數據庫編程基本練習題
1、用一條SQL語句查詢出每門課都大於80分的學生姓名
準備數據的sql代碼：sql
create table score( id int primary key auto_increment, name varchar(20), subject varchar(20), score int); insert into score values (null,'張三','語文',81), (null,'張三','數學',75), (null,'李四','語文',76), (null,'李四','數學',90), (null,'王五','語文',81), (null,'王五','數學',100), (null,'王五 ','英語',90);
答案：數據庫
select distinct name from score where name not in (select distinct name from score where score<=80)
二、每月份的發生額都比101科目多的科目
請用SQL語句實現：從TestDB數據表中查詢出全部月份的發生額都比101科目相應月份的發生額高的科目。請注意：TestDB中有不少科目，都有1－12月份的發生額。 AccID：科目代碼，Occmonth：發生額月份，DebitOccur：發生額。 數據庫名：JcyAudit，數據集：Select * from TestDB函數
準備數據的sql代碼： drop table if exists TestDB; create table TestDB( id int primary key auto_increment, AccID varchar(20), Occmonth date, DebitOccur bigint); insert into TestDB values (null,'101','1988-1-1',100), (null,'101','1988-2-1',110), (null,'101','1988-3-1',120), (null,'101','1988-4-1',100), (null,'101','1988-5-1',100), (null,'101','1988-6-1',100), (null,'101','1988-7-1',100), (null,'101','1988-8-1',100); --複製上面的數據，故意把第一個月份的發生額數字改小一點 insert into TestDB values (null,'102','1988-1-1',90), (null,'102','1988-2-1',110), (null,'102','1988-3-1',120), (null,'102','1988-4-1',100), (null,'102','1988-5-1',100), (null,'102','1988-6-1',100), (null,'102','1988-7-1',100), (null,'102','1988-8-1',100); --複製最上面的數據，故意把全部發生額數字改大一點 insert into TestDB values (null,'103','1988-1-1',150), (null,'103','1988-2-1',160), (null,'103','1988-3-1',180), (null,'103','1988-4-1',120), (null,'103','1988-5-1',120), (null,'103','1988-6-1',120), (null,'103','1988-7-1',120), (null,'103','1988-8-1',120); --複製最上面的數據，故意把全部發生額數字改大一點 insert into TestDB values (null,'104','1988-1-1',130), (null,'104','1988-2-1',130), (null,'104','1988-3-1',140), (null,'104','1988-4-1',150), (null,'104','1988-5-1',160), (null,'104','1988-6-1',170), (null,'104','1988-7-1',180), (null,'104','1988-8-1',140); --複製最上面的數據，故意把第二個月份的發生額數字改小一點 insert into TestDB values (null,'105','1988-1-1',100), (null,'105','1988-2-1',80), (null,'105','1988-3-1',120), (null,'105','1988-4-1',100), (null,'105','1988-5-1',100), (null,'105','1988-6-1',100), (null,'105','1988-7-1',100), (null,'105','1988-8-1',100);
答案：post
select distinct AccID from TestDB where AccID not in (select TestDB.AccIDfrom TestDB, (select * from TestDB where AccID='101') as db101 where TestDB.Occmonth=db101.Occmonth and TestDB.DebitOccur<=db101.DebitOccur );
三、統計每一年每個月的信息
year month amount 1991 1 1.1 1991 2 1.2 1991 3 1.3 1991 4 1.4 1992 1 2.1 1992 2 2.2 1992 3 2.3 1992 4 2.4 查成這樣一個結果： year m1 m2 m3 m4 1991 1.1 1.2 1.3 1.4 1992 2.1 2.2 2.3 2.4
準備sql語句： drop table if exists sales; create table sales( id int auto_increment primary key, year varchar(10), month varchar(10), amount float(2,1)); insert into sales values (null,'1991','1',1.1), (null,'1991','2',1.2), (null,'1991','3',1.3), (null,'1991','4',1.4), (null,'1992','1',2.1), (null,'1992','2',2.2), (null,'1992','3',2.3), (null,'1992','4',2.4);
