我有一個錯誤在我的數據庫MySQL錯誤在創建表
Error: Invalid structure on line 18. Refer to our Manual (PHPMYADMIN)
我使用(WAMPSERVER 2 32位) -PHPMYADMIN - MYSQL 5.5.6 - PHP 5
雖然我需要使用InnoDB(ENGINE=InnoDB
)
感謝幫助我。看看在結構 ...變量名的含義並不重要。
這裏是我的代碼:
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;
感謝您的幫助! :d
那麼如果我們沒有錯誤。其種類無用...... – Neal
同意。請發佈您遇到的錯誤,以幫助將某人指向正確的方向。 – RobB
錯誤:行18無效的結構,請參考我們的手冊(PHPMYADMIN) – Hulk