I'm trying to create a table and am getting an error telling me there's something wrong around line 9. This is the code.
CREATE TABLE shirts_link (
adult VARCHAR(1) NOT NULL,
kids VARCHAR(1) NOT NULL,
babies VARCHAR(1) NOT NULL,
shirt_id INT(4) NOT NULL,
size_id INT(4) NOT NULL,
price_id INT(4) NOT NULL,
PRIMARY KEY (shirt_id,size_id,price_id),
FOREIGN KEY (shirt_id) REFERENCES shirts(id),
FOREIGN KEY (size_id) REFERENCES shirt_sizes(id),
FOREIGN KEY (price_id) REFERENCES shirt_prices(id)
)ENGINE=INNODB;
here's the other tables I'm trying to link to..
CREATE TABLE shirts (
id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY NOT NULL,
shirt_name VARCHAR(20) NOT NULL,
men VARCHAR(10) NULL,
women VARCHAR(10) NULL,
boys VARCHAR(10) NULL,
girls VARCHAR(10) NULL,
babies VARCHAR(10) NULL,
)ENGINE=INNODB;
CREATE TABLE shirt_sizes (
id INT UNSIGNED AUTO_INCREMENT NOT NULL PRIMARY KEY,
size_name VARCHAR(10) NOT NULL
)ENGINE=INNODB;
CREATE TABLE shirt_prices (
id INT UNSIGNED AUTO_INCREMENT NOT NULL PRIMARY KEY,
price_cat VARCHAR(10) NOT NULL,
price NUMERIC(6,2) NOT NULL
)ENGINE=INNODB;
the problem is that you are wrapping table names and column names with single quotes. it shouldn't be or use backtick instead
FOREIGN KEY (shirt_id) REFERENCES shirts(id),
FOREIGN KEY (size_id) REFERENCES shirt_sizes(id),
FOREIGN KEY (price_id) REFERENCES shirt_prices(id)
or
FOREIGN KEY (`shirt_id`) REFERENCES `shirts`(`id`),
FOREIGN KEY (`size_id`) REFERENCES `shirt_sizes`(`id`),
FOREIGN KEY (`price_id`) REFERENCES `shirt_prices`(`id`)
but in this case, they are optional since non of they are MySQL reserved Keywords.
MySQL Reserved Keywords List
UPDATE 1
The data type of the keys must be the same with each other, declare these columns as UNSIGNED
shirt_id INT(4) UNSIGNED NOT NULL,
size_id INT(4) UNSIGNED NOT NULL,
price_id INT(4) UNSIGNED NOT NULL,
SQLFiddle Demo
CREATE TABLE shirts_link (
adult VARCHAR(1) NOT NULL,
kids VARCHAR(1) NOT NULL,
babies VARCHAR(1) NOT NULL,
shirt_id INT(4) NOT NULL,
size_id INT(4) NOT NULL,
price_id INT(4) NOT NULL,
PRIMARY KEY (shirt_id,size_id,price_id),
FOREIGN KEY (shirt_id) REFERENCES shirts(id),
FOREIGN KEY (size_id) REFERENCES shirt_sizes(id),
FOREIGN KEY (price_id) REFERENCES shirt_prices(id)
)ENGINE=INNODB;
it worked for me on sqlfiddle
Here is my code:
CREATE TABLE shirts(
id INT NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE shirts_link (
adult VARCHAR(1) NOT NULL,
kids VARCHAR(1) NOT NULL,
babies VARCHAR(1) NOT NULL,
shirt_id INT(4) NOT NULL,
size_id INT(4) NOT NULL,
price_id INT(4) NOT NULL,
PRIMARY KEY (shirt_id,size_id,price_id),
FOREIGN KEY (shirt_id) REFERENCES shirts(id)
);
Related
This is the parent table
CREATE TABLE Customer(
Customer_ID INT(5) not null,
CustName VARCHAR(50) not null,
CustSurname VARCHAR(50) not null,
CustEmail VARCHAR(100) unique not null,
CustMobileNo INT(12) not null,
HomeAddress VARCHAR(255) not null,
Password VARCHAR(10) not null,
constraint c_cuid_pk PRIMARY KEY (Customer_ID))
ENGINE = InnoDB;
When I try to create the property table it shows me
"#1072 - Key column 'Customer_ID' doesn't exist in table"
this is the child table
CREATE TABLE Property(
Property_ID INT(5) not null,
PhouseNumber INT(4) not null,
PstreetName VARCHAR(50) not null,
PpostCode VARCHAR(7) not null,
Pcity VARCHAR(20) not null,
constraint p_phn_pk PRIMARY KEY (Property_ID),
constraint p_cuid_fk FOREIGN KEY (Customer_ID)
references IndividualCustomer(Customer_ID))
ENGINE = InnoDB;
In table Property, you must define a column named Customer_ID (well it could be named anything!), then you define the FOREIGN KEY on it.
