diff --git a/create.sql b/create.sql
new file mode 100644
index 0000000..3067182
--- /dev/null
+++ b/create.sql
@@ -0,0 +1,55 @@
+ CREATE DATABASE IF NOT EXISTS lab_mysql;
+
+USE lab_mysql;
+
+CREATE DATABASE lab_mysql;
+
+USE lab_mysql;
+
+CREATE TABLE Cars (
+    id INT,
+    vin VARCHAR(50),
+    manufacturer VARCHAR(50),
+    model VARCHAR(50),
+    year INT,
+    color VARCHAR(50)
+);
+
+DROP TABLE IF EXISTS Cars;
+
+CREATE TABLE cars (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    vin VARCHAR(50),
+    manufacturer VARCHAR(50),
+    model VARCHAR(50),
+    year INT,
+    color VARCHAR(50)
+);		
+
+CREATE TABLE customers (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    cust_name VARCHAR(50),
+    cust_phone  INT,
+    cust_email VARCHAR(50),
+    cust_address VARCHAR(50),
+    cust_city VARCHAR(50),
+    cust_state VARCHAR(50),
+    cust_country VARCHAR(50),
+    cust_zipcode INT
+);	
+
+CREATE TABLE salespersons (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    staff_id  INT,
+	name VARCHAR(50),
+    store VARCHAR(50)
+);	
+
+CREATE TABLE invoices (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    invoice_number  INT,
+	date INT,
+	car INT,
+    customer INT,
+	salesperson INT
+);	
diff --git a/lab_mysql.sql b/lab_mysql.sql
new file mode 100644
index 0000000..e14ed03
--- /dev/null
+++ b/lab_mysql.sql
@@ -0,0 +1,539 @@
+-- ==================================================
+-- SETTING UP THE DATABASE
+-- ==================================================
+
+-- Setting the working database
+USE bank;
+
+-- ==================================================
+-- BASIC QUERIES
+-- ==================================================
+
+-- Selecting all the data from table trans
+SELECT * FROM trans;
+
+-- It's good practice to use the database name when referencing tables 
+-- especially when working with multiple databases on a server.
+SELECT * FROM bank.trans;
+
+-- Selecting specific columns
+SELECT trans_id, account_id, date, type FROM bank.trans;
+
+-- Fully qualifying columns with database and table names
+SELECT bank.trans.trans_id, bank.trans.account_id, bank.trans.date, bank.trans.type 
+FROM bank.trans;
+
+-- Limiting the result set
+SELECT trans_id, account_id, date, type FROM bank.trans 
+LIMIT 10;
+
+-- Counting the number of records in a table
+SELECT COUNT(*) FROM bank.trans;
+
+-- Getting unique values of a column
+SELECT DISTINCT status FROM bank.loan; 
+
+-- Aliasing columns for better readability
+SELECT trans_id AS Transaction_ID, 
+       account_id AS Account_ID, 
+       date AS Date, 
+       type AS Type_of_account
+FROM bank.trans;
+
+-- Aliasing columns and tables
+-- While aliasing tables isn't necessary here, it's useful in JOIN operations.
+SELECT bt.trans_id AS Transaction_ID, 
+       bt.account_id AS Account_ID, 
+       bt.date AS Date, 
+       bt.type AS Type
+FROM bank.trans AS bt;
+
+-- ==================================================
+-- CHECK FOR UNDERSTANDING
+-- ==================================================
+
+-- 1. Use an appropriate `SELECT` statement to retrieve a list of unique card types from the `bank` database. (Hint: Use DISTINCT function here.)
+-- 2. Get a list of all the district names in the bank database. Suggestion is to use the `files_for_activities/case_study_extended` here to work out which column is required here because we are looking for the names of places, not just the IDs. It would be great to see the results under the heading `district_name`. (Hint: Use an alias.). You should have 77 rows.
+-- 3. Bonus: Revise your query to also show the Region, and limit the results to just 30 rows. Sort the results alphabetically by district name A>Z before selecting the 30.
+-- 4. Get again the district names and the regions but use aliases to display the corresponding columns as `district_name` and `region_name` respectively. Limit the query to the first 30 rows.
