рассчитать опыт сотрудника

Question

общее количество сотрудников, а также показать работника, имеющего более 5 лет опыта работы ...

Пожалуйста, помогите мне. Благодарим вас в Advance.

SELECT tbl_Employee.EFirstName, 
     tbl_Employee.EMiddleName, 
     tbl_Employee.ELastName, 
     tbl_Empl‌​oyee.EmployeeID, 
     tbl_BranchMaster.BranchName, 
     tbl_DepartmentMaster.DepartmentName, 
     tbl_StateMaster.‌​StateName, 
     tbl_Employee.EmployeementDate, 
     dbo.FN_Basic(tbl_Employee.EmployeeID) AS basic, 
     YEAR(CURDATE())-YEAR(tbl_Employee.EmployeementDate) AS Age 
FROM tbl_StateMaster 
INNER JOIN tbl_DepartmentMaster 
INNER JOIN tbl_EmployeeStructure 
INNER JOIN tbl_Employee ON tbl_EmployeeStructure.EmployeeID = tbl_Employee.EmployeeID 
INNER JOIN tbl_BranchMaster ON tbl_EmployeeStructure.BranchID = tbl_BranchMaster.BranchID 
          ON tbl_DepartmentMaster.DepartmentID = tbl_EmployeeStructure.DepartmentID 
          ON tbl_StateMaster.StateID =tbl_EmployeeStructure.StateID 

Answer 1

DROP TABLE IF EXISTS employee; 

CREATE TABLE employee(id INT , joining_date DATETIME); 

INSERT INTO `employee` (`id`, `joining_date`) VALUES ('1', NOW()),('2',NOW() - INTERVAL 6 YEAR); 

SELECT * FROM employee; 
+------+---------------------+ 
| id | joining_date  | 
+------+---------------------+ 
| 1 | 2014-02-17 17:56:53 | 
| 2 | 2008-02-17 17:56:53 | 
+------+---------------------+ 
2 rows in set (0.00 sec) 

SELECT id,joining_date FROM employee WHERE DATEDIFF(DATE(NOW()),DATE(joining_date))/365 > 5; 
+------+---------------------+ 
| id | joining_date  | 
+------+---------------------+ 
| 2 | 2008-02-17 17:56:53 | 
+------+---------------------+ 
1 row in set (0.01 sec) 

Вы также можете попробовать

SELECT id,joining_date,DATE_FORMAT(FROM_DAYS(TO_DAYS(NOW())-TO_DAYS(joining_date)),'%Y')+0 AS YearsOfExp FROM employee WHERE DATE_FORMAT(FROM_DAYS(TO_DAYS(NOW())-TO_DAYS(joining_date)),'%Y')+0 > 5; 
+------+---------------------+------------+ 
| id | joining_date  | YearsOfExp | 
+------+---------------------+------------+ 
| 2 | 2008-02-17 17:56:53 |   6 | 
+------+---------------------+------------+ 
1 row in set (0.00 sec) 

Для лучшего использования индексов вы можете записать его как

SELECT * FROM employee WHERE joining_date <= DATE_ADD(NOW(), INTERVAL -5 YEAR);