Your code currently tries to link IndificualCustomer(Customer_ID) to Property(Customer_ID), but Property(Customer_ID) does not exist.
Declaring the constraint just links the columns, but it does not create them.
CREATE TABLE Property(
Property_ID INT(5) not null,
PhouseNumber INT(4) not null,
PstreetName VARCHAR(50) not null,
PpostCode VARCHAR(7) not null,
Pcity VARCHAR(20) not null,
Customer_ID INT(5) not null,
constraint p_phn_pk PRIMARY KEY (Property_ID),
constraint p_cuid_fk FOREIGN KEY (Customer_ID)
references IndividualCustomer(Customer_ID))
ENGINE = InnoDB;
In the first place, you should add a column: Customer_ID to your Property table to create your foreign key constraint:
Property.Customer_ID →　IndividualCustomer.Customer_ID
You can not make a constraint on something that does not exist.
Or change the +++ +++ part of your table creation query to a column that exists.
constraint p_cuid_fk FOREIGN KEY (+++Customer_ID+++) --to be changed accordingly
references IndividualCustomer(Customer_ID)
Example: adding the Customer_ID column to your table declaration would give you the following code
CREATE TABLE Property(
Property_ID INT(5) not null,
Customer_ID INT(5) not null,
PhouseNumber INT(4) not null,
PstreetName VARCHAR(50) not null,
PpostCode VARCHAR(7) not null,
Pcity VARCHAR(20) not null,
constraint p_phn_pk PRIMARY KEY (Property_ID),
constraint p_cuid_fk FOREIGN KEY (Customer_ID)
references IndividualCustomer(Customer_ID))
ENGINE = InnoDB;
CREATE TABLE Property(
Property_ID INT(5) not null,
PhouseNumber INT(4) not null,
PstreetName VARCHAR(50) not null,
PpostCode VARCHAR(7) not null,
Pcity VARCHAR(20) not null,
Customer_ID INT(5) not null,
PRIMARY KEY (Property_ID),
KEY p_cuid_fk (Customer_ID),
constraint p_cuid_fk FOREIGN KEY (Customer_ID)
references Customer(Customer_ID))
ENGINE = InnoDB;
You have to add Customer_ID in the table PROPERTY.
