![5Expressions(32) List the last names of all current employees, their position titles, the last names of their supervisors and the departments in which they work. The view should have 4 columns named “Supervisee”, “Position”, “Supervisor”, and “Department” and be in alphabetical order by supervisor’s last name.WHERE EmpRec.endDate IS NULLSELECT Person.lName AS Supervisee, Position.posTitle AS Position, Person.lName AS Supervisor, Dept.deptName AS DepartmentORDER BY Person.lName [Supervisor]](https://image.slidesharecdn.com/complexinnerjoins-110417140039-phpapp02/85/Complex-inner-joins-5-320.jpg)
![8Another ExampleSELECT A.lName AS Designer, B.lName AS [Design Assistant], SUM(H.desHr) AS [Designer Hours], SUM(H.desAstHr) AS [Design Assistant Hours], D.projNo AS [Project Number], C.lName AS ClientFROM ProjDetail D, ProjHr H, Project P, Person A, Person B, Person CWHERE D.dtID=H.dtID AND P.projNo=D.projNo AND P.dID=A.pID AND P.daID=B.pID AND P.cID=C.pIDGROUP BY D.projNo, A.lName, B.lName, C.lName;](https://image.slidesharecdn.com/complexinnerjoins-110417140039-phpapp02/85/Complex-inner-joins-8-320.jpg)
Complex inner joins
- 2. 2Strategy for Complex Joins(A) Identify which table columns contain the data needed.(B) Use the PK-FK pairs that make each join to plot the join “pathway”. Identify expressions to perform column and/or row operations needed and the clauses in which they belong.
- 3. 3Table Columns(32) List the last names of all current employees, their position titles, the last names of their supervisors and the departments in which they work. The view should have 4 columns named “Supervisee”, “Position”, “Supervisor”, and “Department” and be in alphabetical order by supervisor’s last name.Person.lNamePosition.posTitlePerson.lNameDept.deptName
- 4. 4Join PathwayEmpRecempIDPerson.pIDDept.deptIDsupIDdeptIDposIDPosition.posIDBecause Person.pID makes 2 joins to EmpRec, we’ll have to make a virtual copy of Person.
- 5. 5Expressions(32) List the last names of all current employees, their position titles, the last names of their supervisors and the departments in which they work. The view should have 4 columns named “Supervisee”, “Position”, “Supervisor”, and “Department” and be in alphabetical order by supervisor’s last name.WHERE EmpRec.endDate IS NULLSELECT Person.lName AS Supervisee, Position.posTitle AS Position, Person.lName AS Supervisor, Dept.deptName AS DepartmentORDER BY Person.lName [Supervisor]
- 6. 6SQL StatementCombine the 3 steps and tweak the syntax to get the complete query.SELECT P.lName AS Supervisee, A.posTitle AS Position, B.LName AS Supervisor, D.deptName AS DepartmentFROM Person P, Position A, Person B, Dept D, EmpRec EWHERE P.pID=E.empID AND B.pID=E.supID AND E.deptID=D.deptID AND E.posID=A.posID AND E.endDate IS NULLORDER BY B.lName;
- 7. 7Another Example(33) List the designers’ and design assistants’ last names, the total numbers of hours each worked on each project, the project numbers, and the clients’ last names. The view should have 6 columns: “Designer”, “Design Assistant”, “Designer Hours”, “Design Assistant Hours”, “Project Number” and “Client”.
- 8. 8Another ExampleSELECT A.lName AS Designer, B.lName AS [Design Assistant], SUM(H.desHr) AS [Designer Hours], SUM(H.desAstHr) AS [Design Assistant Hours], D.projNo AS [Project Number], C.lName AS ClientFROM ProjDetail D, ProjHr H, Project P, Person A, Person B, Person CWHERE D.dtID=H.dtID AND P.projNo=D.projNo AND P.dID=A.pID AND P.daID=B.pID AND P.cID=C.pIDGROUP BY D.projNo, A.lName, B.lName, C.lName;