2016-04-28 94 views
3

我有問題試圖讓這個sql語句工作。差距應該像這樣計算。主要GPA的計算方式與GPA相同,只是計算中只包括學生主修的課程。例如,假設一個CS學生在CS課程中獲得A,在Math類中獲得B,則該學生的主要GPA爲4.0。查詢處理平均成績點平均數

列出學生及其主要GPA。結果不需要包括沒有參加過任何課程的學生。

到目前爲止,我有下面的查詢,它不完全返回我想要的。

select sum(value*units)/sum(units) as GPA ,f.name from faculty f 
inner join sections s on s.instructor_id = f.id 
inner join courses c on c.id=s.id 
inner join grades g on g.value =c.units 
group by f.name 

表我得到的是

name gpa 
Art  4.00 
Bob  4.00 
Pat  3.30 
Amy  3.30 
Kim  3.30 
Sue  3.30 
Joe  2.70 
Lee  2.70 
Max 2.70 

這是我想要什麼差不多,但它顯示誰甚至沒有采取任何班級的學生的GPA。

數據庫看起來像

create table departments (
    id  integer primary key, 
    name varchar(255) 
); 

insert into departments (id, name) values (10, 'Computer Science'); 
insert into departments (id, name) values (20, 'Math'); 
insert into departments (id, name) values (30, 'Drama'); 

create table students (
    id    integer primary key, 
    name   varchar(255), 
    graduation_date date, 
    major_id  integer references departments(id) 
); 

insert into students (id, name, graduation_date, major_id) values 
    (1, 'Joe', null, 10); 
insert into students (id, name, graduation_date, major_id) values 
    (2, 'Amy', '2009-04-22', 20); 
insert into students (id, name, graduation_date, major_id) values 
    (3, 'Max', null, 10); 
insert into students (id, name, graduation_date, major_id) values 
    (4, 'Sue', '2009-01-10', 20); 
insert into students (id, name, graduation_date, major_id) values 
    (5, 'Bob', '2009-03-05', 30); 
insert into students (id, name, graduation_date, major_id) values 
    (6, 'Kim', null, 20); 
insert into students (id, name, graduation_date, major_id) values 
    (7, 'Art', null, 30); 
insert into students (id, name, graduation_date, major_id) values 
    (8, 'Pat', '2005-07-11', 20); 
insert into students (id, name, graduation_date, major_id) values 
    (9, 'Lee', null, 10); 

create table grades (
    id   integer primary key, 
    letter  varchar(255) not null unique, 
    value  real 
); 

insert into grades (id, letter, value) values (1, 'A', 4.0); 
insert into grades (id, letter, value) values (2, 'A-', 3.7); 
insert into grades (id, letter, value) values (3, 'B+', 3.3); 
insert into grades (id, letter, value) values (4, 'B', 3.0); 
insert into grades (id, letter, value) values (5, 'B-', 2.7); 
insert into grades (id, letter, value) values (6, 'C+', 2.3); 
insert into grades (id, letter, value) values (7, 'C', 2.0); 
insert into grades (id, letter, value) values (8, 'C-', 1.7); 
insert into grades (id, letter, value) values (9, 'D+', 1.3); 
insert into grades (id, letter, value) values (10, 'D', 1.0); 
insert into grades (id, letter, value) values (11, 'D-', 0.7); 
insert into grades (id, letter, value) values (12, 'F', 0.0); 
insert into grades (id, letter) values (13, 'CR'); 
insert into grades (id, letter) values (14, 'NC'); 

create table courses (
    id    integer primary key, 
    title   varchar(255), 
    units   integer, 
    department_id integer references departments(id) 
); 

insert into courses (id, title, units, department_id) values 
    (12, 'Databases', 4, 10); 
insert into courses (id, title, units, department_id) values 
    (22, 'Compilers', 4, 10); 
insert into courses (id, title, units, department_id) values 
    (32, 'Calculus 1', 4, 20); 
insert into courses (id, title, units, department_id) values 
    (42, 'Algebra', 4, 20); 
insert into courses (id, title, units, department_id) values 
    (52, 'Acting', 4, 30); 
insert into courses (id, title, units, department_id) values 
    (62, 'Elocution', 2, 30); 
insert into courses (id, title, units, department_id) values 
    (72, 'Calculus 2', 4, 20); 
insert into courses (id, title, units, department_id) values 
    (82, 'Topology', 4, 20); 

create table sections (
    id    integer primary key, 
    course_id  integer not null references courses(id), 
    instructor_id integer references faculty(id), 
    year   integer 
); 

insert into sections (id, course_id, instructor_id, year) values 
    (12, 12, 6, 2007); 
insert into sections (id, course_id, instructor_id, year) values 
    (13, 12, 1, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (14, 22, 1, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (23, 12, 6, 2009); 
insert into sections (id, course_id, instructor_id, year) values 
    (24, 22, 1, 2009); 
insert into sections (id, course_id, instructor_id, year) values 
    (32, 32, 2, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (33, 32, 2, 2009); 
insert into sections (id, course_id, instructor_id, year) values 
    (34, 82, 2, 2009); 
insert into sections (id, course_id, instructor_id, year) values 
    (43, 32, 3, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (51, 62, 4, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (52, 52, 4, 2008); 
insert into sections (id, course_id, instructor_id, year) values 
    (53, 62, 4, 2009); 
insert into sections (id, course_id, instructor_id, year) values 
    (54, 52, 4, 2009); 

create table enrollment (
    id   integer primary key, 
    student_id integer not null references students(id), 
    section_id integer not null references sections(id), 
    grade_id integer references grades(id) 
); 

insert into enrollment (id, student_id, section_id, grade_id) values 
    (14, 1, 12, 8); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (15, 1, 13, 3); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (16, 1, 14, 5); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (17, 1, 32, 1); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (18, 1, 34, 2); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (19, 1, 53, 13); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (24, 3, 12, 2); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (25, 3, 14, 5); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (26, 3, 32, 1); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (27, 3, 34, 2); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (28, 3, 54, 7); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (34, 2, 43, 3); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (44, 4, 33, 4); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (54, 4, 53, 1); 
insert into enrollment (id, student_id, section_id, grade_id) values 
    (64, 6, 53, 1) 
+0

不相關,但是:對於長度爲255(或更小)的'varchar'列,沒有魔術或隱藏優化。實際空間使用情況**僅取決於存儲在列中的值,而不取決於可容納的最大長度。 –

+0

你怎麼知道哪些學生沒有在他們的專業上任何課程? –

+0

*主要課程沒有限制*。您可能需要將其添加到連接條件:「內部連接課程c上c.id = s.id和s.major_id = c.department_id」 – dnoeth

回答

0

我想你的樣品,但不能管理有相同的結果你。

我嘗試這樣做:

​​

凡我對學生和入學表篩選,這意味着,如果學生沒有課程,他不會在結果

是否適合你?

+0

不,該表應該顯示所有的學生,並從那裏得到嘖嘖稱讚專業。所以我們不計算專業之外的成績。所以她只有1個班級(拓撲結構是她的專業),並得到了B + ..所以她的GPA是4(單位)* 3.3/4 = 3.3 – henryzo

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@henryzo:如果你想向學生展示他們是否他們沒有采取任何主要課程,但是,您必須切換到外部連接。 – dnoeth