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私は、データベースを作成しようとしていますが、私はこのコードを実行すると、それは：「外部キー制約を追加することはできません」
私に次のエラーを与えますcode 1215: “Cannot add foreign key constraint”
drop database if exists kat_db;
create database if not exists kat_db;
use kat_db;
create table if not exists Patients(
PatientID int not null PRIMARY KEY,
PatientName varchar(50),
TypeOfAnimal varchar(50),
DateOfLastApt date,
PatientAge int,
Microchipped boolean,
OwnerID int not null,
DoctorID int not null,
foreign key (OwnerID) references Owners(OwnerID),
foreign key (DoctorID) references Doctors(DoctorID)
);
create table if not exists Owners(
OwnerID int not null PRIMARY KEY,
OwnerFirstName varchar(50),
OwnerLastName varchar(50),
PatientID int,
OwnerPhone int,
OwnerZip int,
PrefferedPaymentMethod varchar(50),
foreign key (PatientID) references Patients(PatientID)
);
create table if not exists Doctors(
DoctorID int not null PRIMARY KEY,
DoctorFirstName varchar(50),
DoctorLastName varchar(50),
PatientID int,
DoctorPhone int,
DoctorZip int,
AnimalSpecialty varchar(50),
foreign key (PatientID) references Patients(PatientID)
);
insert into Patients
(PatientID, TypeOfAnimal, PatientAge, PatientName, DateOfLastApt, Microchipped, DoctorID, OwnerID)
values
(01, 'dog', 6, 'Dima', '2017-02-05', 1, 101, 1001),
(02, 'dog', 3, 'Misha', '2017-02-05', 1, 102, 1002),
(03, 'cat', 2, 'Pistol', '2017-06-09', 0, 103, 1003),
(04, 'cat', 1, 'Nessie', '2017-09-04', 0, 104, 1004),
(05, 'cat', 2, 'Charlie', '2016-06-04', 1, 105, 1005),
(06, 'cat', 12, 'Francis', '2015-05-07', 1, 106, 1006),
(07, 'rabbit', 5, 'Bunny', '2015-05-06', 0, 107, 1007),
(08, 'turtle', 6, 'Aqua', '2017-06-08', 0, 108, 1008),
(09, 'dog', 16, 'Sammie', '2012-09-07', 1, 109, 1009),
(10, 'dog', 14, 'Dog', '2016-04-06', 0, 110, 1010)
;
insert into Owners
(OwnerID, OwnerFirstName, OwnerLastName, OwnerPhone, OwnerZip, PrefferedPaymentMethod, PatientID)
values
(101, 'Kat', 6, 'Dima', '2017-02-05', 1, 101, 1001),
(102, 'Hunter', 3, 'Misha', '2017-02-05', 1, 102, 1002),
(103, 'Vicky', 2, 'Pistol', '2017-06-09', 0, 103, 1003),
(104, 'Vasnessa', 1, 'Nessie', '2017-09-04', 0, 104, 1004),
(105, 'Weston', 2, 'Charlie', '2016-06-04', 1, 105, 1005),
(106, 'Jonas', 12, 'Francis', '2015-05-07', 1, 106, 1006),
(107, 'Devin', 5, 'Bunny', '2015-05-06', 0, 107, 1007),
(108, 'Grego', 6, 'Aqua', '2017-06-08', 0, 108, 1008),
(109, 'Jackson', 16, 'Sammie', '2012-09-07', 1, 109, 1009),
(110, 'dog', 14, 'Dog', '2016-04-06', 0, 110, 1010)
;
insert into Patients
(PatientID, TypeOfAnimal, PatientAge, PatientName, DateOfLastApt, Microchipped, DoctorID, OwnerID)
values
(01, 'dog', 6, 'Dima', '2017-02-05', 1, 101, 1001),