USE ccva1;
CREATE TABLE art_piece(
art_inventory_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
artist_id INT(11) NOT NULL,
gallery_id INT(11) NOT NULL,
art_collection VARCHAR(4) NOT NULL,
art_genre VARCHAR(50)NOT NULL,
art_title VARCHAR(50)NOT NULL,
art_medium VARCHAR(50)NOT NULL,
art_acquired VARCHAR(200)NOT NULL,
art_date_acquired INT(10)NOT NULL,
art_appraisal_value VARCHAR(20),
art_size_inches VARCHAR(65),
art_size_centimeters VARCHAR(50),
art_image VARCHAR(500)NOT NULL,
art_modified_by VARCHAR(30) NOT NULL,
art_modified_date VARCHAR(30),
INDEX art_piece_ind(artist_id),
FOREIGN KEY (artist_id)
REFERENCES artist(artist_id)
ON DELETE CASCADE,
INDEX art_piece1_ind(gallery_id),
FOREIGN KEY (gallery_id)
REFERENCES gallery(gallery_id)
ON DELETE CASCADE,
INDEX art_piece2_ind(artist_id),
FOREIGN KEY (art_modified_by)
REFERENCES admin(admin_user_name)
ON DELETE CASCADE )
ENGINE = INNODB;
CREATE TABLE art_exhibition(exhibition_id INT(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
gallery_id INT(11) NOT NULL,
exhibition_name VARCHAR(200) NOT NULL,
exhibition_date VARCHAR(60) NOT NULL,
exhibition_time VARCHAR(30) NOT NULL,
exhibition_description VARCHAR(3000) NOT NULL,
exhibition_image VARCHAR(65) NOT NULL,
exhibition_modified_by VARCHAR(30)NOT NULL,
exhibition_modified_date VARCHAR(30),
INDEX gal_ind(gallery_id),
FOREIGN KEY (gallery_id)
REFERENCES gallery(gallery_id),
INDEX artist_ind(artist_id),
FOREIGN KEY (artist_id)
REFERENCES artist(artist_id),
CONSTRAINT admin_unique UNIQUE (admin_user_name)
) ENGINE = INNODB;
You don't have a column called artist_id in art_exhibition.
Assuming you have a table called artist, define a column with the appropriate type, which is presumably int, before defining the foreign key constraint.
Here is the code below, I'm not sure what this error means...
All I know is that it has something to do with the employeeshifts table.
Does it have something to do with foreign keys?
(insterting this text as I have nothing else to say, but stack requires a lot of text)
CREATE TABLE Customer (
CustomerID INT AUTO_INCREMENT,
Name VARCHAR(90) NOT NULL,
Phone VARCHAR(45) NULL,
CustomerAddress VARCHAR(45) NOT NULL,
PRIMARY KEY (CustomerID));
CREATE TABLE Location (
Address VARCHAR(100) NOT NULL,
Latitude VARCHAR(45) NULL DEFAULT ' ',
Longitude VARCHAR(45) NULL,
PRIMARY KEY (Address));
CREATE TABLE Employee (
EmployeeID INT AUTO_INCREMENT,
Name VARCHAR(90) NOT NULL,
PRIMARY KEY (EmployeeID));
CREATE TABLE Truck (
LicensePlate CHAR(20) NOT NULL,
color VARCHAR(45) NULL,
capacity VARCHAR(45) NULL,
PRIMARY KEY (LicensePlate));
CREATE TABLE Shifts (
ShiftTime DATETIME NOT NULL,
PRIMARY KEY (ShiftTime));
CREATE TABLE EmployeeShifts (
DesiredShift DATETIME NOT NULL,
EmployeeWorking VARCHAR(90) NOT NULL,
DateOfShift DATE,
PRIMARY KEY(DesiredShift, EmployeeWorking, DateOfShift),
FOREIGN KEY (EmployeeWorking) REFERENCES Employee(Name),
