JavaFX program Create a JavaFX GUI that allows the user to retrieve records from the product table in the example database. The GUI should allow the user to specify desired fields, desired order, and a where condition. Display only the desired fields in the desired order for the desired where condition. You may display the records in the GUI in any way you wish. database below
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JavaFX program
Create a JavaFX GUI that allows the user to retrieve records from the product table in the example
You may display the records in the GUI in any way you wish.
database below
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- CREATE TABLE EMPLOYEE(ENUM DECIMAL(12) NOT NULL,/* Employee number */FNAME VARCHAR(50) NOT NULL,/* First name */LNAME VARCHAR(50) NOT NULL,/* Last name */DOB DATE NULL,/* Date of birth */CONSTRAINT EMPLOYEE_PKEY PRIMARY KEY(ENUM) );CREATE TABLE DRIVER(ENUM DECIMAL(12) NOT NULL,/* Employee number */LNUM DECIMAL(8) NOT NULL,/* Driving license number */STATUS VARCHAR(10) NOT NULL,/* Driver status */CONSTRAINT DRIVER_PKEY PRIMARY KEY(ENUM),CONSTRAINT DRIVER_UNIQUE UNIQUE(LNUM),CONSTRAINT DRIVER_FKEY FOREIGN KEY(ENUM) REFERENCES EMPLOYEE(ENUM),CONSTRAINT DRIVER_STATUS CHECK (STATUS IN ('AVAILABLE', 'BUSY', 'ON LEAVE')) );CREATE TABLE TRUCK(REGNUM VARCHAR(10) NOT NULL,/* Registration number */CAPACITY DECIMAL(7) NOT NULL,/* Capacity */WEIGHT DECIMAL(7) NOT NULL,/* Weight */STATUS VARCHAR(10) NOT NULL,/* Present status */CONSTRAINT TRUCK_PKEY PRIMARY KEY(REGNUM),CONSTRAINT TRUCK_STATUS CHECK( STATUS IN ('AVAILABLE', 'USED', 'MAINTAINED')),CONSTRAINT TRUCK_WEIGHT CHECK( WEIGHT > 0.0 AND…CREATE TABLE EMPLOYEE(ENUM DECIMAL(12) NOT NULL,/* Employee number */FNAME VARCHAR(50) NOT NULL,/* First name */LNAME VARCHAR(50) NOT NULL,/* Last name */DOB DATE NULL,/* Date of birth */CONSTRAINT EMPLOYEE_PKEY PRIMARY KEY(ENUM) );CREATE TABLE DRIVER(ENUM DECIMAL(12) NOT NULL,/* Employee number */LNUM DECIMAL(8) NOT NULL,/* Driving license number */STATUS VARCHAR(10) NOT NULL,/* Driver status */CONSTRAINT DRIVER_PKEY PRIMARY KEY(ENUM),CONSTRAINT DRIVER_UNIQUE UNIQUE(LNUM),CONSTRAINT DRIVER_FKEY FOREIGN KEY(ENUM) REFERENCES EMPLOYEE(ENUM),CONSTRAINT DRIVER_STATUS CHECK (STATUS IN ('AVAILABLE', 'BUSY', 'ON LEAVE')) );CREATE TABLE TRUCK(REGNUM VARCHAR(10) NOT NULL,/* Registration number */CAPACITY DECIMAL(7) NOT NULL,/* Capacity */WEIGHT DECIMAL(7) NOT NULL,/* Weight */STATUS VARCHAR(10) NOT NULL,/* Present status */CONSTRAINT TRUCK_PKEY PRIMARY KEY(REGNUM),CONSTRAINT TRUCK_STATUS CHECK( STATUS IN ('AVAILABLE', 'USED', 'MAINTAINED')),CONSTRAINT TRUCK_WEIGHT CHECK( WEIGHT > 0.0 AND…Description The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id. +--- | Id | Name | Salary | ManagerId | +---- | 1 | 2 | 3 | 4 +---- | Joe | 70000 | 3 | Henry | 80000 | 4 | Sam | Max Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager. + | Employee | + | Joe + | 60000 | NULL | 90000 | NULL I
- Computer Science please please any help ASAP Query #5: Code a query that will make use of SQL Server intrinsic functions and reference the Instructors table. Must have alias names for each column. First column: Display the instructors name as first name initial followed by a period and space and then last name such as B. Brown Second column: Display the year only for the hire date. Third column: Difference in months between today’s date and the hire date. Last column: Display the string literal ‘Full-time instructor’ if an instructor has a Status of F or return the string literal of ‘Adjunct instructor’ if they have a status of P. You can use the Case function or the IIF function.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…What is a NULL value?
