SQL

Ace your homework & exams now with Quizwiz!

Write the correct SQL statement to create a new table called Persons. PersonID, LastName, FirstNAme, Address, City

CREATE TABLE Persons (PersonID int, LastName varchar(255), FirstName varchar(255), Address varchar(255), City varchar(255) );

Write the correct SQL statement to create a new database called testDB.

CREATE DATABASE testDB;

Add a column of type DATE called Birthday.

ALTER TABLE Persons ADD Birthday DATE;

Delete the column Birthday from the Persons table.

ALTER TABLE Persons DROP COLUMN Birthday;

Delete all the records from the Customers table where the Country value is 'Norway'.

DELETE FROM Customers WHERE Country = 'Norway';

Delete all the records from the Customers table.

DELETE FROM Customers;

Write the correct SQL statement to delete a database named testDB.

DROP DATABASE testDB;

Write the correct SQL statement to delete a table called Persons.

DROP TABLE Persons;

Insert a new record in the Customers table. CustomerName - Hekkan Burger Address - Gateveien 15 City - Sandnes PostalCode - 4306 Country - Norway

INSERT INTO Customers (CustomerName, Address, City, PostalCode, Country) VALUES ('Hekkan Burger', 'Gateveien 15', 'Sandnes', '4306', 'Norway');

Select all records from the Customers table, sort the result alphabetically by the column City.

SELECT * FROM Customers ORDER BY City;

Select all records from the Customers table, sort the result alphabetically, first by the column Country, then, by the column City.

SELECT * FROM Customers ORDER BY Country, City;

Select all records from the Customers where the PostalCode column is empty.

SELECT * FROM Customers WHERE PostalCode IS NULL;

Select all records where the value of the City column contains the letter "a".

SELECT * FROM Customers WHERE CITY LIKE '%a%';

Select all records where the value of the City column ends with the letter "a".

SELECT * FROM Customers WHERE CITY LIKE '%a';

Select all records where the value of the City column starts with letter "a" and ends with the letter "b".

SELECT * FROM Customers WHERE CITY LIKE 'a%b';

Select all records where the value of the City column does NOT start with the letter "a".

SELECT * FROM Customers WHERE CITY NOT LIKE 'a%';

Use an SQL function to calculate the sum of all the Price column values in the Products table.

SELECT SUM(Price) FROM Product;

Select all records where the City column has the value 'Berlin' or 'London'.

SELECT * FROM Customers WHERE City = 'Berlin' OR City = 'London';

Select all records where the City column has the value 'Berlin' and the PostalCode column has the value 12209.

SELECT * FROM Customers WHERE City = 'Berlin' and PostalCode = '12209';

Select all records where the City column has the value "Berlin".

SELECT * FROM Customers WHERE City = 'Berlin';

Select all records where the value of the City column starts with the letter "a".

SELECT * FROM Customers WHERE City LIKE 'a%';

Use the IN operator to select all the records where Country is either "Norway" or "France".

SELECT * FROM Customers WHERE Country IN ('Norway','France');

Select all records where the CustomerID column has the value 32.

SELECT * FROM Customers WHERE CustomerID = 32;

Use the NOT keyword to select all records where City is NOT "Berlin".

SELECT * FROM Customers WHERE NOT City = 'Berlin';

Select all records from the Customers where the PostalCode column is NOT empty.

SELECT * FROM Customers WHERE PostalCode IS NOT NULL;

Select all records from the Customers table, sort the result reversed alphabetically by the column City.

SELECT * FROM Customers ORDER BY City DESC;

When displaying the Customers table, refer to the table as Consumers instead of Customers.

SELECT * FROM Customers AS Consumers;

Write the missing statement to get all the columns from the Customers table.

SELECT * FROM Customers;

Choose the correct JOIN clause to select all records from the two tables where there is a match in both tables.

SELECT * FROM Orders INNER JOIN Customers ON Orders.CustomerID=Customers.CustomerID;

Choose the correct JOIN clause to select all the records from the Customers table plus all the matches in the Orders table.

SELECT * FROM Orders RIGHT JOIN Customers ON Orders.CustomerID=Customers.CustomerID;

Insert the missing parts in the JOIN clause to join the two tables Orders and Customers, using the CustomerID field in both tables as the relationship between the two tables.

SELECT * FROM Orders LEFT JOIN Customers ON Orders.CustomerID=Customers.CustomerID;

Use the BETWEEN operator to select all the records where the value of the Price column is between 10 and 20.

SELECT * FROM Products WHERE Price BETWEEN 10 AND 20;

Use the BETWEEN operator to select all the records where the value of the Price column is NOT between 10 and 20.

SELECT * FROM Products WHERE Price NOT BETWEEN 10 AND 20;

Use the BETWEEN operator to select all the records where the value of the ProductName column is alphabetically between 'Geitost' and 'Pavlova'.

SELECT * FROM Products WHERE ProductName BETWEEN 'Geitost' AND 'Pavlova';

Use an SQL function to calculate the average price of all products.

SELECT AVG(Price) FROM Product;

Use the correct function to return the number of records that have the Price value set to 18.

SELECT COUNT(*) FROM Products WHERE Price = 18;

List the number of customers in each country.

SELECT COUNT(CustomerID), Country FROM Customers GROUP BY Country;

Write a statement that will select the City column from the Customers table.

SELECT City FROM Customers;

When displaying the Customers table, make an ALIAS of the PostalCode column, the column should be called Pno instead.

SELECT CustomerName, Address, PostalCode AS Pno FROM Customers;

Select all the different values from the Country column in the Customers table.

SELECT DISTINCT Country FROM Customers;

Use an SQL function to select the record with the highest value of the Price column.

SELECT MAX(Price) FROM Products;

Use the MIN function to select the record with the smallest value of the Price column.

SELECT MIN(Price) FROM Products;

List the number of customers in each country, ordered by the country with the most customers first.

SELECTY COUNT(CustomerID), Country, FROM Customers GROUP BY Country ORDER BY COUNT(CustomerID) DESC;

Use the TRUNCATE statement to delete all data inside the Persons table

TRUNCATE TABLE Persons;

Set the value of the City columns to 'Oslo', but only the ones where the Country column has the value "Norway".

UPDATE Customers SET City = 'Oslo' WHERE Country = 'Norway';

Update the City column of all records in the Customers table to 'Oslo'

UPDATE Customers SET City = 'Oslo';

Update the City value and the Country value.

UPDATE Customers SET City = 'Oslo', Country = 'Norway' WHERE CustomerID = 32;

Select all records where the first letter of the City is NOT an "a" or a "c" or an "f".

WHERE City LIKE '[!acf]%';

Select all records where the first letter of the City is an "a" or a "c" or an "s".

WHERE City LIKE '[acs]%';

Select all records where the second letter of the City is an "a".

WHERE City LIKE '_a%';

Select all records where the first letter of the City starts with anything from an "a" to an "f".

WHERE City LIKE 'a-f%';

Use the IN operator to select all the records where Country is NOT "Norway" and NOT "France".

WHERE Country NOT IN ('Norway', 'France');


Related study sets

Investment Planning - Measures of Investment Returns

View Set

Practice Quiz for Block Final #2

View Set

Chapter 45 Tissue Integrity/Imtegumentary Prep U

View Set

Nutrition Chapter 6 - Amino Acids

View Set

Chapter 16: The Microcirculation and Lymphatic System

View Set

KS3 Year 7 Biology Specialised cells - function

View Set