Describe what is wrong with the following code: CREATE TABLE CUSTOMER (ID NUMERIC (2) PRIMARY KEY, NAME CHAR (5), ADDRESS VARCHAR (100)); INSERT INTO CUSTOMER VALUES (123, JOHN', 'MONTREAL) INSERT INTO CUSTOMER VALUES (123, JOHN SMITH', 'MONTREAL")
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- Delete the index named ITEM_INDEX3.Course Title: Data Structure and Algorithms Question A table stores the data for all the members of an exotic resort. Members are identified by these numbers 1,5,30,70,80,20,15,100,40,25,7,3,8,33,22. All members are to be invited in a meeting to review the membership policies. What will be the fastest way to call all the members? Write algorithm to justify your answer.DROP TABLE IF EXISTS Worker; CREATE TABLE Worker ( ); 0 INSERT ); WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST NAME TEXT, LAST NAME TEXT, SALARY INTEGER (15), JOINING DATE DATETIME, DEPARTMENT CHAR (25) INTO Worker (WORKER_ID, FIRST_NAME, LAST_NAME, SALARY, JOINING_DATE, DEPARTMENT) VALUES 'Arora', 100000, '14-02-20 09.00.00', 'HR'), (001, 'Monika', (002, 'Niharika', 'Verma' 80000, '14-06-11 09.00.00', 'Admin'), 'HR'), 'Admin'), Admin'), 'Account'), 'Kumar', 75000, '14-01-20 09.00.00', 'Account') DROP TABLE IF EXISTS Bonus; CREATE TABLE Bonus ( ); (003, 'Vishal' 'Singhal', 300000, '14-02-20 09.00.00', (004, 'Amitabh 'Singh', 500000, '14-02-20 09.00.00', (005, 'Vivek' 'Bhati', 500000, '14-06-11 09.00.00'. (006, 'Vipul', 'Diwan' 200000, '14-06-11 09.00.00', (007, 'Satish' (008, 'Geetika', 'Chauhan', 90000, '14-04-11 09.00.00', 'Admin'); WORKER_REF_ID INTEGER, BONUS AMOUNT INTEGER(10), BONUS_DATE DATETIME, FOREIGN KEY (WORKER_REF_ID) REFERENCES Worker (WORKER_ID) ON DELETE…
- Sales Database: Customers(custId, lastName, firstName, address, phone, creditLimit) Orders(ordNumber, itemNumber, qtyOrdered.) Items(itemNumber, itemName, price) For the Sales Database referenced above, write the SQL command to create the LineItem table, assuming the Orders table and items table already exist.Update rows in Horse table The Horse table has the following columns: ID - integer, auto increment, primary key RegisteredName - variable-length string Breed - variable-length string, must be one of the following: Egyptian Arab, Holsteiner, Quarter Horse, Paint, Saddlebred Height - decimal number, must be ≥ 10.0 and ≤ 20.0 BirthDate - date, must be ≥ Jan 1, 2015 Make the following updates: Change the height to 15.6 for horse with ID 2. Change the registered name to Lady Luck and birth date to May 1, 2015 for horse with ID 4. Change every horse breed to NULL for horses born on or after December 22, 2016.1) Write the SQL code that will create the table structure for a table named EMP. This table is a subset of the EMPLOYEE table. The basic EMP table structure is summarized in the table below. (Note that the EMP_NUM is the PK)? ATTRIBUTE(FIELD) NAME DATADECLARATION Constraints EMP_NUM INT Auto generated Identifier, start from 101 EMP_LNAME VARCHAR(15) Not null EMP_FNAME VARCHAR(15) Not Null EMP_INITIAL CHAR(1) EMP_HIREDATE DATE Default value is system date JOB_CODE CHAR(3) Make sure input is in the range500,501,502
- Task 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Create a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the three fields. The CAR_ID and MAINTENACNE_TYPE_ID should be foreign keys to their original tables. Cascade update and cascade delete the foreign keys. Answer in MYSQL pleaseindividual characters using their 0-based index. Hint: You will need to do this for checking all the rules. However, when you access a character .Alter - Add Referential Integrity Constraint You have to only use Alter table SQL statement Identify the common key between the two tables given below and establish referential integrity constraint between them. Pointofinterest PK Town pointID Number describe Varchar(30) opentime varchar(10) closetime varchar(10) Number Varchar(30) varchar(30) varchar(30)- varchar(30) PK townID townname state longitude latitude FK townID Number summertemp Number wintertemp Number Number sealevel
- THIS MODULE IS ABOUT SUBQUERIES, SO YOU MUST USE SUBQUERIES INSTEAD OF TABLE JOINS IN THISASSIGNMENT.LAB - Select horses with logical operators The Horse table has the following columns: • ID integer, primary key • RegisteredName - variable-length string • Breed - variable-length string • Height-decimal number • BirthDate - date Write a SELECT statement to select the registered name, height, and birth date for only horses that have a height between 15.0 and 16.0 (inclusive) or have a birth date on or after January 1, 2020.MySql Workbench CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 1. List last name of all students whose first name is longer than 4 letters in ascending order accordingto the last name. Duplicated rows should be removed from the output.2. Count the total number of rooms in Location.3. Find the number of students in each major.4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...