+-- 5. Get all the types of the card as `type_of_card` and the issue date as `issue_date` from card and alias the table as `bc`.
+
+-- ==================================================
+
+
+-- ==================================================
+-- ARITHMETIC OPERATORS
+-- ==================================================
+
+
+-- Display the difference between columns
+SELECT *, amount-payments AS balance
+FROM bank.loan;
+
+-- Displaying specific columns with calculated values
+SELECT loan_id, account_id, date, duration, status, 
+       (amount-payments)/1000 AS 'balance in Thousands'
+FROM bank.loan;
+
+-- Example of modulus operator to get remainder
+SELECT duration%2
+FROM bank.loan;
+
+-- ==================================================
+-- COMPARISON OPERATORS AND FILTERING
+-- ==================================================
+
+-- Filtering rows based on a specific condition
+SELECT * FROM bank.loan
+WHERE status = 'B';
+
+-- The operators <> and != are equivalent
+SELECT * FROM bank.loan
+WHERE status <> 'B';
+
+-- Using IN clause for multiple possible values
+SELECT * FROM bank.loan
+WHERE status IN ('B','b');
+
+-- Combining multiple conditions
+SELECT * FROM bank.loan
+WHERE status IN ('B','b') AND amount > 100000;
+
+-- Sorting and limiting results
+SELECT * FROM bank.account
+ORDER BY account_id DESC
+LIMIT 10;
+
+
+-- ==================================================
+-- CHECK FOR UNDERSTANDING
+-- ==================================================
+
+-- 1. Select districts and salaries (from the `district` table) where salary is greater than 10000. Return columns as `district_name` and `average_salary`.
+-- 2. Select those loans whose contract finished and were not paid. **Hint**: You should be looking at the `loan` column and you will need the extended case study information to tell you which value of status is required.
+-- 3. Select cards of type `junior`. Return just first 10 in your query.
+-- 4. Select those loans whose contract finished and not paid back. Return the debt value from the status you identified in the last activity, together with loan id and account id.
+-- 5. Calculate the urban population for all districts. Hint: You are looking for the number of inhabitants and the % of urban inhabitants in each district. Return columns as district_name and urban_population.
+-- 6. On the previous query result - rerun it but filtering out districts where the rural population is greater than 50%.
+
+-- ==================================================
+
+-- ==================================================
+-- LOGICAL OPERATORS 
+-- ==================================================
+
+-- Two conditions applied on the table
+SELECT *
+FROM bank.loan
+WHERE status = 'B' AND amount > 100000;
+
+-- We can have as many conditions as we need
+SELECT *
+FROM bank.loan
+WHERE status = 'B' AND amount > 100000 AND duration <= 24;
+
+-- Status B and D are the clients that were bad for business for the bank
+SELECT *
+FROM bank.loan
+WHERE status = 'B' OR status = 'D';
+
+SELECT *
+FROM bank.loan
+WHERE (status = 'B' OR status ='D') AND amount > 200000;
+
+-- Logical NOT operator - it negates the boolean expression that we are evaluating
+SELECT *
+FROM bank.order
+WHERE NOT k_symbol = 'SIPO';
+
+SELECT *
+FROM bank.order
+WHERE NOT k_symbol = 'SIPO' AND NOT amount < 1000;
+
+-- ==================================================
+-- NUMERIC FUNCTIONS
+-- ==================================================
+
+SELECT order_id, ROUND(amount/1000,2) FROM bank.order;
+
+-- Checking the number of rows in the table, both methods give the same result
+-- given that there are no nulls in the column in the second case:
+SELECT COUNT(*) FROM bank.order;
+
+SELECT COUNT(order_id) FROM bank.order;
+
+SELECT MAX(amount) FROM bank.order;
+SELECT MIN(amount) FROM bank.order;
+
+-- Rounding numbers to the greatest integer smaller than the value can be done using the function `floor()`
+SELECT FLOOR(AVG(amount)) FROM bank.order;
+
+-- Rounding numbers to the smallest integer bigger than the value can be done using the function `ceiling()`
