这两天比较懒，没有学习，这个是我问一个学java的小伙伴要的sql的总结资料，大体语句全在上面了，复制到博客上，以后忘记可以查看
#1命令行连接MySQL
msyql -u root -proot;
#2创建数据库
DATABASE `school`;
#3查看数据库列表
SHOW DATABASES;
#4选择数据库
USE `school`;
#5删除数据库
DROP DATABASE `school`;
#6创建表 学生表 （字段：学生编号 学生名）
CREATE TABLE IF NOT EXISTS student( studentId INT(4) PRIMARY KEY, studentName VARCHAR(20));
#7创建学生表 字段的约束及属性 主键 注释 字符集
CREATE TABLE IF NOT EXISTS student ( studentId INT(4) PRIMARY KEY, studentName VARCHAR(20)) COMMENT="学生表" DEFAULT CHARSET=utf8;
#8查看表
SHOW TABLES;
SELECT * FROM 表名;
#9删除表
DROP TABLE student;
#10修改表表名
ALTER TABLE new_student RENAME student;
#11修改表字段
ALTER TABLE student MODIFY studentname VARCHAR(10) NOT NULL;
ALTER TABLE 表名 CHANGE 旧字段名 新字段名[属性或约束];
#12添加表字段
ALTER TABLE student ADD newC INT(4);
#13删除字段
ALTER TABLE student DROP COLUMN newc;
#14添加主键
ALTER TABLE student ADD PRIMARY KEY `id`;
#15添加外键
ALTER TABLE student ADD CONSTRAINT g_s FOREIGN KEY (gradeId) REFERENCES `grade`(`gradeId`);
#16编写SQL语句实现从学生表提取姓名、手机号两列数据存储到通讯录表中
CREATE TABLE tongxunlu (SELECT studentname,phone, FROM student);
#17把成绩都降低10%后加5分，再查询及格成绩，并按照成绩从高到低排序
SELECT studentresult*0.9+5 FROM result WHERE studentresult*0.9+5>=60 ORDER BY studentresult*0.9+5 DESC;
#18查询所有年级编号为1的学员信息，按学号升序排序
#显示前4条记录
#每页4条，显示第2页，即从第5条记录开始显示4条数据
SELECT * FROM student WHERE gradeId = 1 ORDER BY studentNo LIMIT 4,4;
#19将学生表中学号为20000的学生的邮箱修改为stu20000@163.com，密码改为000
UPDATE student SET email='stu20000@163.com' WHERE studentNO = 20000;
#20将科目表中课时数大于200且年级编号为1的科目的课时减少10
UPDATE `subject` SET classhour=classhour-10 WHERE classHour>200 AND gradeId = 1;
#21将所有年级编号为1的学员姓名、性别、出生日期、手机号码信息保存到新表student_grade1中
CREATE TABLE student_grade1(SELECT studentName,sex,bornDate,phone FROM student);
#22查询2016年2月17日考试前5名的学员的学号和分数
SELECT studentNO,studentresult FROM result WHERE examdate = '2016-02-17' ORDER BY studentresult DESC LIMIT 5;
#23将所有女学生按年龄从大到小排序，从第2条记录开始显示6名女学生的姓名、年龄、出生日期、手机号信息
SELECT studentname,borndate,phone FROM student WHERE sex = '女' ORDER BY bornDate LIMIT 1,6;
#24查询参加2016年2月17日考试的所有学员的最高分、最低分、平均分
SELECT MAX(studentresult),MIN(studentresult),AVG(studentresult) FROM result WHERE examdate='2016-02-17';
#25编写SQL语句，查看年龄比“李斯文”小的学生，要求显示这些学生的信息
SELECT * FROM student WHERE bornDate < (SELECT borndate FROM student WHERE studentName = '李斯文');
#26查询参加最近一次Logic Java考试成绩的学生的最高分和最低分
SELECT MAX(studentResult),MIN(studentResult) FROM result WHERE examDate = (SELECT MAX(examdate) FROM result WHERE subjectno = (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava'));
#27查询“Logic Java”课程考试成绩为60分的学生名单
SELECT studentName FROM result,student WHERE result.`studentNo`=student.`studentNo` AND studentResult=60;
#28检查“Logic Java”课程最近一次考试成绩,#如果有 80分以上的成绩，显示分数排在前5名的学员学号和分数
SELECT studentResult FROM result WHERE EXISTS (SELECT studentNo FROM result WHERE studentResult>=80 AND examDate = (SELECT MAX(examDate) FROM result WHERE subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')) AND
subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')
) AND examDate = (SELECT MAX(examDate) FROM result WHERE subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')) AND
subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava') ORDER BY studentResult DESC LIMIT 5;
#29检查“Logic Java”课程最近一次考试成绩,#如果全部未通过考试（60分及格），认为本次考试偏难，计算的该次考试平均分加5分
SELECT AVG(studentResult)+5 FROM result WHERE NOT EXISTS (SELECT studentNo FROM result WHERE studentResult<60 AND examDate = (SELECT MAX(examDate) FROM result WHERE subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')) AND
subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')
) AND examDate = (SELECT MAX(examDate) FROM result WHERE subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava')) AND
subjectNo IN (SELECT subjectno FROM `subject` WHERE subjectName = 'logicjava');
#30根据课程编号分组，求每个小组平均分
SELECT AVG(studentResult) FROM result GROUP BY subjectNo;
#31根据课程编号分组，求每个小组平均分,由小到大排序
SELECT AVG(studentResult) FROM result GROUP BY subjectNo ORDER BY AVG(studentResult);
#32以年级，性别分组，求人数
SELECT COUNT(0) '人数' FROM student GROUP BY gradeId,sex;
#33求每个平均分数大于60的年级编号
SELECT gradeId FROM result,`subject` WHERE result.`subjectNo`=`subject`.`subjectNo` GROUP BY gradeId HAVING AVG(studentResult)>60;
#34查询学生成绩信息及个人信息显示
SELECT * FROM student JOIN result USING (studentNo);
#35查询u1所在的年级课程
SELECT sb.* FROM `subject` sb JOIN grade g USING (gradeid)
WHERE gradeName = 'u1';