INNER JOIN execution in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from two tables joined on the "id" column.

PostgreSQL

SELECT * FROM employees [1] departments ON employees.id = departments.id;

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AINNER JOIN

BRIGHT JOIN

CLEFT JOIN

DFULL JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN instead of INNER JOIN returns unmatched rows from the left table.

Using FULL JOIN returns all rows from both tables, not just matches.

The INNER JOIN returns rows when there is a match in both tables based on the join condition.

2fill in blank

medium

Complete the code to join "orders" and "customers" tables on the "customer_id" column.

PostgreSQL

SELECT orders.order_id, customers.name FROM orders [1] customers ON orders.customer_id = customers.customer_id;

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ALEFT JOIN

BCROSS JOIN

CRIGHT JOIN

DINNER JOIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using CROSS JOIN returns all combinations, not just matches.

LEFT JOIN includes unmatched rows from the left table.

Using INNER JOIN ensures only orders with matching customers are selected.

3fill in blank

hard

Fix the error in the join condition to correctly join "products" and "categories" on "category_id".

PostgreSQL

SELECT * FROM products INNER JOIN categories ON products.[1] = categories.category_id;

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Acategory

Bcategory_id

Cid

Dproduct_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong column like 'category' or 'product_id' causes join errors or wrong results.

Using 'id' instead of 'category_id' mismatches the join keys.

The join condition must use the correct column category_id from the products table to match categories.

4fill in blank

hard

Fill both blanks to create a join that selects employee names and their department names.

PostgreSQL

SELECT employees.name, departments.[1] FROM employees [2] departments ON employees.department_id = departments.id;

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Aname

BLEFT JOIN

CINNER JOIN

Dsalary

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN returns employees without departments.

Selecting 'salary' instead of 'name' shows wrong data.

The query selects the department name and uses an INNER JOIN to get matching employees and departments.

5fill in blank

hard

Fill all three blanks to join "sales" and "stores" tables and select store name and total sales.

PostgreSQL

SELECT stores.[1], SUM(sales.amount) AS total_sales FROM sales [2] stores ON sales.[3] = stores.id GROUP BY stores.name;

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Aname

BINNER JOIN

Cstore_id

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Using LEFT JOIN includes sales without matching stores.

Joining on 'date' instead of 'store_id' causes wrong matches.

The query selects the store name, joins with INNER JOIN on store_id to match sales to stores, and sums sales amounts.