+SELECT CEILING(AVG(amount)) FROM bank.order;
+
+-- Rounding digits can be done with `round( , number_of_digits)`
+SELECT ROUND(AVG(amount),2) FROM bank.order;
+
+-- ==================================================
+-- STRING FUNCTIONS
+-- ==================================================
+
+-- To obtain the length of a string we can use the function `length()`
+SELECT LENGTH('himanshu');
+SELECT *, LENGTH(k_symbol) AS 'Symbol_length' FROM bank.order;
+
+-- We can concatenate two strings using the function `concat()`
+SELECT *, CONCAT(order_id, account_id) AS 'concat' FROM bank.order;
+
+-- Functions to lowercase `lower()` and uppercase `upper()` a string
+SELECT *, LOWER(A2), UPPER(A3) FROM bank.district;
+
+-- Function to retrieve a given number of characters of a string starting from the left: `left()`
+-- There is an equivalent function to retrieve a given number of characters of a string starting from the right
+-- `right()`
+SELECT A2, LEFT(A2,5) FROM bank.district;
+
+-- Functions to drop white spaces of a string `ltrim()` and `rtrim()`
+SELECT LTRIM(' Hello ');
+SELECT RTRIM(' Hello ');
+
+
+-- ==================================================
+-- Check for Understanding: More Simple queries
+-- ==================================================
+
+-- 1. Get all junior cards issued last year. Hint: Use the numeric value (980000).
+
+-- 2. Get the first 10 transactions for withdrawals that are not in cash. You will need the extended case study information to tell you which values are required here, and you will need to refer to conditions on two columns.
+
+-- 3. Refine your query from last activity on loans whose contract finished and not paid back - filtered to loans where they were left with a debt bigger than 1000.
+--   Return the debt value together with loan id and account id. Sort by the highest debt value to the lowest.
+SELECT *   from bank.loan 
+WHERE status = 'B';
+
+SELECT loan_id,account_id,amount-payments as debt from bank.loan where status = 'B' and amount-payments > 1000 ORDER  BY debt DESC;
+-- 4. Get the biggest and the smallest transaction with non-zero values in the database (use the trans table in the bank database).
+SELECT * FROM bank.trans;
+SELECT max(amount) as max,  min(amount) as min from bank.trans where amount  <> 0;
+-- 5. Get account information with an extra column year showing the opening year as 'YY'. Eg., 1995 will show as 95. Hint: Look at the first two characters of the string date in the account table. You would have to use function `substr`. Google is your friend. 
+SELECT * from bank.account;
+select * , substring(date,3,2) AS year from bank.account;
+-- ==================================================
+
+-- ==================================================
+-- DATE AND TIME FUNCTIONS
+-- ==================================================
+
+-- Go through the case study pdf document to understand the format in which dates have been recorded in these tables - account, card, and loan.
+-- In the first table, the column date is of type integer. So we will convert the column into date format. For now, this change will only be for display purposes as we not altering the values of the table where the column has been defined.
+-- To make data type conversions, we can make use of the function convert() as shown below.
+
+-- Using the function `convert()` to change the displayed type of a column
+-- We don't alter the values themselves, we change how they are displayed.
+
+-- ==================================================
+-- DATE AND TIME FUNCTIONS
+-- ==================================================
+
+-- Date conversion and formatting
+SELECT account_id, district_id, frequency, 
+       CONVERT(account.date, DATE) 
+FROM bank.account;
+
+-- Note: In the previous example, one of the columns has the same name 'date' as the variable type 'date'. 
+-- This shouldn't be a problem as long as you differentiate them. 
+-- In the CONVERT() function, the first argument is the name of the column and the second is the type to which you want to convert.