FOREIGN KEY (DesiredShift) REFERENCES Shifts(ShiftTime));
CREATE TABLE Reservation (
ReservNum INT NOT NULL,
ReserveDate DATE NULL,
PickupTime VARCHAR(45) NOT NULL,
NumOfPassengers INT NULL,
sheduledTime VARCHAR(45) NULL,
ActualPickupTime VARCHAR(45),
ActualTime VARCHAR(45),
PricePaid VARCHAR(45),
DriverHourlyRate DECIMAL(7,2) NOT NULL,
PassEmployeeHourlyRate DECIMAL (7,2) NOT NULL,
DriverSalary VARCHAR(10),
PassEmployeeSalary VARCHAR(10),
Customer_CustomerID INT,
Truck_LicensePlate char(20) NOT NULL,
Employee_EmployeeID_Driver INT,
Location_Address_Pickup VARCHAR(100),
Employee_EmployeeID_Passenger INT,
Location_Address_Drop VARCHAR(100),
PRIMARY KEY (ReservNum),
FOREIGN KEY (Customer_CustomerID) REFERENCES Customer (CustomerID),
FOREIGN KEY (Truck_LicensePlate) REFERENCES Truck (LicensePlate),
FOREIGN KEY (Employee_EmployeeID_Driver) REFERENCES Employee (EmployeeID),
FOREIGN KEY (Location_Address_Pickup) REFERENCES Location (Address),
FOREIGN KEY (Employee_EmployeeID_Passenger) REFERENCES Employee (EmployeeID),
FOREIGN KEY (Location_Address_Drop) REFERENCES Location (Address));
You need to reference UNIQUE or PRIMARY KEY column:
CREATE TABLE EmployeeShifts (
DesiredShift DATETIME NOT NULL,
EmployeeWorking VARCHAR(90) NOT NULL,
DateOfShift DATE,
PRIMARY KEY(DesiredShift, EmployeeWorking, DateOfShift),
FOREIGN KEY (EmployeeWorking) REFERENCES Employee(Name), -- name is not UNIQUE/PK
FOREIGN KEY (DesiredShift) REFERENCES Shifts(ShiftTime)
);
to:
CREATE TABLE EmployeeShifts (
DesiredShift DATETIME NOT NULL,
EmployeeID INT NOT NULL, -- here
DateOfShift DATE,
PRIMARY KEY(DesiredShift, EmployeeID, DateOfShift),
FOREIGN KEY (EmployeeID) REFERENCES Employee(EmployeeID),
FOREIGN KEY (DesiredShift) REFERENCES Shifts(ShiftTime)
);
SqlFiddleDemo
sorry for my stupid question,but I do database in MySQL and was so happy,when i finished.Unfortunately,I had found a lot of mistakes in my database n when i will properly fix a bug,so i've another. Also,this is my Database and that guy what really boring right now,please write about my bugs and solution for them.I novice as a programmer and this database i do about 4 days !
CREATE TABLE`Skoleni` (
`sk_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`Cena` MEDIUMINT NOT NULL,
`Obsazenost` text NOT NULL,
`Kapacita` datetime NOT NULL,
`Popis` text,
`Prerekvizity` text,
`Certifikat` MEDIUMINT NOT NULL,
PRIMARY KEY (`sk_id`),
FOREIGN KEY (`sk_id`) REFERENCES Skoleni_has_Termin(`Skoleni_sk_id`),
FOREIGN KEY (`sk_id`) REFERENCES Skoleni_has_Uzivatel(`Skoleni_sk_id`)
);
CREATE TABLE `Skoleni_has_Termin`(
`Skoleni_sk_id`INT NOT NULL UNIQUE,
`Termin_ter_id` INT NOT NULL UNIQUE,
PRIMARY KEY (`Skoleni_sk_id`)
);
CREATE TABLE `Termin` (
`ter_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`Datum_cas` DATETIME NOT NULL,
`Misto_mo_id` INT NOT NULL,
PRIMARY KEY (`ter_id`),
