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Hier ist mein SQL-Code (den ich mit SQL Fiddle getestet habe) für eine Zuweisung. Der Fehler, den ich bekomme, sagt, es gibt ein Problem mit ) in Zeile 14 ("REFERENCES Department(Dnumber)"), aber es sagt, dass, selbst wenn dort keine Klammern da sind. Was stimmt nicht mit der Syntax dieses Codes?MySQL: "Überprüfen Sie das Handbuch, das Ihrer MySQL-Server-Version für die richtige Syntax in der Nähe ')' in Zeile 14 entspricht"
Die Fehlermeldung lautet:
„Sie haben einen Fehler in der SQL-Syntax, das Handbuch, dass zu Ihrer MySQL-Server-Version für die richtige Syntax entspricht zu verwenden, in der Nähe von‚)‘in Zeile 14 "
#Create the table definitions
CREATE TABLE Employee(
Fname VARCHAR(20) NOT NULL,
Minit VARCHAR(1),
Lname VARCHAR(20) NOT NULL,
Ssn CHAR(9) PRIMARY KEY,
Bdate DATE,
Address VARCHAR(30),
Sex CHAR(1),
Salary NUMERIC(10,2),
Super_ssn CHAR(9) REFERENCES Department(Mgr_ssn),
Dno INTEGER REFERENCES Department(Dnumber),
);
CREATE TABLE Department(
Dname VARCHAR(15) NOT NULL,
Dnumber INTEGER PRIMARY KEY,
Mgr_ssn CHAR(9),
Mgr_start_date DATE,
);
CREATE TABLE Dept_Locations(
Dnumber INTEGER REFERENCES Department,
Dlocation VARCHAR(20),
PRIMARY KEY (Dnumber, Dlocation)
);
CREATE TABLE Project(
Pname VARCHAR(20) NOT NULL,
Pnumber INTEGER PRIMARY KEY,
Plocation VARCHAR(20),
Dnum INTEGER REFERENCES Department,
);
CREATE TABLE Works_On(
Essn CHAR(9) REFERENCES Employee(Ssn),
Pno INTEGER REFERENCES Project(Pnumber),
Hours NUMERIC(2,1),
PRIMARY KEY(Essn, Pno)
);
CREATE TABLE Dependent(
Essn CHAR(9) REFERENCES Employee(Ssn),
Dependent_name VARCHAR(20) NOT NULL,
Sex CHAR(1),
Bdate DATE,
Relationship VARCHAR(20),
PRIMARY KEY(Essn, Dependent_name)
);
#Insert records into the tables
INSERT INTO Employee
Values
('John', 'B','Smith', '123456789','1965-01-09','731 Fondren, Houston, TX','M',30000,'333445555',5),
('Franklin','T','Wong', '333445555','1955-12-08','638 Voss, Houston, TX', 'M',40000,'888665555',5),
('Alicia', 'J','Zelaya', '999887777','1968-01-19','3321 Castle, Spring, TX', 'F',25000,'987654321',4),
('Jennifer','S','Wallace','987654321','1941-06-20','291 Berry, Bellaire, TX', 'F',43000,'888665555',4),
('Ramesh', 'K','Narayan','666884444','1962-09-15','975 Fire Oak, Humble, TX','M',38000,'333445555',5),
('Joyce', 'A','English','453453453','1972-07-31','5631 Rice, Houston, TX', 'F',25000,'333445555',5),
('Ahmad', 'V','Jabbar', '987987987','1969-03-29','980 Dallas, Houston, TX', 'M',25000,'987654321',4),
('James', 'E','Borg', '888665555','1937-11-10','450 Stone, Houston, TX', 'M',55000, NULL, 1);
INSERT INTO Department
Values
('Research', 5,'333445555','1988-05-22'),
('Administration',4,'987654321', '1995,01,01'),
('Headquarters', 1,'888665555','1981-06-19');
INSERT INTO Dept_Locations
Values
(1,'Houston'),
(4,'Stafford'),
(5,'Bellaire'),
(5,'Sugarland'),
(5,'Houston');
INSERT INTO Project
Values