+
+-- Applying the same conversion for the 'loan' table
+SELECT CONVERT(date, DATE) 
+FROM bank.loan;
+
+-- In the next example, we perform a two-step process for date conversion:
+-- 1. Extract the date string using SUBSTRING_INDEX
+-- 2. Convert the string to a date format
+
+SELECT SUBSTRING_INDEX(issued, ' ', 1) 
+FROM bank.card;
+
+SELECT CONVERT(SUBSTRING_INDEX(issued, ' ', 1), DATE) 
+FROM bank.card;
+
+-- Converting the original format to a desired date format for displaying purposes
+SELECT DATE_FORMAT(CONVERT(date, DATE), '%Y-%M-%D') 
+FROM bank.loan;
+
+-- Extracting the year from the date
+SELECT DATE_FORMAT(CONVERT(date, DATE), '%Y') 
+FROM bank.loan;
+
+
+-- ==================================================
+-- Check For Understanding
+-- ==================================================
+
+-- 1. Get `card_id` and `year_issued` for all **gold** cards.
+select * from bank.card;
+select card_id, type, DATE_FORMAT(CONVERT(SUBSTRING_INDEX(issued,' ',1), date),'%Y') AS year FROM bank.card where type = 'gold';
+-- 2. When was the first gold card **issued**? (Year)
+select card_id, DATE_FORMAT(CONVERT(SUBSTRING_INDEX(issued, ' ',1),date), '%y-%m-%d') AS year
+from bank.card where type = 'gold' order by year asc limit 1; 
+-- 3. Get issue date as:
+select card_id, DATE_FORMAT(CONVERT(SUBSTRING_INDEX(issued, ' ', 1) date), '%M,%D,%Y') AS American, DATE_FORMAT(CONVERT(SUBSTRING_INDEX(issued, ' ', 1) date), '%d of %M of %Y') AS Spanish;
+--    - date_issued: 'November 7th, 1993'( American format)
+--    - fecha_emision: '07 of November of 1993' (Spanish format)
+
+-- ==================================================
+
+-- ==================================================
+-- NULL HANDLING
+-- ==================================================
+-- Null values are highly confusing in SQL.
+
+-- A column with a NULL value is a field with no value. It is very important to understand that a NULL value is different than a zero value or a field that contains spaces.
+
+-- We can check if we have null values in our table using the function `isnull()`
+-- which return either 0 or 1, where 0 means not null, 1 means null
+
+-- Identifying NULL values
+SELECT ISNULL(card_id) 
+FROM bank.card;
+
+-- With this information in hand, we can compute how many nulls there are in one column just adding the output
+SELECT sum(isnull(card_id)) FROM bank.card;
+
+-- Filtering out rows with NULL values
+SELECT * FROM bank.order
+WHERE k_symbol IS NULL;
+
+-- Filtering out rows with non-null and non-empty values
+SELECT * FROM bank.order
+WHERE k_symbol IS NOT NULL AND k_symbol NOT IN ('', ' ');
+
+
+-- ==================================================
+-- CASE STATEMENTS
+-- ==================================================
+
+-- In the 'loan' table, there's a column named 'status' with values A, B, C, and D. 
+-- Using the CASE statement, we will try to replace these values with a more descriptive status.
+
+SELECT loan_id, account_id,
+CASE 
+    WHEN status = 'A' THEN 'Good - Contract Finished'
+    WHEN status = 'B' THEN 'Defaulter - Contract Finished'
+    WHEN status = 'C' THEN 'Good - Contract Running'
+    ELSE 'In Debt - Contract Running'
+END AS 'Status_Description'
+FROM bank.loan;
+
+
+-- ==================================================
+-- Check For Understanding
+-- ==================================================
+
+-- - Null or empty values are usually a problem. Think about why those null values can appear
+SELECT SUM(ISNULL(amount)) FROM bank.trans;
+-- - Check for transactions with null or empty values for the column `amount`.
+-- - Count how many transactions have empty and non-empty `k_symbol` (in one query). Hint: consider using the function sum() with a condition inside.
+SELECT COUNT(ISNULL(amount)) FROM bank.trans;
+SELECT SUM(ISNULL(amount)) FROM bank.trans;
+SELECT SUM(ISNULL(k_symbol)=1 OR k_symbol IN ("", " "))  FROM bank.trans;
+SELECT SUM(NOT ISNULL(k_symbol) OR k_symbol IN ("", " "))  FROM bank.trans
+WHERE k_symbol IN ("", " ");
+SELECT SUM(amount IS NULL OR amount IN (" ", "")) AS Transactions FROM bank.trans;
+
+-- ==================================================
+
+-- ==================================================
+-- ORDERING VALUES
+-- ==================================================
+
+-- Displaying transactions without sorting.