(02, 'dog', 3, 'Misha', '2017-02-05', 1, 102, 1002),
(03, 'cat', 2, 'Pistol', '2017-06-09', 0, 103, 1003),
(04, 'cat', 1, 'Nessie', '2017-09-04', 0, 104, 1004),
(05, 'cat', 2, 'Charlie', '2016-06-04', 1, 105, 1005),
(06, 'cat', 12, 'Francis', '2015-05-07', 1, 106, 1006),
(07, 'rabbit', 5, 'Bunny', '2015-05-06', 0, 107, 1007),
(08, 'turtle', 6, 'Aqua', '2017-06-08', 0, 108, 1008),
(09, 'dog', 16, 'Sammie', '2012-09-07', 1, 109, 1009),
(10, 'dog', 14, 'Dog', '2016-04-06', 0, 110, 1010)
;
Iは異なる順序でテーブルを作成しようとしました。私も別のデータベースを作成しました。実際は、これを使ってデータベースを作成しました。
drop database if exists kat3_db;
create database if not exists kat3_db;
use kat3_db;
create table if not exists Book(
BookNumber int not null,
BookName varchar(50),
BookPrice decimal(5,2),
CoverType varchar(15) default 'hardcover',
PublicationDate date,
primary key (BookNumber)
);
create table if not exists Course(
CourseNo int not null,
CourseName varchar(20),
Semester varchar(10),
BookNumber int not null,
primary key (CourseNo),
foreign key (BookNumber) references Book(BookNumber)
);
insert into Book
(BookNumber, BookName, BookPrice, CoverType, PublicationDate)
values
(1, 'Math', 023.64, 'paperback', '1999-02-05'),
(3, 'English', 999.36, 'e-book', '2013-06-04'),
(5, 'Calc', 056.69, 'paperback', '1964-05-05'),
(6, 'C++', 053.98, 'paperback', '2016-03-04'),
(7, 'Programming', 113.02, 'paperback', '2001-10-11'),
(9, 'Art', 250.99, 'paperback', '1996-11-11'),
(10, 'Networking', 036.64, 'e-book', '2014-05-06'),
(12, 'Drawing', 111.36, 'paperback', '2013-06-04'),
(13, 'Robotics', 012.03, 'e-book', '2016-06-05'),
(14, 'Computer', 25.03, 'e-book', '2001-06-04')
;
insert into Book
(BookNumber, BookName, BookPrice, PublicationDate)
values
(2, 'Reading', 364.20, '2016-06-05'),
(4, 'Database', 036.64, '2017-06-05'),
(8, 'Wellness', 050.00, '1999-10-10'),
(11, 'Linux', 55.36, '2013-06-08'),
(15, 'FYE', 03.01, '1991-02-05')
;
insert into Course
(CourseNo, CourseName, Semester, BookNumber)
values
(111, 'IntroToMath', 'Spring', 1),
(222, 'IntroToReading', 'Fall', 2),
(333, 'IntroToEnglish', 'Spring', 3),
(444, 'IntroToDatabase', 'Fall', 4),
(555, 'IntroToCalc', 'Spring', 5),
(666, 'IntroToC++', 'Spring', 6),
(777, 'IntroToProgramming', 'Fall', 7),
(888, 'IntroToWellness','Spring', 8),
(999, 'IntroToArt', 'Fall', 9),
(010, 'IntroToNetworking', 'Spring', 10),
(011, 'IntroToLinux', 'Fall', 11),
(012, 'IntroToDrawing', 'Spring', 12),
(013, 'IntroToRobotics', 'Fall', 13),
(014, 'IntroToComputer', 'Spring', 14),
(015, 'IntroToFYE', 'Summer', 15)
;
私は異なったやり方を見ません。どんな提案も大歓迎です。誰かが上の最初のコードを実行し、それが彼らのために働くかどうかを教えてくれれば、それも大きな助けになるでしょう。ありがとうございました！
テーブルを作成する前に、外部キー制約を作成することはできません。 '患者 'の前に'医者 'と'所有者'を作りなさい。 –