TABLE- TITLE WORKER_REF_ID WORKER_TITLE AFFECTED_FROM 1 Manager 2016-02-20 00:00:00 2 Executive 2016-06-11 00:00:00 8 Executive 2016-06-11 00:00:00 5 Manager 2016-06-11 00:00:00 4 Asst. Manager 2016-06-11 00:00:00 7 Executive 2016-06-11 00:00:00 6. Lead 2016-06-11 00:00:00 3 Lead 2016-06-11 00:00:00
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- Changing the. EmployeeID = 1 to 10 in the table Employee. If it is rejected, explain EMPLOYEE( EmployeeID, EmployeeName, SupervisorID, DepartmentID) PROJECT (ProjectID, EmployeeID) DEPARTMENT( Department ID, DepartmentName) CREATE TABLE EMPLOYEE ( EmployeeID INT PRIMARY KEY, EmployeeName VARCHAR(50) NOT NULL, SupervisorID INT DEFAULT 9, DepartmentID INT. DEFAULT 6, FOREIGN KEY (SupervisorID) REFERENCES EMPLOYEE (EmployeeID) ON DELETE SET DEFAULT , FOREIGN KEY (DepartmentID) REFERENCES DEPARTMENT(DepartmentID) ON UPDATE SET NULL ); CREATE TABLE PROJECT ( ProjectID INT PRIMARY KEY, EmployeeID INT DEFAULT 9, FOREIGN KEY (EmployeeID) REFERENCES EMPLOYEE (EmployeeID) ON DELETE SET NULL ON UPDATE CASCADE ); CREATE TABLE DEPARTMENT( DepartmentID INT PRIMARY KEY, DepartmentName VARCHAR(50) EMPLOYEE EmployeeID EmployeeName SupervisorID DepartmentID 6 A 7 6 7 B 8 6…Access Assignment Problem: JMS TechWizards is a local company that provides technical services to several small businesses in the area. The company currently keeps its technicians and clients’ records on papers. The manager requests you to create a database to store the technician and clients’ information. The following table contains the clients’ information. Client Number Client Name Street City State Postal Code Telephone Number Billed Paid Technician Number AM53 Ashton-Mills 216 Rivard Anderson TX 78077 512-555-4070 $315.50 $255.00 22 AR76 The Artshop 722 Fisher Liberty Corner TX 78080 254-555-0200 $535.00 $565.00 23 BE29 Bert's Supply 5752 Maumee Liberty Corner TX 78080 254-555-2024 $229.50 $0.00 23 DE76 D & E Grocery 464 Linnell Anderson TX 78077 512-555-6050 $485.70…Problem: JMS TechWizards is a local company that provides technical services to several small businesses in the area. The company currently keeps its technicians and clients’ records on papers. The manager requests you to create a database to store the technician and clients’ information. The following table contains the clients’ information. Client Number Client Name Street City State Postal Code Telephone Number Billed Paid Technician Number AM53 Ashton-Mills 216 Rivard Anderson TX 78077 512-555-4070 $315.50 $255.00 22 AR76 The Artshop 722 Fisher Liberty Corner TX 78080 254-555-0200 $535.00 $565.00 23 BE29 Bert's Supply 5752 Maumee Liberty Corner TX 78080 254-555-2024 $229.50 $0.00 23 DE76 D & E Grocery 464 Linnell Anderson TX 78077 512-555-6050 $485.70 $400.00 29 GR56 Grant Cleaners 737 Allard Kingston TX 78084 512-555-1231 $215.00 $225.00 22…
- 1.6) Constraints should be used to force valid data entry / modification to the databaseTrue False 1.7) A primary key can contain only single row with null value since duplicate values are not allowedTrue False 1.8) Comparison with null can be meaningfully done with equals and not equals operatorTrue False 8. Create a trigger for the Invoices table that automatically inserts the vendor name and address for a paid invoice into a table named ShippingLabels. The trigger should fire any time the PaymentTotal column of the Invoices table is updated. The structure of the ShippingLabels table is as follows: CREATE TABLE ShippingLabels (VendorName varchar(50), VendorAddress1 varchar(50), VendorAddress2 varchar(50), VendorCity VendorState VendorZipCode varchar(50), char(2), varchar(20)); Use this UPDATE statement to test the trigger: UPDATE Invoices SET PaymentTotal = 67.92, PaymentDate = '2020-02-23' WHERE InvoiceID = 100;Topic: Database Software: SQLite Schema Table: CREATE TABLE name_basics (nconst VARCHAR(45) NOT NULL, primaryName VARCHAR(45) NULL, birthYear YEAR(4) NULL, deathYear YEAR(4) NULL, age INT NULL, movieNum INT NULL, PRIMARY KEY (nconst)); CREATE TABLE title_episode (episodeID VARCHAR(45) NOT NULL,parentTconst VARCHAR(45) NULL, seasonNumber INT NULL,episodeNumber INT NULL, PRIMARY KEY (episodeID), CONSTRAINT parentTconst FOREIGN KEY (parentTconst) REFERENCES title_basics (tconst) ON DELETE NO ACTION ON UPDATE NO ACTION); CREATE TABLE title_principals (tconst VARCHAR(45) NOT NULL, ordering INT NULL, nconst VARCHAR(45) NULL, category VARCHAR(45) NULL,job VARCHAR(45) NULL, characters VARCHAR(45) NOT NULL, PRIMARY KEY (tconst, characters), CONSTRAINT nconst FOREIGN KEY (nconst)REFERENCES name_basics (nconst) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT tconst FOREIGN KEY (tconst) REFERENCES title_basics (tconst) ON DELETE NO ACTION ON UPDATE NO ACTION); CREATE TABLE title_ratings…
