The document describes a sample SQL screening test containing tables, questions, and solutions. It includes tables for event logs and employees. The questions ask about analyzing the event logs to determine the number of unique events, dates with most customers serviced, employee responsible for most events, average salaries by position, and longest job by duration. The solutions provide the SQL queries and analyses to answer each question.

1. Creation Date: March 28th 2023
Author: Magnus Skonberg
SAMPLE SQL SCREENING
Purpose: inspect the tables and write SQL queries to answer the questions provided below.
…
Tables
Event Log
event_id employee_id customer_id event timestamp
1 1 1 Check in 2023-03-01 08:03:54
2 2 2 Check In 2023-03-01 09:17:32
3 1 1 Complete 2023-03-01 11:33:00
4 1 3 Check in 2023-03-01 12:27:00
5 1 1 Issue PO 2023-03-01 13:15:33
6 1 3 Complete 2023-03-01 16:11:11
7 2 2 Complete 2023-03-01 17:35:11
8 2 2 Issue PO 2023-03-02 06:52:22
9 1 3 Issue PO 2023-03-02 07:58:23
10 1 4 Check in 2023-03-02 08:17:16
11 1 4 Complete 2023-03-02 15:56:45
12 1 4 Issue PO 2023-03-02 16:22:12
Employees
emp_id name position salary
1 Willy Washington Analyst 20500
2 Avery Adams Analyst 18600
3 Jenny Jefferson Analyst 19900
4 Laura Lincoln Consultant 15700
5 Ricky Roosevelt Consultant 14550
6 Ellis Eisenhower Consultant 17800
7 Jimmy Johnson Sales 12000
8 Barbara Bush Sales 256000
9 Orville Obama Sales 14800
…

2. Creation Date: March 28th 2023
Author: Magnus Skonberg
Questions
Q1 How many different events have been logged? And what’s the most popular `event`?
Q2 On which date were the most customers serviced?
Q3 What’s the name of the employee responsible for the most events? (How) would our results
differ if we used a left vs. inner join?
Q4 What’s the average employee salary per position? And where does each employee rank
with regard to salary for their position?
Q5 What job (employee_id + customer_id) took the longest to complete? And how long did it
take?
…

3. Creation Date: March 28th 2023
Author: Magnus Skonberg
Solutions
Q1 How many different events have been logged? And what’s the most popular `event`?
SELECT COUNT(*)
FROM event;
SELECT
event,
COUNT(*)
FROM event
GROUP BY 1
ORDER BY 2 DESC;
Answer: 12 events have been logged. All events have occurred at the same frequency. Thus,
there is no “most popular” event.
…
Q2 On which date were the most customers serviced?
SELECT
DATE_TRUNC('day', start_time) as start_date,
COUNT(DISTINCT customer_id) as customers_serviced
FROM event
GROUP BY 1
ORDER BY 2 DESC;
Answer: I took “serviced” to mean that some work function, including issuing the PO, was
completed. As such, the dates had an equivalent number of customers serviced. 3 customers
were serviced on the 1st and 2nd of March 2023.
…
Q3 What’s the name of the employee responsible for the most events? (How) would our results
differ if we used a left vs. inner join?
SELECT
e1.name,
COUNT(*) as events
FROM employees e1
INNER JOIN event e2
ON e1.emp_id = e2.employee_id
GROUP BY 1

4. Creation Date: March 28th 2023
Author: Magnus Skonberg
ORDER BY 2 DESC;
Answer: Willy Washington was responsible for the most events with 9.
In this particular instance, a LEFT JOIN would return all names from the employees table
whereas an INNER JOIN would only return names with entries in the events table. While the
outcome / top entry would be the same with either join, the subsequent entries would differ and
could be misinterpreted. As such, we go with and recommend an INNER JOIN.
…
Q4 What’s the average employee salary per position? And where does each employee rank
with regard to salary for their position?
SELECT *,
ROUND(AVG(salary) OVER(PARTITION BY position),2) AS average_position_salary,
RANK() OVER(PARTITION BY position ORDER BY salary DESC) AS
position_salary_rank
FROM employees;
…
Q5 What job (employee_id + customer_id) took the longest to complete? And how long did it
take?
WITH duration AS (
SELECT
*,
LEAD(start_time) OVER(PARTITION BY employee_id, customer_id ORDER BY
start_time) as end_time,
LEAD(start_time) OVER(PARTITION BY employee_id, customer_id ORDER BY
start_time) - start_time as duration
FROM event

5. Creation Date: March 28th 2023
Author: Magnus Skonberg
)
SELECT
employee_id,
customer_id,
SUM(duration) as total_duration
FROM duration
GROUP BY 1,2
ORDER BY 3 DESC;
Answer: the job with employee_id 2 and customer_id 2 took the longest, with a total duration of
21 hours 34 minutes and 50 seconds.