+SELECT * FROM bank.trans;
+
+-- Sorting transactions by amount in ascending order.
+SELECT * FROM bank.trans ORDER BY amount;
+-- The following query produces the same result.
+SELECT * FROM bank.trans ORDER BY amount ASC;
+
+-- Sorting transactions by amount in descending order.
+SELECT * FROM bank.trans ORDER BY amount DESC;
+
+-- We can also sort the values according to several columns. To do this, we need to provide the column names separated by comma alongside the sorting modifier. If we provide several columns for sorting, then the sorting is done first by the first column values and then by the second, and so on.
+
+
+-- Here in this example we are going to see how to sort the transactions by date and amount
+
+-- Sorting transactions by multiple columns - date and amount.
+SELECT * FROM bank.trans ORDER BY date, amount;
+SELECT * FROM bank.trans ORDER BY amount, date;
+
+-- Specifying a different ordering for each column.
+SELECT * FROM bank.trans ORDER BY date DESC, amount DESC;
+
+-- Applying filters and sorting.
+SELECT * FROM bank.trans 
+WHERE operation <> '' 
+ORDER BY date DESC, amount DESC;
+
+-- We can't exchange the last two rows in the previous query because of the `order of processing`in SQL
+-- Check out the previous material.
+
+
+
+-- ==================================================
+-- FILTERING BY RANGE
+-- ==================================================
+
+-- Now we could be interested in adding filters based on ranges instead of values. For this purpose, we can use the BETWEEN clause with the following syntax:
+
+-- between min_value and max_value
+
+-- Here in this example we are going to see how to filter the values of the previous queries by date or amount ranges.
+
+-- We need to be very careful when dealing with dates. In our case, the `date` values of table trans
+-- are integers! Then we can use a simple condition to refine our query but base on integers.
+
+-- Filtering transactions by date range using integer values.
+SELECT * 
+FROM bank.trans 
+WHERE operation <> '' AND date BETWEEN 971231 AND 981231 
+ORDER BY date DESC, amount DESC;
+
+-- Finally, we are going to see how to look for basic and simple pattern matchings (regex is left for Python). To look for patterns in strings we can use the clause like alongside pattern identifiers. Those pattern identifiers are specified using % and/or _.
+
+-- ==================================================
+-- PATTERN MATCHING
+-- ==================================================
+
+-- Using LIKE for pattern matching
+-- Finding district names starting with 'K'
+SELECT * FROM bank.district
+WHERE A2 LIKE 'K%';
+
+-- Finding district names ending with 'K'
+SELECT * FROM bank.district
+WHERE A2 LIKE '%K';
+
+-- Finding district names containing a 'K'
+SELECT * FROM bank.district
+WHERE A2 LIKE '%K%';
+
+-- Using _ for specific character positions
+SELECT * FROM bank.district
+WHERE A2 LIKE '____';
+
+SELECT * FROM bank.district
+WHERE A2 regexp 'M+';
+
+SELECT * FROM bank.district
+WHERE A2 regexp 'M*';
+
+SELECT * FROM bank.district
+WHERE A2 regexp 'm*';
+
+SELECT * FROM bank.district
+WHERE A2 regexp '^B';
+
+SELECT * FROM bank.district
+WHERE A2 regexp '[a-yA-Z]{5}';
+
+SELECT * FROM bank.district
+WHERE A2 regexp 'v$';  -- + -> after letter means it appears at least once, * -> appears any amount 
+
+
+
+
+-- ==================================================
+-- Check For Understanding
+-- ==================================================
+
+-- 1. Get transactions in the first 15 days of 1993.
+-- 2. Find the different values from the field `A2` that start with the letter 'M'.
+-- 3. Find the different values from the field `A2` that end with the letter 'M'.
+
+-- ==================================================
+
+-- ==================================================
+-- AGGREGATIONS
+-- ==================================================
+
+-- Aggregate data from the entire table.