- Topic 1: The 5 JoinsRun each of the following statements in APEX. Explain the differences in results.SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES EINNER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ELEFT OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ERIGHT OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES EFULL OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ECROSS JOIN HOL_JOBS J; DBMS 130 Worksheet: JOINS and SET Operations2Topic 2: Write an SQL cartesian product between Regions and Countries.Copy your APEX here.How many rows are in the result set?How could you have calculated the number of rows in the result set BEFORE running yourquery?Topic 3: Briefly explain how set operators handle NULL values.Run the following 3 SQL queries in APEX which use set operators over these…A company provides reimbursement of mobile phone subscription charges toits employees belonging to the managerial cadre and above. The following recordstructure captures the details. employee number, which is designated as theprimary key, is a numerical 3-digit key. type refers to a post-paid or pre-paid classof subscription to the mobile service. subscription charges refer to thecharges incurred by the employee at the end of every month.employeenumber designation mobilenumber type SubscriptionchargesFor a sample set of records implement the file asa) an array of records (block size = 1), andb) an array of pointers to records (assume that each pointer to a record is a linkedlist of two nodes, each representing a record. In other words, each block is a linkedlist of two nodes (block size = 2)).Topic: Database Software: SQLite Schema Table: CREATE TABLE name_basics (nconst VARCHAR(45) NOT NULL, primaryName VARCHAR(45) NULL, birthYear YEAR(4) NULL, deathYear YEAR(4) NULL, age INT NULL, movieNum INT NULL, PRIMARY KEY (nconst)); CREATE TABLE attributes (attributeID INT NOT NULL,attribute VARCHAR(45) NOT NULL,PRIMARY KEY (attributeID)); CREATE TABLE attributes_collection (akasID INT NOT NULL,attributeID INT NOT NULL,PRIMARY KEY (akasID, attributeID),CONSTRAINT attributeIDFOREIGN KEY (attributeID)REFERENCES attributes (attributeID)ON DELETE NO ACTIONON UPDATE NO ACTION);CREATE TABLE crew_directors (crewID INT NOT NULL,nconst VARCHAR(45) NOT NULL,PRIMARY KEY (crewID, nconst),CONSTRAINT crewIDFOREIGN KEY (crewID)REFERENCES title_crew (crewID)ON DELETE NO ACTIONON UPDATE NO ACTION,CONSTRAINT nconstFOREIGN KEY (nconst)REFERENCES name_basics (nconst)ON DELETE NO ACTIONON UPDATE NO ACTION);CREATE TABLE crew_writers (crewID INT NOT NULL,nconst VARCHAR(45) NOT NULL,PRIMARY KEY…
- Task 4: The InstantRide Driver Relationship team wants to create groups for drivers according to their ratings such as 3+ or 4+. For instance, a driver with the rating 3.8 will be 3+; whereas a driver with the rating 4.2 will be 4+. You need to return a two column output with DRIVER_ID and DRIVER_RATING which has first FLOOR applied and then CONCAT with + sign for all drivers with a rating greater than 0. Task: Create groups of drivers according to their ratings. (SQL Database Test)3.9 LAB - Select lesson schedule with inner join The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule. Hint: Perform a join on the Student and LessonSchedule tables, matching the student IDs.The Driver Relationship team wants to update the driving license id of an active driver: Driver ID: 2003, New Driving License ID: 1735488 In addition, the team wants to do the update over the VIEW and also want to see the actual change in the DRIVERS table. Task Update a driver's record using the ACTIVE_DRIVERS VIEW.