FOREIGN KEY (`ter_id`) REFERENCES Skoleni_has_Termin(`Termin_ter_id`)
);
CREATE TABLE `Misto` (
`mo_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`ulice` MEDIUMINT NOT NULL,
`cislo_popisne` MEDIUMINT NOT NULL,
`lat` FLOAT (10,6) NOT NULL,
`lng` FLOAT (10,6) NOT NULL,
PRIMARY KEY (`mo_id`)
)ENGINE = MYISAM;
SELECT TABLE.Misto(
INSERT (`ulice`, `cislo_popisne`, `lat`, `lng`),
VALUES (`dr_Zikmunda_Wintra_376_5``16000 Praha 6 Bubenec``14.407438``50.101049`)
);
CREATE TABLE `Skoleni_has_Uzivatel` (
`Skoleni_sk_id` INT NOT NULL UNIQUE,
`Uzivatel_uziv_id` INT NOT NULL UNIQUE,
PRIMARY KEY (`Skoleni_sk_id`)
);
CREATE TABLE `Uzivatel` (
`uziv_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`Jmeno` VARCHAR(30) NOT NULL,
`Typ` MEDIUMINT UNIQUE,
`Heslo` VARCHAR(32) NOT NULL,
`Potvrzeni` VARCHAR(1) NOT NULL,
PRIMARY KEY (`uziv_id`),
FOREIGN KEY (`uziv_id`) REFERENCES Skoleni_has_Uzivatel(`Uzivatel_uziv_id`)
);
CREATE TABLE `Skoleni_has_Lektor` (
`Skoleni_sk_id` INT NOT NULL UNIQUE,
`Lektor_lek_id` INT NOT NULL UNIQUE,
PRIMARY KEY (`Lektor_lek_id`)
);
CREATE TABLE `Lektor` (
`lek_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`Titul'pred'` VARCHAR(10) NOT NULL,
`Jmeno` VARCHAR(20) NOT NULL,
`Prijmeni` VARCHAR(20) NOT NULL,
`Titul'za'` VARCHAR(10),
`Firma` VARCHAR(30),
`Rodne cislo` CHAR(11) NOT NULL,
`Datum narozeni` DATE NOT NULL,
PRIMARY KEY (`lek_id`)
);
CREATE TABLE `Firma` (
`fir_id` INT NOT NULL UNIQUE AUTO_INCREMENT,
`E-mail` VARCHAR(15) NOT NULL,
`Telefon` VARCHAR(15) NOT NULL,
`Web` VARCHAR(30),
`IC` CHAR(8) NOT NULL,
`DIC` VARCHAR(12) NOT NULL,
`Misto_mo_id` INT NOT NULL,
PRIMARY KEY (`fir_id`),
PRIMARY KEY (`Misto_mo_id`),
FOREIGN KEY (`Misto_mo_id`) REFERENCES Firma(`Misto_mo_id`)
);
The tables `skoleni_has_termin' and 'skoleni_has_lektor' are most likely not needed here. I would bring dates to trainings table and leave just lektor_id in skoleni table as relation between trainings and lecturers is 1:n not m:n (unless one training can have more than one lecturer)
I have an error in my database
Error: Invalid structure on line 18. Refer to our Manual (PHPMYADMIN)
I use (WAMPSERVER 2 32bits)
-PHPMYADMIN
- MYSQL 5.5.6
- PHP 5
Although i need to use InnoDB (ENGINE=InnoDB)
thanks to help me. Take a look at the structure... the variable name meaning is not important.
Here's my code:
DROP TABLE IF EXISTS Adresse;
DROP TABLE IF EXISTS Telephone;
DROP TABLE IF EXISTS Personne;
DROP TABLE IF EXISTS TelPers;
DROP TABLE IF EXISTS Specialiste;
DROP TABLE IF EXISTS Patient;
DROP TABLE IF EXISTS ListePatient;
DROP TABLE IF EXISTS Produit;
DROP TABLE IF EXISTS Medicament;
DROP TABLE IF EXISTS Materiel;
DROP TABLE IF EXISTS Panier;
CREATE TABLE Adresse(
idAdresse INT(100) NOT NULL PRIMARY KEY AUTO_INCREMENT,
num INT(5) NOT NULL,
rue VARCHAR(30) NOT NULL,
ville VARCHAR(15) NOT NULL,
postal VARCHAR(6) NOT NULL
)ENGINE=InnoDB;