答案：學習
select sales.year, (select t.amount from sales as t where t.month='1' and t.year = sales.year) as 'm1', (select t.amount from sales as t where t.month='2' and t.year = sales.year) as 'm2', (select t.amount from sales as t where t.month='3' and t.year = sales.year) as 'm3', (select t.amount from sales as t where t.month='4' and t.year = sales.year) as 'm4' from sales group by year
四、顯示文章標題，發帖人、最後回覆時間
表：id,title,postuser,postdate,parentidthis
準備sql語句： drop table if exists articles; create table articles( id int auto_increment primary key, title varchar(50), postuser varchar(10), postdate datetime, parentid int references articles(id) ); insert into articles values (null,'第一條','張三','1998-10-10 12:32:32',null), (null,'第二條','張三','1998-10-10 12:34:32',null), (null,'第一條回覆1','李四','1998-10-10 12:35:32',1), (null,'第二條回覆1','李四','1998-10-10 12:36:32',2), (null,'第一條回覆2','王五','1998-10-10 12:37:32',1), (null,'第一條回覆3','李四','1998-10-10 12:38:32',1), (null,'第二條回覆2','李四','1998-10-10 12:39:32',2), (null,'第一條回覆4','王五','1998-10-10 12:39:40',1);
答案：spa
select zhu.title,zhu.postuser, (select max(fu.postdate) from articles as fu where zhu.id = fu.parentid) as postdata from articles as zhu where zhu.parentid is null
五、刪除除了id號不一樣,其餘都相同的學生冗餘信息
學生表 以下: id號 學號 姓名 課程編號 課程名稱 分數 1 2005001 張三 0001 數學 69 2 2005002 李四 0001 數學 89 3 2005001 張三 0001 數學 69
準備數據： create table student2( id int auto_increment primary key, code varchar(20), name varchar(20) ); insert into student2 values (null,'2005001','張三'), (null,'2005002','李四'), (null,'2005001','張三');
答案：code
delete from student2 where id not in(select min(id) from (select * from student2) as t group by t.name);
六、航空網的幾個航班查詢題：
實驗環境： create table city( cityID int auto_increment primary key, cityName varchar(20)); create table flight ( flightID int auto_increment primary key, StartCityID int references city(cityID), endCityID int references city(cityID), StartTime timestamp); //航班原本應該沒有日期部分纔好，可是下面的題目當中涉及到了日期 insert into city values (null,'北京'), (null,'上海'), (null,'廣州'); insert into flight values (null,1,2,'9:37:23'), (null,1,3,'9:37:23'), (null,1,2,'10:37:23'), (null,2,3,'10:37:23');
(1) 查詢起飛城市是北京的全部航班，按到達城市的名字排序blog
答案：排序
select * from flight as f,city as c where f.endCityID = c.cityID and StartCityID = (select cityID from city where cityName = "北京") order by endCityID desc
(2)查詢北京到上海的全部航班紀錄（起飛城市，到達城市，起飛時間，航班號）
答案：
select c1.cityName as '起飛城市',c2.cityName as '到大城市',f.StartTime as '到大城市',f.flightID as '航班號' from city as c1,city as c2,flight f where c1.cityID = f.StartCityID and c2.cityID = f.endCityID and c1.cityName = "北京" and c2.cityName = "上海"
(3)查詢具體某一天（2005-5-8）的北京到上海的的航班次數
答案：
select count(*) from city as c1,city as c2,flight as f where c1.cityID = f.StartCityID and c2.cityID = f.endCityID and c1.cityName = "北京" and c2.cityName = "上海" and 查幫助得到的某個日期處理函數(startTime) like '2005-5-8%'
