﻿ SQL Query on Multiple Tables - Exercises, Practice, Solution - w3resource

# SQL Exercises, Practice, Solution - Query on Multiple Tables

## SQL [ 8 exercises with solution]

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1. From the following tables, write a SQL query to find the salespeople and customers who live in the same city. Return customer name, salesperson name and salesperson city.

Sample table: salesman

``` salesman_id |    name    |   city   | commission
-------------+------------+----------+------------
5001 | James Hoog | New York |       0.15
5002 | Nail Knite | Paris    |       0.13
5005 | Pit Alex   | London   |       0.11
5006 | Mc Lyon    | Paris    |       0.14
5007 | Paul Adam  | Rome     |       0.13
5003 | Lauson Hen | San Jose |       0.12
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample Output:

```cust_name	name		city
Nick Rimando	James Hoog	New York
Brad Davis	James Hoog	New York
Julian Green	Pit Alex	London
Fabian Johnson	Mc Lyon		Paris
Fabian Johnson	Nail Knite	Paris
```

Click me to see the solution with pictorial presentation

2. From the following tables, write a SQL query to locate all the customers and the salesperson who works for them. Return customer name, and salesperson name.

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample table: salesman

``` salesman_id |    name    |   city   | commission
-------------+------------+----------+------------
5001 | James Hoog | New York |       0.15
5002 | Nail Knite | Paris    |       0.13
5005 | Pit Alex   | London   |       0.11
5006 | Mc Lyon    | Paris    |       0.14
5007 | Paul Adam  | Rome     |       0.13
5003 | Lauson Hen | San Jose |       0.12
```

Sample Output:

```cust_name	name
Nick Rimando	James Hoog
Graham Zusi	Nail Knite
Julian Green	Nail Knite
Fabian Johnson	Mc Lyon
Geoff Cameron	Lauson Hen
```

Click me to see the solution with pictorial presentation

3. From the following tables, write a SQL query to find those salespeople who generated orders for their customers but are not located in the same city. Return ord_no, cust_name, customer_id (orders table), salesman_id (orders table).

Sample table: salesman

``` salesman_id |    name    |   city   | commission
-------------+------------+----------+------------
5001 | James Hoog | New York |       0.15
5002 | Nail Knite | Paris    |       0.13
5005 | Pit Alex   | London   |       0.11
5006 | Mc Lyon    | Paris    |       0.14
5007 | Paul Adam  | Rome     |       0.13
5003 | Lauson Hen | San Jose |       0.12
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample table: orders

```ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001
```

Sample Output:

```ord_no	cust_name	customer_id	salesman_id
70004	Geoff Cameron	3009		5003
70003	Geoff Cameron	3009		5003
70011	Jozy Altidor	3003		5007
70001	Graham Zusi	3005		5002
70007	Graham Zusi	3005		5002
70012	Julian Green	3008		5002
```

Click me to see the solution with pictorial presentation

4. From the following tables, write a SQL query to locate the orders made by customers. Return order number, customer name.

Sample table: orders

```ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample Output:

```ord_no	cust_name
70002	Nick Rimando
70004	Geoff Cameron
70008	Nick Rimando
70010	Fabian Johnson
70003	Geoff Cameron
70011	Jozy Altidor
70013	Nick Rimando
70001	Graham Zusi
70007	Graham Zusi
70012	Julian Green
```

Click me to see the solution with pictorial presentation

5. From the following tables, write a SQL query to find those customers where each customer has a grade and is served by a salesperson who belongs to a city. Return cust_name as "Customer", grade as "Grade".

Sample table: orders

```ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample Output:

```Customer      |Grade|Order No|
--------------|-----|--------|
Nick Rimando  |  100|   70002|
Geoff Cameron |  100|   70004|
Nick Rimando  |  100|   70008|
Fabian Johnson|  300|   70010|
Geoff Cameron |  100|   70003|
Jozy Altidor  |  200|   70011|
Nick Rimando  |  100|   70013|
Graham Zusi   |  200|   70001|
Graham Zusi   |  200|   70007|
Julian Green  |  300|   70012|
```

Click me to see the solution with pictorial presentation

6. From the following table, write a SQL query to find those customers who are served by a salesperson and the salesperson earns commission in the range of 12% to 14% (Begin and end values are included.). Return cust_name AS "Customer", city AS "City".

Sample table: salesman

``` salesman_id |    name    |   city   | commission
-------------+------------+----------+------------
5001 | James Hoog | New York |       0.15
5002 | Nail Knite | Paris    |       0.13
5005 | Pit Alex   | London   |       0.11
5006 | Mc Lyon    | Paris    |       0.14
5007 | Paul Adam  | Rome     |       0.13
5003 | Lauson Hen | San Jose |       0.12
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample Output:

```Customer	City		Salesman	commission
Graham Zusi	California	Nail Knite	0.13
Julian Green	London		Nail Knite	0.13
Fabian Johnson	Paris		Mc Lyon		0.14
Geoff Cameron	Berlin		Lauson Hen	0.12
Jozy Altidor	Moscow		Paul Adam	0.13
```

Click me to see the solution with pictorial presentation

7. From the following tables, write a SQL query to find all orders executed by the salesperson and ordered by the customer whose grade is greater than or equal to 200. Compute purch_amt*commission as “Commission”. Return customer name, commission as “Commission%” and Commission.

Sample table: salesman

``` salesman_id |    name    |   city   | commission
-------------+------------+----------+------------
5001 | James Hoog | New York |       0.15
5002 | Nail Knite | Paris    |       0.13
5005 | Pit Alex   | London   |       0.11
5006 | Mc Lyon    | Paris    |       0.14
5007 | Paul Adam  | Rome     |       0.13
5003 | Lauson Hen | San Jose |       0.12
```

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample table: orders

```ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001
```

Sample Output:

```ord_no	cust_name	Commission%	Commission
70010	Fabian Johnson	0.14		277.6802
70011	Jozy Altidor	0.13		9.7877
70001	Graham Zusi	0.13		19.5650
70007	Graham Zusi	0.13		123.3050
70012	Julian Green	0.13		32.5585
```

Click me to see the solution with pictorial presentation

8. From the following table, write a SQL query to find those customers who placed orders on October 5, 2012. Return customer_id, cust_name, city, grade, salesman_id, ord_no, purch_amt, ord_date, customer_id and salesman_id.

Sample table: customer

``` customer_id |   cust_name    |    city    | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando   | New York   |   100 |        5001
3007 | Brad Davis     | New York   |   200 |        5001
3005 | Graham Zusi    | California |   200 |        5002
3008 | Julian Green   | London     |   300 |        5002
3004 | Fabian Johnson | Paris      |   300 |        5006
3009 | Geoff Cameron  | Berlin     |   100 |        5003
3003 | Jozy Altidor   | Moscow     |   200 |        5007
3001 | Brad Guzan     | London     |       |        5005
```

Sample table: orders

```ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001
```

Sample Output:

```customer_id	cust_name	city		grade	salesman_id	ord_no	purch_amt	ord_date	customer_id	salesman_id
3002		Nick Rimando	New York	100	5001		70002	65.26		2012-10-05	3002		5001
3005		Graham Zusi	California	200	5002		70001	150.50		2012-10-05	3005		5002
```

Click me to see the solution with pictorial presentation

## Practice Online

More to Come !

Query visualizations are generated using Postgres Explain Visualizer (pev).

Do not submit any solution of the above exercises at here, if you want to contribute go to the appropriate exercise page.

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