Metrics collection in HLD - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from the metrics table.

HLD

SELECT [1] FROM metrics;

Drag options to blanks, or click blank then click option'

A*

BALL

Ccolumns

Dmetrics

Attempts:

3 left

💡 Hint

Common Mistakes

Using the table name instead of '*' to select all columns.

Using 'ALL' which is not valid in this context.

Using * selects all columns from the table.

2fill in blank

medium

Complete the code to count the number of entries in the metrics table.

HLD

SELECT COUNT([1]) FROM metrics;

Drag options to blanks, or click blank then click option'

A*

Ball

C1

Dmetric_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using a specific column name which may exclude NULL values.

Using 'all' which is not valid syntax.

Using COUNT(*) counts all rows in the table.

3fill in blank

hard

Fix the error in the query to get the average value of the metric column.

HLD

SELECT AVG([1]) FROM metrics;

Drag options to blanks, or click blank then click option'

Ametric_value

Bmetric

Cvalue

Davg

Attempts:

3 left

💡 Hint

Common Mistakes

Using the function name as a column name.

Using a column name that does not exist.

The column storing the metric values is named value. Using AVG(value) calculates the average.

4fill in blank

hard

Fill both blanks to filter metrics collected after 2023-01-01 and order by value descending.

HLD

SELECT * FROM metrics WHERE collected_date [1] '2023-01-01' ORDER BY value [2];

Drag options to blanks, or click blank then click option'

A>

B<

CDESC

DASC

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' to filter dates after a date.

Ordering ascending when descending is needed.

Use > to filter dates after 2023-01-01 and DESC to order values from highest to lowest.

5fill in blank

hard

Fill all three blanks to create a dictionary-like result with metric name as key, average value as value, filtering values above 50.

HLD

SELECT metric_name AS [1], AVG(value) AS [2] FROM metrics WHERE value [3] 50 GROUP BY metric_name;

Drag options to blanks, or click blank then click option'

Akey

Bavg_value

C>

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect aliases that don't describe the columns.

Using '<' instead of '>' in the WHERE clause.

Use name as the key alias, avg_value as the average value alias, and > to filter values greater than 50.

Practice

1. What is the main purpose of using aggregate functions like COUNT, SUM, and AVG in metrics collection?

easy

A. To create new tables for storing data

B. To change the structure of the database tables

C. To calculate totals, averages, and counts from data

D. To delete unwanted data from the database

Solution

1. Step 1: Understand aggregate functions

   Aggregate functions like COUNT, SUM, and AVG are used to calculate summary values from multiple rows of data.

2. Step 2: Identify their purpose in metrics

   These functions help get totals, averages, and counts which are useful for metrics and reports.

3. Final Answer:

   To calculate totals, averages, and counts from data -> Option C

4. Quick Check:

   Aggregate functions = summary numbers [OK]

Hint: Aggregate functions summarize data into numbers fast [OK]

Common Mistakes:

• Thinking aggregate functions modify data structure
• Confusing aggregate functions with data deletion
• Believing aggregate functions create tables

2. Which of the following SQL queries correctly counts the number of orders in a table named Orders?

easy

A. SELECT COUNT(Orders) FROM *;

B. SELECT COUNT FROM Orders;

C. SELECT SUM(*) FROM Orders;

D. SELECT COUNT(*) FROM Orders;

Solution

1. Step 1: Recall correct COUNT syntax

   The correct syntax to count rows is COUNT(*) inside SELECT.

2. Step 2: Check each option

   SELECT COUNT(*) FROM Orders; uses COUNT(*) correctly. Others have syntax errors or wrong functions.

3. Final Answer:

   SELECT COUNT(*) FROM Orders; -> Option D

4. Quick Check:

   COUNT(*) counts rows correctly [OK]

Hint: Use COUNT(*) to count all rows in a table [OK]

Common Mistakes:

• Omitting parentheses after COUNT
• Using COUNT without specifying columns or *
• Confusing SUM with COUNT

3. Given the table Sales with columns amount and region, what will this query return?

SELECT AVG(amount) FROM Sales WHERE region = 'East';

medium

A. The average sales amount for the East region

B. The total sales amount for all regions

C. The count of sales records in the East region

D. An error because AVG cannot be used with WHERE

Solution

1. Step 1: Understand the WHERE clause effect

   The WHERE clause filters rows to only those where region is 'East'.

2. Step 2: Understand AVG function

   AVG calculates the average of the filtered amount values.

3. Final Answer:

   The average sales amount for the East region -> Option A

4. Quick Check:

   AVG with WHERE filters = average for filtered data [OK]

Hint: WHERE filters rows before aggregate functions run [OK]

Common Mistakes:

• Thinking AVG returns total sum
• Confusing COUNT with AVG
• Believing WHERE cannot be used with aggregates

4. Identify the error in this SQL query that tries to get the total sales for the 'West' region:

SELECT SUM(amount) FROM Sales WHERE region == 'West';

medium

A. Using double equals (==) instead of single equals (=) in WHERE clause

B. SUM cannot be used with WHERE clause

C. Missing GROUP BY clause for aggregation

D. amount column cannot be summed

Solution

1. Step 1: Check WHERE clause syntax

   SQL uses single equals (=) for comparison, not double equals (==).

2. Step 2: Confirm aggregate and WHERE usage

   SUM works with WHERE filtering; GROUP BY is not needed here.

3. Final Answer:

   Using double equals (==) instead of single equals (=) in WHERE clause -> Option A

4. Quick Check:

   Use = for comparisons in SQL WHERE [OK]

Hint: Use single = for comparisons in SQL WHERE [OK]

Common Mistakes:

• Using == instead of = in WHERE
• Thinking GROUP BY is always required with SUM
• Assuming SUM cannot be used with WHERE

5. You want to find the average order amount per customer but only for customers who placed more than 5 orders. Which SQL approach correctly achieves this?

hard

A. Use WHERE COUNT(*) > 5 and SELECT AVG(amount) without GROUP BY

B. Use GROUP BY customer_id with HAVING COUNT(*) > 5 and SELECT AVG(amount)

C. Use WHERE AVG(amount) > 5 and GROUP BY customer_id

D. Use HAVING AVG(amount) > 5 without GROUP BY

Solution

1. Step 1: Understand grouping by customer

   To get average per customer, group rows by customer_id.

2. Step 2: Filter groups with more than 5 orders

   Use HAVING with COUNT(*) > 5 to keep only customers with more than 5 orders.

3. Step 3: Select average order amount

   Use AVG(amount) in SELECT to get average per filtered customer group.

4. Final Answer:

   Use GROUP BY customer_id with HAVING COUNT(*) > 5 and SELECT AVG(amount) -> Option B

5. Quick Check:

   GROUP BY + HAVING filters groups before AVG [OK]

Hint: Use HAVING to filter groups after GROUP BY [OK]

Common Mistakes:

• Using WHERE to filter aggregated counts
• Forgetting GROUP BY when using aggregates per group
• Using HAVING without GROUP BY