('ProductX', 1,'Bellaire', 5),
('ProductY', 2,'Sugarland',5),
('ProductZ', 3,'Houston', 5),
('Computerization',10,'Stafford', 4),
('Reorganization', 20,'Houston', 1),
('Newbenefits', 30,'Stafford', 4);
INSERT INTO Works_On
Values
('123456789', 1,32.5),
('123456789', 2, 7.5),
('666884444', 3,40.0),
('453453453', 1,20.0),
('453453453', 2,20.0),
('333445555', 2,10.0),
('333445555', 3,10.0),
('333445555',10,10.0),
('333445555',20,10.0),
('999887777',30,30.0),
('999887777',10,10.0),
('987987987',10,35.0),
('987987987',30, 5.0),
('987654321',30,20.0),
('987654321',20,15.0),
('888665555',20,NULL);
INSERT INTO Dependent
Values
('333445555','Alice', 'F','1986-04-05','Daughter'),
('333445555','Theodore', 'M','1983-10-25','Son'),
('333445555','Joy', 'F','1958-05-03','Spouse'),
('987654321','Abner', 'M','1942-02-28','Spouse'),
('123456789','Michael', 'M','1988-01-04','Son'),
('123456789','Alice', 'F','1988-12-30','Daughter'),
('123456789','Elizabeth','F','1967-05-05','Spouse');
#Show values in all tables
SELECT *
FROM Employee;
SELECT *
FROM Department;
SELECT *
FROM Dept_Locations;
SELECT *
FROM Project;
SELECT *
FROM Works_On;
SELECT *
FROM Dependent;
#Implement queries from Exercise 8.16
#a)
SELECT E.Fname, E.Lname
FROM Employee E, Works_On W, Project P
WHERE E.Ssn = W.Essn and
P.Pnum = W.Pno and
E.Dno = 5 and
P.Pname = 'ProjectX' and
W.Hours > 10.0;
#b)
SELECT E.Fname, E.Lname
FROM Employee E, Dependent D
WHERE E.Ssn = D.Essn and
E.Fname = D.Dependent_name;
#c)
SELECT Fname, Lname
FROM Employee
WHERE Super_ssn = (SELECT Ssn
FROM Employee
WHERE Fname = 'Franklin' and
Lname = 'Wong');
#d)
SELECT SUM(W.Hours)
FROM Project P, Works_On W
WHERE P.Pnumber = W.Pno
GROUP BY P.Pname;
#e)
SELECT E.Fname, E.Lname
FROM Employee E
WHERE NOT EXISTS
(SELECT P.Pnumber
FROM Project P
WHERE P.Pnumber NOT IN
(SELECT W.Pno
FROM Works_On W
WHERE W.Essn = E.Ssn));
#f)
SELECT E.Fname, E.Lname
FROM Employee E
WHERE EXISTS
(E.Essn
MINUS
SELECT [Distinct] W.Essn
FROM Works_On W
);
#g)
SELECT AVG(E.Salary)
FROM Employee E, Department D
WHERE E.Dno = D.Dnumber
GROUP BY D.Dname;
#h)
SELECT AVG(Salary)
FROM Employee
WHERE Sex = 'F'
GROUP BY Sex;
#i)
SELECT E.Fname, E.Lname, E.Address
FROM Employee E
WHERE EXISTS
(SELECT P.Pnumber
FROM Project P, Works On W
WHERE P.Pnumber = W.Pno and
W.Essn = E.Ssn and
Plocation = 'Houston' and
NOT EXISTS
(SELECT Dl.Dnumber
FROM Dept_Locations Dl
WHERE Dl.Dlocation = 'Houston' and
E.Dno = Dl.Dnumber));
#j)
SELECT E.Lname
FROM Employee E, Department De
WHERE De.Mgr_ssn = E.Ssn and
NOT EXISTS
(SELECT Dp.Essn
FROM Dependent Dp
WHERE Dp.Essn = E.Ssn);
#Drop all tables
DROP TABLE Dependent RESTRICT;
DROP TABLE Works_On RESTRICT;
DROP TABLE Project RESTRICT;
DROP TABLE Dept_Locations RESTRICT;
DROP TABLE Department RESTRICT;
DROP TABLE Employee CASCADE;
Wenn Sie eine Tabelle referenzieren, müssen Sie sie vorher definieren. Setzen Sie "Abteilung" auf die Oberseite. –