Oraclecase函数如何使用介绍

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

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