE WHEN sc.CId = '01' THEN sc.score ELSE NULL END) AS 语文, MAX(CASE WHEN sc.CId = '02' THEN sc.score ELSE NULL END) AS 数学, MAX(CASE WHEN sc.CId = '03' THEN sc.score ELSE NULL END) AS 英语, ROUND(AVG(sc.score), 1) AS 平均成绩 FROM Student s LEFT JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname ORDER BY 平均成绩 DESC;sql
SELECT c.CId, c.Cname, COUNT(sc.SId) AS 选修人数, MAX(sc.score) AS 最高分, MIN(sc.score) AS 最低分, ROUND(AVG(sc.score), 1) AS 平均分, -- 及格率(>=60) ROUND(SUM(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS 及格率, -- 中等率(70-80) ROUND(SUM(CASE WHEN sc.score BETWEEN 70 AND 80 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS 中等率, -- 优良率(80-90) ROUND(SUM(CASE WHEN sc.score BETWEEN 80 AND 90 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS 优良率, -- 优秀率(>=90) ROUND(SUM(CASE WHEN sc.score >= 90 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS 优秀率 FROM Course c JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname ORDER BY 选修人数 DESC, c.CId ASC;sql
-- 使用RANK()函数(保留空缺) SELECT CId, 平均成绩, RANK() OVER (ORDER BY 平均成绩 DESC) AS 排名 FROM ( SELECT CId, ROUND(AVG(score), 1) AS 平均成绩 FROM SC GROUP BY CId ) AS t;sql
-- 使用DENSE_RANK()函数(不保留空缺) SELECT CId, 平均成绩, DENSE_RANK() OVER (ORDER BY 平均成绩 DESC) AS 排名 FROM ( SELECT CId, ROUND(AVG(score), 1) AS 平均成绩 FROM SC GROUP BY CId ) AS t;sql
SELECT s.SId, s.Sname, IFNULL(SUM(sc.score), 0) AS 总成绩, RANK() OVER (ORDER BY IFNULL(SUM(sc.score), 0) DESC) AS 排名 FROM Student s LEFT JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname;sql
SELECT s.SId, s.Sname, IFNULL(SUM(sc.score), 0) AS 总成绩, DENSE_RANK() OVER (ORDER BY IFNULL(SUM(sc.score), 0) DESC) AS 排名 FROM Student s LEFT JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname;sql
SELECT c.CId, c.Cname, -- [100-85] SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS '100-85人数', ROUND(SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS '100-85百分比', -- [85-70] SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS '85-70人数', ROUND(SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS '85-70百分比', -- [70-60] SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS '70-60人数', ROUND(SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS '70-60百分比', -- [60-0] SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END) AS '60-0人数', ROUND(SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END)/COUNT(sc.SId)*100, 1) AS '60-0百分比' FROM Course c JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname;sql
-- 方法:使用窗口函数ROW_NUMBER() SELECT * FROM ( SELECT sc.SId, sc.CId, sc.score, ROW_NUMBER() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS 排名 FROM SC sc ) AS t WHERE 排名 <= 3;sql
SELECT c.CId, c.Cname, COUNT(DISTINCT sc.SId) AS 选修人数 FROM Course c LEFT JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname;sql
SELECT s.SId, s.Sname FROM Student s JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname HAVING COUNT(DISTINCT sc.CId) = 2;sql
SELECT Ssex AS 性别, COUNT(*) AS 人数 FROM Student GROUP BY Ssex;sql
SELECT * FROM Student WHERE Sname LIKE '%风%';sql
SELECT Sname, Ssex, COUNT(*) AS 同名人数 FROM Student GROUP BY Sname, Ssex HAVING COUNT(*) > 1;sql