CREATE TABLE Telephone(
idTel INT(100) NOT NULL PRIMARY KEY AUTO_INCREMENT,
typeTel VARCHAR(15) NOT NULL,
ind INT(3) NOT NULL,
quartier INT(3) NOT NULL,
num INT(4) NOT NULL,
)ENGINE=InnoDB;
CREATE TABLE Personne(
idPersonne INT(100) PRIMARY KEY NOT NULL AUTO_INCREMENT,
nom VARCHAR(15) NOT NULL,
prenom VARCHAR(15) NOT NULL,
idTel INT(100) NOT NULL,
idAdresse INT(100) NOT NULL,
FOREIGN KEY(idAdresse) REFERENCES Adresse(idAdresse),
FOREIGN KEY(idTel) REFERENCES Telephone(idTel)
)ENGINE=InnoDB;
CREATE TABLE TelPers(
idPersonne INT(100) PRIMARY KEY NOT NULL,
idTel INT(100) PRIMARY KEY NOT NULL,
FOREIGN KEY(idPersonne) REFERENCES Personne(idPersonne),
FOREIGN KEY(idTel) REFERENCES Telephone(idTel)
)ENGINE=InnoDB;
CREATE TABLE Specialiste(
login VARCHAR(10) PRIMARY KEY NOT NULL PRIMARY KEY,
password VARCHAR(10) NOT NULL,
profession VARCHAR(20) NOT NULL,
idListeP INT(5),
idPanier INT(5),
idPersonne INT(100),
FOREIGN KEY(idPersonne) REFERENCES Personne(idPersonne)
)ENGINE=InnoDB;
CREATE TABLE Patient(
idPatient INT(10) NOT NULL PRIMARY KEY AUTO_INCREMENT,
sexe CHAR NOT NULL,
anniv DATE,
assurance INT(3) NOT NULL,
idPersonne INT(100),
FOREIGN KEY(idPersonne) REFERENCES Personne(idPersonne)
)ENGINE=InnoDB;
CREATE TABLE ListePatient(
idListeP INT(5) NOT NULL PRIMARY KEY,
idPatient INT(10)NOT NULL PRIMARY KEY,
FOREIGN KEY(idListeP) REFERENCES Specialiste(idListeP),
FOREIGN KEY(idPatient) REFERENCES Patient(idPatient)
)ENGINE=InnoDB;
CREATE TABLE Produit(
idProduit INT(100) NOT NULL PRIMARY KEY AUTO_INCREMENT,
nom VARCHAR(50) NOT NULL,
descr VARCHAR(255) NOT NULL,
prix DECIMAL(5,2) NOT NULL,
qte INT(100) NOT NULL
)ENGINE=InnoDB;
CREATE TABLE Medicament(
idMedic INT(100)NOT NULL PRIMARY KEY AUTO_INCREMENT,
marque VARCHAR(10) NOT NULL,
typeMed VARCHAR(10) NOT NULL,
idProduit INT(100) NOT NULL,
FOREIGN KEY(idProduit) REFERENCES Produit(idProduit)
)ENGINE=InnoDB;
CREATE TABLE Materiel(
idMateriel INT(100) NOT NULL PRIMARY KEY AUTO_INCREMENT,
rabais INT(99) NOT NULL,
idProduit INT(100) NOT NULL,
FOREIGN KEY(idProduit) REFERENCES Produit(idProduit)
)ENGINE=InnoDB;
CREATE TABLE Panier(
idPanier INT(5) NOT NULL PRIMARY KEY,
idProduit INT(100) NOT NULL PRIMARY KEY,
FOREIGN KEY(idPanier) REFERENCES Specialiste(idPanier),
FOREIGN KEY(idProduit) REFERENCES Produit(idProduit)
)ENGINE=InnoDB;
THANKS FOR YOUR HELP! :D
In Telephone declaration there is a comma (,) before closing )
In TelPers there are two primary keys
In Specialiste there is PRIMARY KEY twice in login column
In ListePatien there are two primary keys
In Panier there are two primary keys
To define a PRIMARY key on two columns use
CREATE TABLE Panier(
idPanier INT(5) NOT NULL,
idProduit INT(100) NOT NULL,
PRIMARY KEY (`idPanier`,`idProduit`),
FOREIGN KEY(idPanier) REFERENCES Specialiste(idPanier),
FOREIGN KEY(idProduit) REFERENCES Produit(idProduit)
)ENGINE=InnoDB;
(same on other tables)