就像命令不工作 Oracle SQL
Like Command not working Oracle SQL
我是数据库新手 PHP。在我的代码中,我试图从 PHP 脚本中创建一个 table,这就是我所拥有的。
create table booktable(BookID INT PRIMARY KEY,
BookName VARCHAR(100),
Published DATE,
Price NUMBER(18,2),
Author1 VARCHAR2(30),
Author2 VARCHAR2(30));
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (1, 'Fundamentals of Digital Logic with VHDL Design','14-APR-08', 190.25,'Stephen Brown','Zvon Ko G.Vranesic');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (2, 'Distributed Systems Principles and Paradigm','26-JUL-13', 197.80,'Andrew S. Tanenbaum','Maarten Van Steen');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (3, 'Eat Real Food The Only Solution to Permanent weight Loss and Disease Prevention','1-APR-15', 29.99,'David Gillespie','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (4, 'Introduction to Computational Science Modeling and Simulation for the sciences','2-MAY-06', 132.75,'Angela B.Shiflet','George W. Shiflet');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (5, 'Live Well on Less A Practical Guide to Running a Lean Household','27-MAY-15', 19.00,'Jody Allen','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (6, 'Middle School: Just My Rotten Luck','1-JUL-15', 15.99,'James Patterson','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (7, 'Clementine Rose and the Birthday Emergency','1-JUL-15', 12.99,'Jacqueline Harvey','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (8, 'My Life It''s a long story','26-MAY-15', 32.99,'Willie Nelson','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (9, 'Sword of Summer Magnus Chase','7-OCT-15', 15.99,'Rick Riordan','');
table 创建成功,但是当我尝试 运行 查询
`SQL> select *
from booktable
where bookname like '%my%';`
它说没有选择行。我不知道我哪里做错了。谢谢
那是因为它区分大小写,并且您在某些字段中包含 My
- 而不是 my
。
试试看:
select *
from booktable
where bookname like '%My%';
@alfasin 是对的。你也可以降低以实现你想要做的事情
select * from booktable where lower(bookname) like '%my%';
我是数据库新手 PHP。在我的代码中,我试图从 PHP 脚本中创建一个 table,这就是我所拥有的。
create table booktable(BookID INT PRIMARY KEY,
BookName VARCHAR(100),
Published DATE,
Price NUMBER(18,2),
Author1 VARCHAR2(30),
Author2 VARCHAR2(30));
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (1, 'Fundamentals of Digital Logic with VHDL Design','14-APR-08', 190.25,'Stephen Brown','Zvon Ko G.Vranesic');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (2, 'Distributed Systems Principles and Paradigm','26-JUL-13', 197.80,'Andrew S. Tanenbaum','Maarten Van Steen');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (3, 'Eat Real Food The Only Solution to Permanent weight Loss and Disease Prevention','1-APR-15', 29.99,'David Gillespie','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (4, 'Introduction to Computational Science Modeling and Simulation for the sciences','2-MAY-06', 132.75,'Angela B.Shiflet','George W. Shiflet');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (5, 'Live Well on Less A Practical Guide to Running a Lean Household','27-MAY-15', 19.00,'Jody Allen','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (6, 'Middle School: Just My Rotten Luck','1-JUL-15', 15.99,'James Patterson','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (7, 'Clementine Rose and the Birthday Emergency','1-JUL-15', 12.99,'Jacqueline Harvey','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (8, 'My Life It''s a long story','26-MAY-15', 32.99,'Willie Nelson','');
INSERT INTO booktable (BookID, BookName, Published, Price, Author1, Author2)
VALUES (9, 'Sword of Summer Magnus Chase','7-OCT-15', 15.99,'Rick Riordan','');
table 创建成功,但是当我尝试 运行 查询
`SQL> select *
from booktable
where bookname like '%my%';`
它说没有选择行。我不知道我哪里做错了。谢谢
那是因为它区分大小写,并且您在某些字段中包含 My
- 而不是 my
。
试试看:
select *
from booktable
where bookname like '%My%';
@alfasin 是对的。你也可以降低以实现你想要做的事情
select * from booktable where lower(bookname) like '%my%';