+SELECT 
+    COUNT(amount) AS 'Total_orders',
+    SUM(amount) AS 'Total_amount', 
+    MIN(amount) AS 'Minimum_amount',
+    MAX(amount) AS 'Maximum_amount',
+    AVG(amount) AS 'Average_amount'
+FROM bank.order;
+
+-- Aggregate data by destination bank.
+SELECT bank_to, ROUND(SUM(amount),2) AS 'Total_amount' 
+FROM bank.order
+GROUP BY bank_to;
+
+-- As you can see, several values of k_symbol column that are empty and we would like to get rid of these rows. The way to do this will be using a filter with WHERE. However, it's not possible to use WHERE AFTER a GROUP BY. If we want to implement a filter AFTER a GROUP BY we need to use a different clause: HAVING. Thus, remember: if you want to apply a filter:
+-- Before GROUP BY use WHERE
+-- After GROUP BY use HAVING
+-- In other words, WHERE can be seen as a pre-filter (filtering BEFORE the aggregation), and HAVING as a post-filter (filtering AFTER the aggregation). The HAVING clause works on aggregated data. Finally, keep in mind that we can't use aggregation function in a WHERE filter.
+
+-- Aggregate data with filters.
+SELECT 
+    bank_to, k_symbol,
+    COUNT(amount) AS 'Total_orders',
+    ROUND(SUM(amount),2) AS 'Total_amount',
+    MIN(amount) AS 'Min_amount',
+    MAX(amount) AS 'Max_amount',
+    ROUND(AVG(amount),2) AS 'Average_amount' 
+FROM bank.order
+WHERE k_symbol NOT IN ('',' ')
+GROUP BY bank_to, k_symbol
+ORDER BY bank_to, k_symbol;
+
+-- Aggregate data using HAVING clause for post-filters.
+SELECT 
+    bank_to, k_symbol,
+    COUNT(amount) AS 'Total_orders',
+    ROUND(SUM(amount),2) AS 'Total_amount',
+    MIN(amount) AS 'Min_amount',
+    MAX(amount) AS 'Max_amount',
+    ROUND(AVG(amount),2) AS 'Average_amount' 
+FROM bank.order
+GROUP BY bank_to, k_symbol
+HAVING k_symbol NOT IN ('',' ')
+ORDER BY bank_to, k_symbol;
+
+-- ==================================================
+-- CHECK FOR UNDERSTANDING
+-- ==================================================
+
+-- 1. Get the total rental revenue using the column `amount` from table `payment`. Display it as `Total_revenue`. 
+-- 2. Get the total rental revenue by customer_id sorted by total benefit in descending order.
+
+-- ==================================================
+-- AGGREGATING BY MULTIPLE COLUMNS
+-- ==================================================
+
+-- Aggregate data by destination bank and k_symbol.
+SELECT 
+    bank_to, k_symbol,
+    COUNT(amount) AS 'Total_orders',
+    ROUND(SUM(amount),2) AS 'Total_amount',
+    MIN(amount) AS 'Min_amount',
+    MAX(amount) AS 'Max_amount',
+    ROUND(AVG(amount),2) AS 'Average_amount' 
+FROM bank.order
+GROUP BY bank_to, k_symbol
+ORDER BY bank_to, k_symbol;
+
+-- Exchanging the aggregation order.
