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TASK 01 (a) RELATIONAL SCHEMA Customer (CustomerID, CustomerName, CustomerAddress, CustomerContactNum) Parcel (ParcelID, ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status) Delivered (DeliverdDate) Returned (ReturnedDate) Lost (LostEntryDate) Invoice (InvoiceID,InvoiceDate, InvoiceAmount) Claim (ClaimID, ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate) Employee (EmployeeID, EmployeeName, EmployeeStartDate)On the basis of relationship that exist between the entities, the table can be further broken down as;Since one customer can order for multiple parcels therefore, 1 to many relationship exists between them.· Parcel (ParcelID, ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status,CustomerID) Foreign Key:CustomerIDSince there is 1 to1 relationship between Parcel and Delivered, Parcel and Returned, Parcel and Lost.· Delivered (ParcelID,DeliverdDate) Primary Key: ParcelID· Returned (ParcelID,ReturnedDate) Primary Key: ParcelID· Lost (ParcelID,LostEntryDate) Primary Key: ParcelIDSince 1 to 1 relation exists between Parcel and Claim.· Claim (ClaimID, ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate,ParcelID) Foreign Key:parcelIDSince 1 to 1 relation exists between LostParcel and Claim.· Lost (LostId,LostEntryDate, ParcelID) Primary Key: LostIDForeign Key: ParcelIDSince there is 1 to 1 relationship between Parcel and Invoice 1. Invoice (InvoiceID,InvoiceDate, InvoiceAmount, ParcelID)Foreign Key:parcelIDSince there is 1 employee can be responsible for multiple Parcel. Therefore,2. Parcel (ParcelID, ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status,CustomerID, EmployeeID)Foreign Key:CustomerID,EmployeeIDSince 1 to many relationship exists between Employee and · Delivered (DeliveredID, ParcelID,DeliverdDate, EmployeeID) Primary Key: DeliveredID· Returned (ReturnedID, ParcelID,ReturnedDate, EmployeeID) Primary Key: ReturnedID· Lost (LostId,ParcelID,LostEntryDate,EmployeeID) Primary Key: LostIdForeign key:EmployeeID,ParcelIDTASK 01 (b) NORMALIZATION1. Parcel (ParcelID, ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status,CustomerID, EmployeeID)In this one customer can order multiple parcels, therefore customerID is not enough to uniquely identify a tuple. There is no partial key dependency. The primary key is the only key that identifies the tuple, therefore, this table is 2NF.For 3NF, there must be a transitive dependency, but in this table here is no attribute that can further identify a tuple uniquely, therefor it is already in 3NF2. Delivered (DeliveredID, ParcelID,DeliverdDate, EmployeeID) In this table, there is no partial key dependency, therefore it is already in 2NF.Every attribute is fully dependent on DeliveredID, there is no transitive dependency among the attributes because no one attribute can uniquely identify any other attribute. Therefore, we say it is already in 3NFTASK 02 INTEGRITY CONSTRAINTS AttributeData typeIntegrity constraint implementedError message DeliveredIDAutonumberPrimary KeyNULL DeliverdDateDate/Time EmployeeIDNumberForeign Key -Indexed ParcelIDNumberForeign Key -Indexed AttributeData typeIntegrity constraint implementedError message ParcelIDAutonumberPrimary Key ParcelWeightInGramsNumberDecimal upto 2 places FromAddressLongText ToAddressLongText DateCollectedDate/Time StatusShortextIn LOOKUP,Display Control=List BoxRow source Type= Value TypeRowSource = “Collected”;”Transit”;”Delivered”;”Returned”;”Lost” EmployeeID,CustomerIDNumber ChargesNumberTASK 03IMPLEMENTATION REPORT1. Any two interesting things/procedures you have learnt by doing this assignment.How to add dropdown menu in status was one interesting thing to learn about. While creating relationship, how to link the primary and foreign key with each other.2. Any one of the assignment tasks that was relatively difficult / complex to solve but has been successfully completed by you.Running queries were the most difficult work.