-- 方法1:YEAR函数 SELECT * FROM Student WHERE YEAR(Sage) = 1990; -- 方法2:LIKE匹配 SELECT * FROM Student WHERE Sage LIKE '1990%';sql
SELECT c.CId, c.Cname, ROUND(AVG(sc.score), 1) AS 平均成绩 FROM Course c JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname ORDER BY 平均成绩 DESC, c.CId ASC;sql
SELECT s.SId, s.Sname, ROUND(AVG(sc.score), 1) AS 平均成绩 FROM Student s JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname HAVING AVG(sc.score) >= 85;sql
SELECT s.Sname, sc.score FROM Student s JOIN SC sc ON s.SId = sc.SId JOIN Course c ON sc.CId = c.CId WHERE c.Cname = '数学' AND sc.score < 60;sql
-- 笛卡尔积生成所有学生-课程组合,再左连接成绩 SELECT s.SId, s.Sname, c.Cname, sc.score FROM Student s CROSS JOIN Course c LEFT JOIN SC sc ON s.SId = sc.SId AND c.CId = sc.CId ORDER BY s.SId, c.CId;sql
SELECT DISTINCT s.Sname, c.Cname, sc.score FROM Student s JOIN SC sc ON s.SId = sc.SId AND sc.score > 70 JOIN Course c ON sc.CId = c.CId;sql
SELECT s.SId, s.Sname, c.Cname, sc.score FROM Student s JOIN SC sc ON s.SId = sc.SId AND sc.score < 60 JOIN Course c ON sc.CId = c.CId;sql
SELECT s.SId, s.Sname FROM Student s JOIN SC sc ON s.SId = sc.SId WHERE sc.CId = '01' AND sc.score > 80;sql
-- 同24题,重复题目 SELECT c.CId, c.Cname, COUNT(DISTINCT sc.SId) AS 学生人数 FROM Course c LEFT JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname;sql
-- 步骤:先找到张三老师的课程,再筛选最高分(无重复) SELECT s.*, sc.score, c.Cname FROM Student s JOIN SC sc ON s.SId = sc.SId JOIN Course c ON sc.CId = c.CId JOIN Teacher t ON c.TId = t.TId WHERE t.Tname = '张三' AND sc.score = ( SELECT MAX(score) FROM SC sc2 JOIN Course c2 ON sc2.CId = c2.CId JOIN Teacher t2 ON c2.TId = t2.TId WHERE t2.Tname = '张三' );sql
-- 方法:使用窗口函数 SELECT * FROM ( SELECT s.*, sc.score, c.Cname, RANK() OVER (ORDER BY sc.score DESC) AS 排名 FROM Student s JOIN SC sc ON s.SId = sc.SId JOIN Course c ON sc.CId = c.CId JOIN Teacher t ON c.TId = t.TId WHERE t.Tname = '张三' ) AS t WHERE 排名 = 1;sql
SELECT DISTINCT sc1.SId, sc1.CId, sc1.score FROM SC sc1 JOIN SC sc2 ON sc1.SId = sc2.SId AND sc1.CId != sc2.CId AND sc1.score = sc2.score ORDER BY sc1.score;sql
-- 同23题(前三名),调整为前两名 SELECT * FROM ( SELECT sc.SId, sc.CId, sc.score, ROW_NUMBER() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS 排名 FROM SC sc ) AS t WHERE 排名 <= 2;sql
SELECT c.CId, c.Cname, COUNT(DISTINCT sc.SId) AS 选修人数 FROM Course c JOIN SC sc ON c.CId = sc.CId GROUP BY c.CId, c.Cname HAVING COUNT(DISTINCT sc.SId) > 5;sql
SELECT SId FROM SC GROUP BY SId HAVING COUNT(DISTINCT CId) >= 2;sql
SELECT s.* FROM Student s JOIN SC sc ON s.SId = sc.SId GROUP BY s.SId, s.Sname HAVING COUNT(DISTINCT sc.CId) = (SELECT COUNT(*) FROM Course);sql
SELECT SId, Sname, YEAR(NOW()) - YEAR(Sage) AS 年龄 FROM Student;sql
SELECT SId, Sname, YEAR(NOW()) - YEAR(Sage) - CASE WHEN DATE_FORMAT(NOW(), '%m%d') < DATE_FORMAT(Sage, '%m%d') THEN 1 ELSE 0 END AS 实际年龄 FROM Student;sql
-- MySQL中WEEKOFYEAR函数:本周(1-53) SELECT * FROM Student WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW());sql
SELECT * FROM Student WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW()) + 1;sql
SELECT * FROM Student WHERE MONTH(Sage) = MONTH(NOW());sql
SELECT * FROM Student WHERE MONTH(Sage) = MONTH(NOW()) + 1;ROUND(...,1)保留 1 位小数,可根据需求调整;