+SELECT 
+    bank_to, k_symbol,
+    COUNT(amount) AS 'Total_orders',
+    ROUND(SUM(amount),2) AS 'Total_amount',
+    MIN(amount) AS 'Min_amount',
+    MAX(amount) AS 'Max_amount',
+    ROUND(AVG(amount),2) AS 'Average_amount' 
+FROM bank.order
+GROUP BY k_symbol, bank_to
+ORDER BY k_symbol, bank_to;
\ No newline at end of file
diff --git a/lab_mysqlERD.png b/lab_mysqlERD.png
new file mode 100644
index 0000000..b838576
Binary files /dev/null and b/lab_mysqlERD.png differ
diff --git a/seeding.sql b/seeding.sql
new file mode 100644
index 0000000..55c27bc
--- /dev/null
+++ b/seeding.sql
@@ -0,0 +1,88 @@
+ CREATE DATABASE IF NOT EXISTS lab_mysql;
+
+USE lab_mysql;
+
+CREATE DATABASE lab_mysql;
+
+USE lab_mysql;
+
+CREATE TABLE Cars (
+    id INT,
+    vin VARCHAR(50),
+    manufacturer VARCHAR(50),
+    model VARCHAR(50),
+    year INT,
+    color VARCHAR(50)
+);
+
+DROP TABLE IF EXISTS Cars;
+
+CREATE TABLE cars (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    vin VARCHAR(50),
+    manufacturer VARCHAR(50),
+    model VARCHAR(50),
+    year INT,
+    color VARCHAR(50)
+);		
+
+CREATE TABLE customers (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    cust_name VARCHAR(50),
+    cust_phone  INT,
+    cust_email VARCHAR(50),
+    cust_address VARCHAR(50),
+    cust_city VARCHAR(50),
+    cust_state VARCHAR(50),
+    cust_country VARCHAR(50),
+    cust_zipcode INT
+);	
+
+CREATE TABLE salespersons (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    staff_id  INT,
+	name VARCHAR(50),
+    store VARCHAR(50)
+);	
+
+CREATE TABLE invoices (
+    id INT AUTO_INCREMENT PRIMARY KEY,
+    invoice_number  INT,
+	date INT,
+	car INT,
+    customer INT,
+	salesperson INT
+);	
+
+INSERT INTO Cars (id, vin, manufacturer, model, year, color) VALUES
+(1, '3K096I98581DHSNUP', 'Volkswagen', 'Tiguan', 2019, 'Blue'),
+(2, 'ZM8G7BEUQZ97IH46V', 'Peugeot', 'Rifter', 2019, 'Red'),
+(3, 'RKXVNNIHLVVZOUB4M', 'Ford', 'Fusion', 2018, 'White'),
+(4, 'HKNDGS7CU31E9Z7JW', 'Toyota', 'RAV4', 2018, 'Silver'),
+(5, 'DAM41UDN3CHU2WVF6', 'Volvo', 'V60', 2019, 'Gray'),
+(6, 'DAM41UDN3CHU2WVF6', 'Volvo', 'V60 Cross Country', 2019, 'Gray');
+
+INSERT INTO Salespersons (id, staff_id, name, store) VALUES
+(1, '00001', 'Petey Cruiser', 'Madrid'),
+(2, '00002', 'Anna Sthesia', 'Barcelona'),
+(3, '00003', 'Paul Molive', 'Berlin'),
+(4, '00004', 'Gail Forcewind', 'Paris'),
+(5, '00005', 'Paige Turner', 'Mimia'),
+(6, '00006', 'Bob Frapples', 'Mexico City'),
+(7, '00007', 'Walter Melon', 'Amsterdam'),
+(8, '00008', 'Shonda Leer', 'São Paulo');
+
+INSERT INTO Invoices (id, invoice_number, date, car, customer, salesperson) VALUES
+(1, '852399038', '2018-08-22', 1, 1, 3),
+(2, '731166526', '2018-12-31', 3, 3, 5),
+(3, '271135104', '2019-01-22', 2, 2, 7);
+
+INSERT INTO Invoices (id, invoice_number, date, car, customer, salesperson) VALUES
+(1, '852399038', '2018-08-22', 1, 1, 3),
+(2, '731166526', '2018-12-31', 3, 3, 5),
+(3, '271135104', '2019-01-22', 2, 2, 7);
+
+INSERT INTO Customers (id, cust_id, cust_name, cust_phone, cust_email, cust_address, cust_city, cust_state, cust_country, cust_zipcode) VALUES
+(0, '10001', 'Pablo Picasso', '+34 636 17 63 82', '-', 'Paseo de la Chopera, 14', 'Madrid', 'Madrid'),
+(1, '20001', 'Abraham Lincoln', '+1 305 907 7086', '-', '120 SW 8th St', 'Miami', 'Florida', 'Spain', '28045'),
+(2, '30001', 'Napoléon Bonaparte', '+33 1 79 75 40 00', '-', '40 Rue du Colisée', 'Paris', 'Île-de-France', 'France', '75008');
\ No newline at end of file