Write a code to the attached photo using Console.WriteLine.
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Write a code to the attached photo using Console.WriteLine.
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- 1.Question Workspace Write an SQL query that displays the therapies and their unit of time for the therapies that include the word bath, hot, or electrical. 2.Question Workspace Write an SQL query to display every therapist's first name and last name as their full name, but only for those instructors not living in zip code 72511.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…Use the following table to answer the question below: Table name: Customers Column Name Role Data Type Length Constraint customerID stores the ID of a customer Char 5 Primary key custFName stores the first name of a customer Varchar2 35 Cannot be null custLName Stores the last name of a customer Varchar2 35 Cannot be null custGender Store the gender of a customer Char 1 It should be ‘M’ or ‘F’ custDOB Stores the date of birth of a customer Date custIncome Store the income of a customer Number 4 Write a SQL query that displays the number of customers who born on the 16th day of a month . Write in SQL a query that displays the ids of the customers who have the highest income . Create a view named ‘customerDetails’ to hold all the details of customers whose first names start with ‘S’ and their incomes are between 500 and 1000 Omani Rials .
- Use the following table to answer the questions below: Table name: Customers Column Name Role Data Type Length Constraint customerID stores the ID of a customer Char 5 Primary key custFName stores the first name of a customer Varchar2 35 Cannot be null custLName Stores the last name of a customer Varchar2 35 Cannot be null custGender Store the gender of a customer Char 1 It should be ‘M’ or ‘F’ custDOB Stores the date of birth of a customer Date custIncome Store the income of a customer Number 4 Write in SQL a command that creates the table Customers according to the above description (2 marks). Write in SQL a statement that adds a new column named accountID of type Char(8). This column should be defined as a foreign key that relates the table Customers to the table Account (1.5 marks). Write in SQL a command that deletes the customers whose date of birth is after 22-Dec-1990 (1.5 Marks).…Which of the following statements retrieves data from the Sales table and stores it in the CompanyDataSet? Me.CompanyDataSet.Fill(Me.Sales) Me.SalesBindingSource.Fill(Me.CompanyDataSet) Me.SalesBindingNavigator.Fill(Me.CompanyDataSet.Sales) Me.SalesTableAdapter.Fill(Me.CompanyDataSet.Sales)Examine this product table's column definitions:pid number primary keypname varchar2(50)You must display column headings as shown:Product_ID Product NameWhich SELECT statement does this? SELECT pid Product_ID, pname Product Name FROM product; SELECT pid Product_ID, pname "Product Name" FROM product; SELECT pid @Product_ID, pname @"Product Name" FROM product; SELECT pid AS Product_ID, pname AS Product Name FROM product
- PLZ HELP WITH THE FOLLOWING: make the following work select MOVIE_NAME, MOVIE_YEAR from MOVIE where MOVIE_COST > (select a.MOVIE_COST from MOVIE a, PRICE b where a.PRICE_CODE = b.PRICE_CODE and b.PRICE_DESC ="weekly Special"); CREATE TABLE MEMBERSHIP ( MEM_NUM CHAR(3) CONSTRAINT MEMBER_MEMNUM_PK PRIMARY KEY, MEM_FNAME VARCHAR(30) NOT NULL, MEM_LNAME VARCHAR(30) NOT NULL, MEM_STREET VARCHAR(30), MEM_CITY VARCHAR(10), MEM_STATE CHAR(2), MEM_ZIP CHAR(5), MEM_BALANCE NUMBER(2) ); CREATE TABLE RENTAL ( RENT_NUM CHAR(4) CONSTRAINT RENTAL_RENTNUM_PK PRIMARY KEY, RENT_DATE DATE, MEM_NUM CHAR(3), CONSTRAINT RENTAL_MEMNUM_FK FOREIGN KEY (MEM_NUM) REFERENCES MEMBERSHIP ); CREATE TABLE PRICE ( PRICE_CODE CHAR(1) CONSTRAINT PRICE_PRICECODE_PK PRIMARY KEY, PRICE_DESC VARCHAR(20), PRICE_RENTFEE NUMBER (3, 1), PRICE_DAILYATFEE NUMBER(3, 1) ); CREATE TABLE MOVIE ( MOVIE_NUM CHAR(4) CONSTRAINT MOVIE_MOVIENUM_PK PRIMARY KEY, MOVIE_NAME VARCHAR(30) NOT NULL,…The tblCities table contains a numeric field named Population. Which of the following LINQ statements selects all cities having a population that exceeds 15000?a. Dim records = From city In CitiesDataSet.tblCities Where Population > 15000 Select cityb. Dim records = From city In CitiesDataSet.tblCities Select city.Population > 15000c. Dim records = From city In CitiesDataSet.tblCities Where city.Population > 15000 Select city d. Dim records = Select city.Population > 15000 From tblCitiesThe Driver Relationship team realized that maintaining driver IDs is difficult and requested an automatic way of incrementing the value when a new driver is added. You need to make the changes on the table to automatically increment the Driver ID. After the change, you need to insert the following driver: First Name: Nursin Last Name: Yilmaz Driving License ID: 4141447 Start Date: 2019-12-28 Driving License Checked: True Rating: 4.0
- Question 6. Identify the customer with lowest number of order placed Create a function ORD_PLACED that will identify the customer with lowest total amount of order. Refer to table BB_BASKET that include IDSHOPPER column which represents customer id. Display the idshopper and the total amount of the order. SQL computer scienceList the order number and order date for every order that was placed by Mary Nelson and that contains an order line for an iron. Use an INTERSECT operator. table name is:ORDER_LINEORLN_ORDER_NUMBER ORLN_PART_NUMBER ORLN_NUMBER_ORDERED ORLN_QUOTED_PRICE tablename is ORDER_TABLE ORDR_ORDER_NUMBER ORDR_ORDER_DATE ORDR_CUSTOMER_NUMBER tablename is CUSTOMER CUST_NUMBERCUST_LAST CUST_FIRST CUST_STREET CUST_CITY CUST_STATE CUST_ZIP_CODE CUST_BALANCE CUST_CREDIT_LIMIT CUST_SALES_REP_NUM tablename is PART PART_NUMBER PART_DESCRIPTION UNITS_ON_HAND ITEM_CLASS WAREHOUSE UNIT_PRICE tablename is SALES_REP SLSR_PKSLSR_LAST SLSR_FIRST SLSR_STREET SLSR_CITY SLSR_STATE SLSR_ZIP_CODE SLSR_TOTAL_COMMISSION SLSR_COMMISSION_RATE SELECT ORDR_ORDER_NUMBER, ORDR_ORDER_DATEFROM ORDER_TABLEWHERE ORDR_ORDER_NUMBER IN (SELECT ORLN_ORDER_NUMBERFROM ORDER_LINEWHERE ORLN_PART_NUMBER IN (SELECT PART_NUMBERFROM PARTWHERE…GOCREATE TABLE dbo.booth(booth_Product char(25) ,booth_Price money);CREATE TABLE dbo.machine(machine_product char(25) ,machine_price money);GOCREATE TABLE dbo.product(product_id int not null,product_description varchar(30) not null ,PRIMARY KEY (product_id)); THis is one way to create a table; but you can also can also right click on databases and then newdatabase is one prefered over the other?