– phpMyAdmin SQL Dump– version 3.3.9– http://www.phpmyadmin.net—- Host: localhost– Generation Time: May 24, 2019 at 11:53 AM– Server version: 5.5.8– PHP Version: 5.3.5SET SQL_MODE=”NO_AUTO_VALUE_ON_ZERO”;/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;/*!40101 SET NAMES utf8 */;—- Database: `db2`—- ———————————————————— Table structure for table `claim`–CREATE TABLE IF NOT EXISTS `claim` ( `ClaimID` int(11) NOT NULL AUTO_INCREMENT, `ClaimLodgedDate` datetime NOT NULL, `ClaimSettledAmount` int(11) NOT NULL, `ClaimPaidDate` date NOT NULL, `ParcelID` int(11) NOT NULL, PRIMARY KEY (`ClaimID`), KEY `ParcelID` (`ParcelID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;—- Dumping data for table `claim`–INSERT INTO `claim` (`ClaimID`, `ClaimLodgedDate`, `ClaimSettledAmount`, `ClaimPaidDate`, `ParcelID`) VALUES(1, ‘2018-09-05 00:00:00’, 1500, ‘2018-09-05’, 5);– ———————————————————— Table structure for table `customer`–CREATE TABLE IF NOT EXISTS `customer` ( `CustomerID` int(11) NOT NULL AUTO_INCREMENT, `CustomerName` varchar(255) NOT NULL, `CustomerAddress` varchar(255) NOT NULL, `CustomerContactNum` int(255) NOT NULL, PRIMARY KEY (`CustomerID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;—- Dumping data for table `customer`–INSERT INTO `customer` (`CustomerID`, `CustomerName`, `CustomerAddress`, `CustomerContactNum`) VALUES(1, ‘Patrick A Evans’, ‘3146 Hood Avenue, San Diego, 92123’, 619),(2, ‘Russell T Foret’, ‘1855 Colony Street,Stamford, 06901’, 203),(4, ‘Byron J Christianson’, ‘4412 Hide A Way Road, San Jose, 95118’, 2147483647),(5, ‘Stanley D Alfonso’, ‘1234 West Virginia Avenue, Colonie, 12205’, 51833438),(6, ‘Ned S Clifford’, ‘3321 Eagles Nest Drive, Woodland,95695’, 53090847),(7, ‘Nelson B Kelley’, ‘2924 Stanton Hollow Road, Randolph,02368’, 78130881),(8, ‘Lawrence M Jaffe’, ‘2426 John Calvin Drive, CULEBRA,00775’, 78745096);– ———————————————————— Table structure for table `delivered`–CREATE TABLE IF NOT EXISTS `delivered` ( `DeliveredID` int(11) NOT NULL AUTO_INCREMENT, `ParcelID` int(11) NOT NULL, `DeliveredDate` date NOT NULL, `EmployeeID` int(11) NOT NULL, PRIMARY KEY (`DeliveredID`), KEY `ParcelID` (`ParcelID`), KEY `EmployeeID` (`EmployeeID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;—- Dumping data for table `delivered`–INSERT INTO `delivered` (`DeliveredID`, `ParcelID`, `DeliveredDate`, `EmployeeID`) VALUES(1, 8, ‘2019-05-22’, 4);– ———————————————————— Table structure for table `employee`–CREATE TABLE IF NOT EXISTS `employee` ( `EmployeeID` int(11) NOT NULL AUTO_INCREMENT, `EmployeeName` varchar(255) NOT NULL, `EmployeeStartDate` date NOT NULL, PRIMARY KEY (`EmployeeID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;—- Dumping data for table `employee`–INSERT INTO `employee` (`EmployeeID`, `EmployeeName`, `EmployeeStartDate`) VALUES(1, ‘Tommy M Horton’, ‘2015-02-24’),(2, ‘Howard V Giddings’, ‘2015-09-25’),(3, ‘Vincent L Figueroa’, ‘2016-02-03’),(4, ‘Nathan M Ross’, ‘2015-08-26’),(5, ‘Jeff C Mixson’, ‘2015-03-14’),(6, ‘Robert G Cornejo’, ‘2016-02-26’);– ———————————————————— Table structure for table `invoice`–CREATE TABLE IF NOT EXISTS `invoice` ( `InvoiceID` int(11) NOT NULL AUTO_INCREMENT, `ParcelID` int(11) NOT NULL, `InvoiceDate` date NOT NULL, `InvoiceAmount` int(11) NOT NULL, PRIMARY KEY (`InvoiceID`), KEY `ParcelID` (`ParcelID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;—- Dumping data for table `invoice`–INSERT INTO `invoice` (`InvoiceID`, `ParcelID`, `InvoiceDate`, `InvoiceAmount`) VALUES(1, 4, ‘2017-08-11’, 1500),(2, 5, ‘2018-12-07’, 2000),(3, 6, ‘2019-05-30’, 2550),(4, 7, ‘2018-05-08’, 2500),(5, 3, ‘2018-02-02’, 1000),(6, 8, ‘2019-05-22’, 1500);– ———————————————————— Table structure for table `lost`–CREATE TABLE IF NOT EXISTS `lost` ( `LostID` int(11) NOT NULL AUTO_INCREMENT, `ParcelID` int(11) NOT NULL, `LostEntryDate` date NOT NULL, `EmployeeID` int(11) NOT NULL, PRIMARY KEY (`LostID`), UNIQUE KEY `ParcelID` (`ParcelID`), KEY `EmployeeID` (`EmployeeID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;—- Dumping data for table `lost`–INSERT INTO `lost` (`LostID`, `ParcelID`, `LostEntryDate`, `EmployeeID`) VALUES(1, 5, ‘2018-09-05’, 5);– ———————————————————— Table structure for table `parcel`–CREATE TABLE IF NOT EXISTS `parcel` ( `ParcelID` int(11) NOT NULL AUTO_INCREMENT, `ParcelWeightInGrams` decimal(11,2) NOT NULL, `FromAddress` varchar(2500) NOT NULL, `ToAddress` varchar(2500) NOT NULL, `DateCollected` date NOT NULL, `ParselValue` int(11) NOT NULL, `Charges` int(11) NOT NULL, `Status` enum(‘Collected’,’Transit’,’Delivered’,’Returned’,’Lost’) NOT NULL, `CustomerID` int(11) NOT NULL, `EmployeeID` int(11) NOT NULL, PRIMARY KEY (`ParcelID`), KEY `CustomerID` (`CustomerID`), KEY `EmployeeID` (`EmployeeID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;—- Dumping data for table `parcel`–INSERT INTO `parcel` (`ParcelID`, `ParcelWeightInGrams`, `FromAddress`, `ToAddress`, `DateCollected`, `ParselValue`, `Charges`, `Status`, `CustomerID`, `EmployeeID`) VALUES(3, ‘200.00’, ‘2376 North Bend River Road, Somerset, 42501’, ‘4748 Bubby Drive,MORGANTOWN, 26501’, ‘2018-02-02’, 220, 1000, ‘Collected’, 4, 3),(4, ‘352.00’, ‘3469 Roosevelt Road,Dodge City,67801’, ‘4269 Short Street,Austin,78723’, ‘2017-08-11’, 356, 1500, ‘Returned’, 6, 4),(5, ‘198.00’, ‘111 Lawman Avenue,Alexandria,22301’, ‘2830 Hide A Way Road,Santa Clara,95050’, ‘2018-12-07’, 469, 2000, ‘Lost’, 2, 6),(6, ‘146.00’, ’34 Sussex Court,Waco,76706′, ‘1855 Hillcrest Circle,Crystal,55429’, ‘2019-05-30’, 789, 2550, ‘Transit’, 7, 5),(7, ‘436.00’, ‘1049 Atha Drive,Bakersfield,93311’, ‘1057 Froe Street,TALMO,30575’, ‘2018-05-08’, 965, 2500, ‘Collected’, 1, 2),(8, ‘716.00’, ‘1057 Froe Street,TALMO,30575’, ‘1049 Atha Drive,Bakersfield,93311’, ‘2019-05-22’, 369, 1500, ‘Delivered’, 5, 5);– ———————————————————— Table structure for table `returned`–CREATE TABLE IF NOT EXISTS `returned` ( `ReturnedID` int(11) NOT NULL AUTO_INCREMENT, `ParcelID` int(11) NOT NULL, `ReturnedDate` date NOT NULL, `EmployeeID` int(11) NOT NULL, PRIMARY KEY (`ReturnedID`), KEY `ParcelID` (`ParcelID`), KEY `EmployeeID` (`EmployeeID`)) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;—- Dumping data for table `returned`–INSERT INTO `returned` (`ReturnedID`, `ParcelID`, `ReturnedDate`, `EmployeeID`) VALUES(1, 4, ‘2017-11-09’, 4);—- Constraints for dumped tables—— Constraints for table `claim`–ALTER TABLE `claim` ADD CONSTRAINT `claim_ibfk_1` FOREIGN KEY (`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE CASCADE ON UPDATE CASCADE;—- Constraints for table `delivered`–ALTER TABLE `delivered` ADD CONSTRAINT `delivered_ibfk_1` FOREIGN KEY (`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE CASCADE ON UPDATE CASCADE, ADD CONSTRAINT `delivered_ibfk_2` FOREIGN KEY (`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON DELETE CASCADE ON UPDATE CASCADE;—- Constraints for table `invoice`–ALTER TABLE `invoice` ADD CONSTRAINT `invoice_ibfk_1` FOREIGN KEY (`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE CASCADE ON UPDATE CASCADE;—- Constraints for table `lost`–ALTER TABLE `lost` ADD CONSTRAINT `lost_ibfk_1` FOREIGN KEY (`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE CASCADE ON UPDATE CASCADE, ADD CONSTRAINT `lost_ibfk_2` FOREIGN KEY (`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON DELETE CASCADE ON UPDATE CASCADE;—- Constraints for table `parcel`–ALTER TABLE `parcel` ADD CONSTRAINT `parcel_ibfk_1` FOREIGN KEY (`CustomerID`) REFERENCES `customer` (`CustomerID`), ADD CONSTRAINT `parcel_ibfk_2` FOREIGN KEY (`EmployeeID`) REFERENCES `employee` (`EmployeeID`);—- Constraints for table `returned`–ALTER TABLE `returned` ADD CONSTRAINT `returned_ibfk_1` FOREIGN KEY (`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE CASCADE ON UPDATE CASCADE, ADD CONSTRAINT `returned_ibfk_2` FOREIGN KEY (`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON DELETE CASCADE ON UPDATE CASCADE;
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