My SQL WORKBENCH**Please note that you may actually use as many tables as you need. You are not limited to use only twotables.Example:SELECT *FROM DEPARTMENTS;SELECT *FROM EMPLOYEES;Now join departments and employees tables.SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTSWHERE EMPLOYEES.DEPARTMENT_ID = DEPARTMENTS.DEPARTMENT_ID;NOTE: If you omit the join condition, the result will be the Cartesian Product of the two tables. A CartesianProduct joins all rows of the first table with all rows of the second table.That is if the first table has n rows and the second table has m rows, the output will have n*m rows.The following query will produce a Cartesian Product:SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTS;**When table names are long, qualifying column names might be very time consuming. Usually peopleprefer to use table aliases for this purpose.Example: List the name of all employees who work for 'Purchasing' departmentSELECT E.FIRST_NAME "EMPLOYEE NAME"FROM EMPLOYEES E, DEPARTMENT DWHERE E.DEPARTMENT_ID = D.DEPARTMENT_ID ANDUPPER(D.DEPARTMENT_NAME) = 'PURCHASING';**Table joins can be performed on columns that are not primary or foreign keys as well.SELECT E.FIRST_NAMEFROM EMPLOYEES A, EMPLOYEES BWHERE A.SALARY = B.SALARY AND A.EMPLOYEE_ID != B.EMPLOYEE_ID; **All these joins are called equijoin because they use an equality to form the join. It is also possible to formnonequijoins. These are joins that use an operator (+) than other.Normally if a row does not satisfy a join condition, it will be left out of the result. You want to see rows of atable that do not satisfy the join condition as well, you have to perform an outer join. Outer join is performedby placing a (+) sign on the right hand side of one of the attribute of join condition.The (+) sign makes it possible to join a 'NULL' row of the table it is close to with rows from the othertable. In other words (+) will allow NULL rows on the side that is placed.Syntax of an outer join:SELECT table1.column11, table1.column12, ..., table2.column21, table2.column22FROM table1, table2WHERE table1.column11(+) = table2.column21(+);**Please be careful, you can use only one (+) in the join condition.Outer Examples:Example1: Display last name of all employees with their department information (department id and departmentname) including those employees who does not work for any department.SELECT e.last_name,d.department_id,d.department_nameFROM employees e, departments dWHERE e.department_id=d.department_id(+);Example2: Display last name of all employees with their department information (department id anddepartment name) including those departments that they do not have any employee.SELECT e.last_name,d.department_id,d.department_nameFROM employees e, departments dWHERE e.department_id(+)=d.department_id;SELF JOINJoining a Table to Itself:Sometimes you need to join a able to itself.To find the name of each employee's manager, you need to join the EMPLOYEES table to itself.Example:Find the name of the managers for all employees.SELECT e.last_name,m.last_nameFROM employees e, employees mWHERE .manager_id=m.employee_id;SQL desc locations Name Null? Type----------------------------------------- ------------------------------------LOCATION_ID NOT NULL NUMBER(6)DEPARTMENT_ID NOT NULL NUMBER(6)ADDRESS VARCHAR2(25)CITY NOT NULL VARCHAR2(25)STATE NOT NULL VARCHAR2(25)COUNTRY VARCHAR2(20) EXERCISESPART A:1-Display the addresses of all the departments. Show locationID, street address,city,state and country name in theoutput. (Use Locations table)2-Display the last name, department number and deparment name for all employees.3-Display the last name, job, department number, and department name for all employees work inToronto. (Employees,Departments,Locations)

Question

My SQL WORKBENCH**Please note that you may actually use as many tables as you need. You are not limited to use only twotables.Example:SELECT *FROM DEPARTMENTS;SELECT *FROM EMPLOYEES;Now join departments and employees tables.SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTSWHERE EMPLOYEES.DEPARTMENT_ID = DEPARTMENTS.DEPARTMENT_ID;NOTE: If you omit the join condition, the result will be the Cartesian Product of the two tables. A CartesianProduct joins all rows of the first table with all rows of the second table.That is if the first table has n rows and the second table has m rows, the output will have n*m rows.The following query will produce a Cartesian Product:SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTS;**When table names are long, qualifying column names might be very time consuming. Usually peopleprefer to use table aliases for this purpose.Example: List the name of all employees who work for 'Purchasing' departmentSELECT E.FIRST_NAME &#34;EMPLOYEE NAME&#34;FROM EMPLOYEES E, DEPARTMENT DWHERE E.DEPARTMENT_ID = D.DEPARTMENT_ID ANDUPPER(D.DEPARTMENT_NAME) = 'PURCHASING';**Table joins can be performed on columns that are not primary or foreign keys as well.SELECT E.FIRST_NAMEFROM EMPLOYEES A, EMPLOYEES BWHERE A.SALARY = B.SALARY AND A.EMPLOYEE_ID != B.EMPLOYEE_ID;  **All these joins are called equijoin because they use an equality to form the join. It is also possible to formnonequijoins. These are joins that use an operator (+) than other.Normally if a row does not satisfy a join condition, it will be left out of the result. You want to see rows of atable that do not satisfy the join condition as well, you have to perform an outer join. Outer join is performedby placing a (+) sign on the right hand side of one of the attribute of join condition.The (+) sign makes it possible to join a 'NULL' row of the table it is close to with rows from the othertable. In other words (+) will allow NULL rows on the side that is placed.Syntax of an outer join:SELECT table1.column11, table1.column12, ..., table2.column21, table2.column22FROM table1, table2WHERE table1.column11(+) = table2.column21(+);**Please be careful, you can use only one (+) in the join condition.Outer Examples:Example1: Display last name of all employees with their department information (department id and departmentname) including those employees who does not work for any department.SELECT e.last_name,d.department_id,d.department_nameFROM employees e, departments dWHERE e.department_id=d.department_id(+);Example2: Display last name of all employees with their department information (department id anddepartment name) including those departments that they do not have any employee.SELECT e.last_name,d.department_id,d.department_nameFROM employees e, departments dWHERE e.department_id(+)=d.department_id;SELF JOINJoining a Table to Itself:Sometimes you need to join a able to itself.To find the name of each employee's manager, you need to join the EMPLOYEES table to itself.Example:Find the name of the managers for all employees.SELECT e.last_name,m.last_nameFROM employees e, employees mWHERE .manager_id=m.employee_id;SQL> desc locations Name Null? Type----------------------------------------- ------------------------------------LOCATION_ID NOT NULL NUMBER(6)DEPARTMENT_ID NOT NULL NUMBER(6)ADDRESS VARCHAR2(25)CITY NOT NULL VARCHAR2(25)STATE NOT NULL VARCHAR2(25)COUNTRY VARCHAR2(20)  EXERCISESPART A:1-Display the addresses of all the departments. Show locationID, street address,city,state and country name in theoutput. (Use Locations table)2-Display the last name, department number and deparment name for all employees.3-Display the last name, job, department number, and department name for all employees work inToronto. (Employees,Departments,Locations)

Accepted Answer

To display the addresses of all the departments with location information:

SELECT l.LOCATION_ID,…