1.學生id ---- s_id 數據庫
2. 學生姓名 ---- s_name 函數
3. 學生生日 ---- s_birth spa
4. 學生性別 ---- s_sex3d
1. 教師id ---- t_id code
2. 教師姓名 ---- t_nameblog
1.課程id ---- c_id 數學
2. 課程名 ---- c_name it
3. 教師id ----- t_id(與教師表對應, 方便創建數據庫鏈接)class
1. 學生id ---- s_id(與學生表對應, 方便創建數據庫鏈接) select
2. 課程id ---- c_id(與課程表對應, 方便創建數據庫鏈接)
3. 學生成績 ----- s_score
1 CREATE TABLE students( 2 s_id VARCHAR(20) PRIMARY KEY, 3 s_name VARCHAR(20) NOT NULL DEFAULT '', 4 s_birth VARCHAR(20) NOT NULL DEFAULT '', 5 s_sex VARCHAR(10) NOT NULL DEFAULT '' 6 ); 7 8 CREATE TABLE course( 9 c_id VARCHAR(20) PRIMARY KEY, 10 c_name VARCHAR(20) NOT NULL DEFAULT '', 11 t_id VARCHAR(20) NOT NULL 12 ); 13 14 CREATE TABLE teacher( 15 t_id VARCHAR(20) PRIMARY KEY, 16 t_name VARCHAR(20) NOT NULL DEFAULT '' 17 ); 18 19 CREATE TABLE score( 20 s_id VARCHAR(20), 21 c_id VARCHAR(20), 22 s_score INT(3), 23 PRIMARY(s_id, c_id) 24 );
1 INSERT INTO students VALUES('01', '趙雷', '1990-01-01', '男'); 2 INSERT INTO students VALUES('02', '錢電', '1990-12-21', '男'); 3 INSERT INTO students VALUES('03', '孫風', '1990-05-20', '男'); 4 INSERT INTO students VALUES('04', '李雲', '1990-12-01', '男'); 5 INSERT INTO students VALUES('05', '周梅', '1991-12-01', '女'); 6 INSERT INTO students VALUES('06', '吳蘭', '1992-03-01', '女'); 7 INSERT INTO students VALUES('07', '鄭竹', '1989-07-01', '女'); 8 INSERT INTO students VALUES('08', '王菊', '1990-01-20', '女'); 9 10 INSERT INTO course VALUES('01', '語文', '02'); 11 INSERT INTO course VALUES('02', '數學', '01'); 12 INSERT INTO course VALUES('03', '英語', '03'); 13 14 INSERT INTO teacher VALUES('01', '張三'); 15 INSERT INTO teacher VALUES('02', '李四'); 16 INSERT INTO teacher VALUES('03', '王五'); 17 18 INSERT INTO score VALUES('01', '01', 80); 19 INSERT INTO score VALUES('01', '02', 90); 20 INSERT INTO score VALUES('01', '03', 99); 21 INSERT INTO score VALUES('02', '01', 70); 22 INSERT INTO score VALUES('02', '02', 60); 23 INSERT INTO score VALUES('02', '03', 80); 24 INSERT INTO score VALUES('03', '01', 80); 25 INSERT INTO score VALUES('03', '02', 80); 26 INSERT INTO score VALUES('03', '03', 80); 27 INSERT INTO score VALUES('04', '01', 50); 28 INSERT INTO score VALUES('04', '02', 30); 29 INSERT INTO score VALUES('04', '03', 20); 30 INSERT INTO score VALUES('05', '01', 76); 31 INSERT INTO score VALUES('05', '02', 87); 32 INSERT INTO score VALUES('06', '03', 31); 33 INSERT INTO score VALUES('06', '01', 34); 34 INSERT INTO score VALUES('07', '02', 89); 35 INSERT INTO score VALUES('07', '03', 98);
想要達到該目標, 咱們就須要使用group by(通常作統計咱們每每會用group by)
課程是求個數, 使用count()函數
總成績求總和, 使用sum()函數
這裏咱們會用到students表和Cource表
使用左鏈接的方式 ---- (inner取的是交際, 若是8個學生只有7個參加考試, 則未參加考試的學生統計不到, 因此得向學生表作左鏈接)
select * from students as a left join score as b on a.s_id = b.s_id;
這樣就獲得了學生的關聯全部信息
使用ifnull將空成績定義爲0
select a.s_id, count(b.c_id), sum(ifnull(b.s_score, 0)) from students as a left join score as b on a.s_id = b.s_id group by a.s_id;
爲了防止之後解決多對多的現象, 咱們能夠再添加一段按s_name分組
select a.s_id, a.s_name, count(b.c_id), sum(ifnull(b.s_score, 0)) from students as a left join score as b on a.s_id = b.s_id group by a.s_id, a.s_name;
select a.s_id, count(b.c_id), sum(case when b.s_score is null then 0 else b.s_score end) from students as a left join score as b on a.s_id = b.s_id group by a.s_id;