七、查出比經理薪水還高的員工信息：
準備數據： Drop table if not exists employees; create table employees( id int primary key auto_increment, name varchar(50), salary int, managerid int references employees(id)); insert into employees values (null,' lhm',10000,null), (null,' zxx',15000,1), (null,'flx',9000,1), (null,'tg',10000,2), (null,'wzg',10000,3); Wzg大於flx,lhm大於zxx
答案：
select e.* from employees as m,employees as e where m.id = e.managerid and m.salary < e.salary
八、求出小於45歲的各個老師所帶的大於12歲的學生人數
實驗數據： drop table if exists tea_stu; drop table if exists teacher; drop table if exists student; create table teacher( teaID int primary key, name varchar(50), age int); create table student( stuID int primary key, name varchar(50), age int); create table tea_stu( teaID int references teacher(teaID), stuID int references student(stuID)); insert into teacher values (1,'zxx',45), (2,'lhm',25) , (3,'wzg',26) , (4,'tg',27); insert into student values (1,'wy',11), (2,'dh',25) , (3,'ysq',26) , (4,'mxc',27); insert into tea_stu values (1,1), (1,2), (1,3); insert into tea_stu values (2,2), (2,3), (2,4); insert into tea_stu values (3,3), (3,4), (3,1); insert into tea_stu values (4,4), (4,1), (4,2), (4,3);
答案：
select teacher.name,count(student.name) as count from teacher,student,tea_stu where teacher.teaID = tea_stu.teaID and student.stuID = tea_stu.stuID and teacher.age < 45 and student.age > 12 group by teacher.name
九、一個用戶表中有一個積分字段，假如數據庫中有100多萬個用戶，若要在每一年第一天凌晨將積分清零，你將考慮什麼，你將想什麼辦法解決?
alter table drop column score; alter table add colunm score int;
可能會很快，可是須要試驗，試驗不能拿真實的環境來操刀，而且要注意，
這樣的操做時沒法回滾的，在個人印象中，只有inert update delete等DML語句才能回滾，
對於create table,drop table ,alter table等DDL語句是不能回滾。
解決方案一， update user set score=0; 解決方案二，假設上面的代碼要執行好長時間，超出咱們的容忍範圍，那我就
alter table user drop column score;alter table user add column score int。
下面代碼實現每一年的那個凌晨時刻進行清零
Runnable runnable = new Runnable(){ public void run(){ clearDb(); schedule(this,new Date(new Date().getYear()+1,0,0)); } }; schedule(runnable,new Date(new Date().getYear()+1,0,1));
十、用一條SQL 語句 查詢出每門課都大於80 分的學生姓名
name course grade 張三 語文 81 張三 數學 75 李四 語文 76 李四 數學 90 王五 語文 81 王五 數學 100 王五 英語 90
答案：
select name from table group by name having min(grade) > 80
11. 現有學生表以下:
自動編號 學號 姓名 課程編號 課程名稱 分數 1 2005001 張三 0001 數學 69 2 2005002 李四 0001 數學 89 3 2005001 張三 0001 數學 69
刪除除了自動編號不一樣, 其餘都相同的學生冗餘信息
答案：
select 自動編號 from 表 where 自動編號 not in (select min(自動編號) from 表 group by 學號,姓名,課程編號,課程名稱,分數)
十二、一個叫 team 的表，裏面只有一個字段name, 一共有4 條紀錄，分別是a,b,c,d, 對應四個球對，如今四個球對進行比賽，用一條sql 語句顯示全部可能的比賽組合
select a.name as '主隊',b.name as '客隊' from team as a,team as b where a.name < b.name
1三、查詢」 01 「課程比」 02 「課程成績高的學生的信息及課程分數
實驗數據：
