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EmployeeID) est la clé primaire.DepartmentID) est la clé primaire.ProjectID) est la clé primaire.#EmployeeID, #ProjectID) est une clé primaire composite (relation plusieurs-à-plusieurs entre Employee et Project).SELECT DepartmentID FROM Employee WHERE COUNT(*) > 5;B) SELECT DepartmentID, COUNT(*) FROM Employee GROUP BY DepartmentID HAVING COUNT(*) > 5;C) SELECT DepartmentID, COUNT(*) FROM Employee HAVING COUNT(*) > 5;D) SELECT DepartmentID FROM Employee GROUP BY DepartmentID;EmployeeID) est la clé primaire.DepartmentID) est la clé primaire.ProjectID) est la clé primaire.#EmployeeID, #ProjectID) est une clé primaire composite (relation plusieurs-à-plusieurs entre Employee et Project).SELECT COUNT(*) FROM Employee;B) SELECT DepartmentID, COUNT(*) FROM Employee;C) SELECT DepartmentID, COUNT(*) FROM Employee GROUP BY DepartmentID;D) SELECT DepartmentID, COUNT(*) FROM Employee HAVING COUNT(*) > 1;Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
Which query selects employees whose salary is greater than all employees in department 2?A) SELECT * FROM Employee WHERE Salary > ALL (SELECT Salary FROM Employee WHERE DepartmentID = 2);B) SELECT * FROM Employee WHERE Salary > (SELECT Salary FROM Employee WHERE DepartmentID = 2);C) SELECT * FROM Employee WHERE Salary > ANY (SELECT Salary FROM Employee WHERE DepartmentID = 2);D) SELECT * FROM Employee WHERE Salary IN (SELECT Salary FROM Employee WHERE DepartmentID = 2);
EmployeeID) est la clé primaire.DepartmentID) est la clé primaire.ProjectID) est la clé primaire.#EmployeeID, #ProjectID) est une clé primaire composite (relation plusieurs-à-plusieurs entre Employee et Project).Quelle requête sélectionne les employés dont le salaire est supérieur à tous les employés du département 2 ?A) SELECT * FROM Employee WHERE Salary > ALL (SELECT Salary FROM Employee WHERE DepartmentID = 2);B) SELECT * FROM Employee WHERE Salary > (SELECT Salary FROM Employee WHERE DepartmentID = 2);C) SELECT * FROM Employee WHERE Salary > ANY (SELECT Salary FROM Employee WHERE DepartmentID = 2);D) SELECT * FROM Employee WHERE Salary IN (SELECT Salary FROM Employee WHERE DepartmentID = 2);
Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
Which query retrieves all employees with a salary greater than 50,000?A) SELECT * FROM Employee WHERE Salary >= 50000;B) SELECT Salary FROM Employee WHERE Salary >= 50000;C) SELECT * FROM Employee HAVING Salary >= 50000;D) SELECT * FROM Employee GROUP BY Salary >= 50000;
Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
SELECT * FROM Employee WHERE DepartmentID NOT IN (3,5);B) SELECT * FROM Employee WHERE DepartmentID != (3,5);C) SELECT * FROM Employee WHERE DepartmentID <> (3,5);D) SELECT * FROM Employee WHERE DepartmentID NOT ANY (3,5);Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
SELECT AVG(Salary) WHERE Employee; B) SELECT MEAN(Salary) FROM Employee;C) SELECT AVG(Salary) FROM Employee GROUP BY Salary;D) SELECT AVG(Salary) FROM Employee;EmployeeID) est la clé primaire.DepartmentID) est la clé primaire.ProjectID) est la clé primaire.#EmployeeID, #ProjectID) est une clé primaire composite (relation plusieurs-à-plusieurs entre Employee et Project).SELECT * FROM Employee INNER JOIN Department;B) SELECT Employee.FirstName, Employee.LastName, Department.DepartmentName FROM Employee JOIN Department ON Employee.DepartmentID = Department.DepartmentID;C) SELECT FirstName, LastName FROM Employee, Department;D) SELECT Employee.FirstName, Employee.LastName, Department.DepartmentName FROM Employee RIGHT JOIN Department;Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
How do you sort employees by salary in descending order?A) SELECT * FROM Employee ORDER Salary DESC;B) SELECT * FROM Employee ORDER BY Salary DESC;C) SELECT * FROM Employee SORT BY Salary DESC;D) SELECT * FROM Employee GROUP BY Salary DESC;
Employee (EmployeeID, FirstName, LastName, Email, Phone, HireDate, Salary, #DepartmentID)Department (DepartmentID, DepartmentName, #ManagerID)Project (ProjectID, ProjectName, StartDate, EndDate, Budget)Employee_Project (#EmployeeID, #ProjectID, Role, AssignmentDate)
SELECT COUNT(*) FROM Employee;B) SELECT DepartmentID, COUNT(*) FROM Employee;C) SELECT DepartmentID, COUNT(*) FROM Employee GROUP BY DepartmentID;D) SELECT DepartmentID, COUNT(*) FROM Employee HAVING COUNT(*) > 1;