- Book Sales database (Just Lee Book sales Company): Which of below SQL option resolves this problem? Produce a list of each customer number and the third and fourth digits of his or her zip code. The query should also display the position of the first occurrence of a 3 in the customer number, if it exists. SELECT customer#, SUBSTR(zip, 3), INSTR(customer#, 3) FROM customers; SELECT customer#, INSTR(zip, 3, 2), INSTR(customer#, 3) FROM customers; SELECT customer#, SUBSTR(zip, 2, 3), INSTR(customer#, 1) FROM customers; SELECT customer#, SUBSTR(zip, 3, 2), INSTR(customer#, 3) FROM customers;NVL to replace null value.PL/SQL Block DECLARE v_weight NUMBER (3) := 600; VARCHAR2 (255) v_message := 'Product 10012'; BEGIN DECLARE v_weight NUMBER ( 3) := 1; v_message VARCHAR2 (255) :- 'Product 11001'; v_new_locn VARCHAR2 (50) := 'Europe'; BEGIN v_weight := v_weight + 1; v_new_locn := 'Western ' || v_new_locn; 1) END; v_weight := v_weight + 1; v_message :- v_message || ' is in stock'; v_new_locn := 'Western ' || v_new_locn; 2) END; racle Academy nly 1. Evaluate the preceding PL/SQL block and determine the data type and value following variables according to the rules of scoping. a. The value of v_weight at position 1 is: each of the b. The value of v_new_locn at position 1 is: c. The value of v_weight at position 2 is: d. The value of v_message at position e. The value of v_new_locn at position .
- INPUT TABLE You're given a orders table and the columns in the orders table are shown below: Orders Order_Id INT Type VARCHAR(10) Red_Shipping_ Days INT Scheduled_Shipping_Days INT Customer_Id INT Order _city VARCHAR(20) Order_Date DATE Order_Region VARCHAR(15) Order_State VARCHAR(20) Order _Status VARCHAR(20) Shipping_Mode VARCHAR(20) QUERY Calculate count of all the orders. Where Order_ State is Maharashtra Note - Use the alias of oc for count of orders Group the results by Type Order them by oc in ascending order. OUTPUT COLUMNS oc, TypeSQL: Create a SQL query that uses an uncorrelated subquery and no joins to display the descriptions for products provided by a vendor in area code 615. Remember, the main difference between a correlated subquery and uncorrelated subquery is that, in a correlated subquery, the inner query is dependent upon the outer query, while, with an uncorrelated query, the inner query does not depend upon the outer query. Instead, it runs just once. You should NOT use union, minus, or intersect in this query. The outer query will use the PRODUCT table and the inner query will use the VENDOR table [hint: you should have 9 records of output]. My code: select p_descript from product where (select v_areacode from vendor where v_areacode = '615' ; I am getting this error, problem is I can't see what is missing any assistance is appreciated. ORA-00936: missing expressiongrade varchar(30), salesman_id int, CONSTRAINT FK_salesman_id FOREIGN KEY (salesman_id) REFERENCES salesman(salesman_id) ); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3002, 'Nick Rimando', 'New York', 100, 5001); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3007, 'Brad Davis', 'New York', 200, 5001); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3005, 'Graham Zusi', "California', 200, 5002); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3008, Julian Green', "London', 300, 5002); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3004, 'Fabian Johnson', 'Paris', 300, 5006); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3009, 'Geoff Cameron', "Berlin', 100, 5003); insert into customer (customer_id, cust_name, city, grade, salesman_id) values (3003, 'Jozy Altidor, 'Moscow', 200, 5007);…