學生表： create table Student( Sid varchar(6), Sname varchar(10), Sage datetime, Ssex varchar(10)); insert into Student values('01' , '趙雷' , '1990-01-01' , '男'); insert into Student values('02' , '錢電' , '1990-12-21' , '男'); insert into Student values('03' , '孫風' , '1990-05-20' , '男'); insert into Student values('04' , '李雲' , '1990-08-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '女'); insert into Student values('06' , '吳蘭' , '1992-03-01' , '女'); insert into Student values('07' , '鄭竹' , '1989-07-01' , '女'); insert into Student values('08' , '王菊' , '1990-01-20' , '女')
成績表 create table SC( Sid varchar(10), Cid varchar(10), score decimal(18,1)); 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)
課程表 create table Course( Cid varchar(10), Cname varchar(10), Tid varchar(10)); insert into Course values ('01' , '語文' , '02'); insert into Course values ('02' , '數學' , '01'); insert into Course values ('03' , '英語' , '03')
教師表 create table Teacher( Tid varchar(10), Tname varchar(10)); insert into Teacher values ('01' , '張三'); insert into Teacher values ('02' , '李四'); insert into Teacher values ('03' , '王五')
1四、 根據（13題）表查詢平均成績大於等於 60 分的同窗的學生編號和學生姓名和平均成績
答案：
select s.Sid,s.Sname,avg(score) as a from Student as s,SC where s.Sid = SC.Sid group by s.Sid having a>60
1五、根據（13題）表查詢在 SC 表存在成績的學生信息
答案：
select s.* from Student as s,SC where s.Sid = SC.Sid group by Sid
1六、查詢全部同窗的學生編號、學生姓名、選課總數、全部課程的總成績(沒成績的顯示爲 null )
答案：
select s.Sid,s.Sname,count(SC.Cid),sum(score) from Student as s left join SC on s.Sid = SC.Sid group by s.Sid
1七、查有成績的學生信息
答案：
select s.Sid,S.Sname,count(score),sum(score), sum(case when SC.Cid=01 then score else null end) as score_1, sum(case when SC.Cid=02 then score else null end) as score_2, sum(case when SC.Cid=03 then score else null end) as score_3 from Student as s,SC,Course where s.Sid = SC.Sid and SC.Cid = Course.Cid group by s.Sid
1八、查詢「李」姓老師的數量
答案：
select count(*) from Teacher where Tname like '李%'
1九、查詢學過「張三」老師授課的同窗的信息
select s.* from Student as s,SC,course,Teacher where s.Sid=SC.Sid and Course.Cid=SC.Cid and Teacher.Tid=Course.Tid and Teacher.Tname = '張三' group by s.Sid
20、查詢沒有學全全部課程的同窗的信息
select * from Student where Sid in (select Sid from SC group by Sid having count(cid) < 3)
2一、查詢至少有一門課與學號爲」 01 「的同窗所學相同的同窗的信息
select s.* from SC,Student as s where Cid in (select Cid from SC where Sid = '01') and s.Sid = SC.Sid group by Sid
2二、查詢沒學過」張三」老師講授的任一門課程的學生姓名
select Sname from Student where Sname not in (select s.Sname from Teacher as t,SC,Course,Student as s where t.Tid = Course.Tid and s.Sid=SC.Sid and SC.Cid=Course.Cid and t.Tname = '張三')
2三、查詢兩門及其以上不及格課程的同窗的學號，姓名及其平均成績
select s.Sid,s.Sname,avg(score) from Student as s,SC where s.Sid = SC.Sid and SC.score < 60 group by s.Sid having count(score > 2)
2四、檢索」 01 「課程分數小於 60，按分數降序排列的學生信息
select s.*,SC.score from Student as s,SC where Cid = '01' and s.Sid=SC.Sid and score < 60 group by score desc
2五、按平均成績從高到低顯示全部學生的全部課程的成績以及平均成績
