通过实例简要介绍case函数的用法。
1.创建测试表:
DROP SEQUENCE student_sequence;
CREATE SEQUENCE student_sequence START WITH 10000 INCREMENT BY 1;
DROP TABLE students;
CREATE TABLE students (
id NUMBER(5) PRIMARY KEY,
first_name VARCHAR2(20),
last_name VARCHAR2(20),
major VARCHAR2(30),
current_credits NUMBER(3),
grade varchar2(2));
INSERT INTO students (id, first_name, last_name, major, current_credits,grade)
VALUES (student_sequence.NEXTVAL, 'Scott', 'Smith', 'Computer Science', 98,null);
INSERT INTO students (id, first_name, last_name, major, current_credits,grade)
VALUES (student_sequence.NEXTVAL, 'Margaret', 'Mason', 'History', 88,null);
INSERT INTO students (id, first_name, last_name, major, current_credits,grade)
VALUES (student_sequence.NEXTVAL, 'Joanne', 'Junebug', 'Computer Science', 75,null);
INSERT INTO students (id, first_name, last_name, major, current_credits,grade)
VALUES (student_sequence.NEXTVAL, 'Manish', 'Murgratroid', 'Economics', 66,null);
commit;
2.查看相应数据
SQL> select * from students;
ID FIRST_NAME LAST_NAME MAJOR CURRENT_CREDITS GR
------ ----------------- ------------------- ----------- ----------------- -----
10000 Scott Smith Computer Science 98
10001 Margaret Mason History 88
10002 Joanne Junebug Computer Science 75
10003 Manish Murgratroid Economics 66
3.更新语句
update students
set grade = (
select grade from
(
select id,
case when current_credits > 90 then 'a'
when current_credits > 80 then 'b'
when current_credits > 70 then 'c'
else 'd' end grade
from students
) a
where a.id = students.id
)
/
4.更新后结果
SQL> select * from students;
ID FIRST_NAME LAST_NAME MAJOR CURRENT_CREDITS GR
------------- ---------------- -------------------- ---------------------------- ----
10000 Scott Smith Computer Science 98 a
10001 Margaret Mason History 88 b
10002 Joanne Junebug Computer Science 75 c
10003 Manish Murgratroid Economics 66 d
|