select Sid, sum(case when Cid=01 then score else null end) as score_01, sum(case when Cid=02 then score else null end) as score_02, sum(case when Cid=03 then score else null end) as score_03, avg(score) from SC group by Sid order by avg(score) desc
2六、查詢各科成績最高分、最低分和平均分，以以下形式顯示：課程 ID，課程 name，最高分，最低分，平均分，及格率，中等率，優良率，優秀率(及格爲>=60，中等爲：70-80，優良爲：80-90，優秀爲：>=90）。 要求輸出課程號和選修人數，查詢結果按人數降序排列，若人數相同，按課程號升序排列
select c.cid as 課程號, c.cname as 課程名稱, count(*) as 選修人數, max(score) as 最高分, min(score) as 最低分, avg(score) as 平均分, sum(case when score >= 60 then 1 else 0 end)/count(*) as 及格率, sum(case when score >= 70 and score < 80 then 1 else 0 end)/count(*) as 中等率, sum(case when score >= 80 and score < 90 then 1 else 0 end)/count(*) as 優良率, sum(case when score >= 90 then 1 else 0 end)/count(*) as 優秀率 from sc, course c where c.cid = sc.cid group by c.cid order by count(*) desc, c.cid asc
2七、統計各科成績各分數段人數：課程編號，課程名稱，[100-85]，[85-70]，[70-60]，[60-0] 及所佔百分比
select c.Cid as 課程編號, c.Cname as 課程名稱, A.* from course as c, (select Cid, sum(case when score >= 85 then 1 else 0 end)/count(*) as 100_85, sum(case when score >= 70 and score < 85 then 1 else 0 end)/count(*) as 85_70, sum(case when score >= 60 and score < 70 then 1 else 0 end)/count(*) as 70_60, sum(case when score < 60 then 1 else 0 end)/count(*) as 60_0 from SC group by Cid) as A where c.Cid = A.Cid
2八、查詢出只選修兩門課程的學生學號和姓名
select s.Sid,s.Sname,count(sc.Cid) from Student as s,SC as sc where s.Sid = sc.Sid group by s.Sid having count(sc.Cid) = 2
2九、查詢名字中含有「風」字的學生信息
select Sname from Student where Sname like '%風%'
30、查詢 1990 年出生的學生名單
select * from Student where Sage like('1990%')
3一、成績不重複，查詢選修「張三」老師所授課程的學生中，成績最高的學生信息及其成績
select s.* from Student as s,Teacher as t,SC as sc,Course as c where s.Sid=sc.Sid and c.Tid=t.Tid and sc.Cid = c.Cid and t.Tname='張三' order by sc.score desc limit 0,1
3二、成績有重複的狀況下，查詢選修「張三」老師所授課程的學生中，成績最高的學生信息及其成績
select s.* from Student as s,SC where SC.Sid=s.Sid and SC.Cid = (select c.Cid from Teacher as t,Course as c where t.Tid = c.Tid and t.Tname = '張三') and SC.score = ( select sc.score from Student as s,Teacher as t,SC as sc,Course as c where s.Sid=sc.Sid and c.Tid=t.Tid and sc.Cid = c.Cid and t.Tname='張三' order by sc.score desc limit 0,1 )
3三、查詢各學生的年齡，只按年份來算
select Sname,year(now())-year(Sage) as age from Student
3四、按照出生日期來算，當前月日 < 出生年月的月日則，年齡減一
select Sname,year(now())-year(Sage)-1 as age from Student
3五、查詢本週過生日的學生
select Sname from Student where week(now()) = week(Sage)
3六、查詢下週過生日的學生
select Sname from Student where (week(now())+1) = week(Sage)
3七、查詢本月過生日的學生
select Sname from Student where month(now()) = month(Sage)
3八、查詢下月過生日的學生
select Sname from Student where (month(now())+1) = month(Sage)
3九、查詢和」 01 「號的同窗學習的課程徹底相同的其餘同窗的信息
select * from Student where sid in (select sid from SC where cid in (select sc.Cid from SC where sc.Sid='01') and sid <>'01' group by sid having count(cid) >= 3 )
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