Solution:Step 1: Understand the scalar subquery usageThe subquery fetches the department name matching each employee's department_id.Step 2: Analyze the outputEach employee's name is paired with their department name from the departments table.Final Answer:Employee names with their department names

[Solved] Given the tables:employees(id, name, department_id)departments(id, name)What will this query return?SELECT name, (SELECT name FROM departments WHERE id = department_id) AS dept_name FROM employees; | PostgreSQL

PostgreSQL - Subqueries in PostgreSQL

Given the tables:
employees(id, name, department_id)
departments(id, name)
What will this query return?

SELECT name, (SELECT name FROM departments WHERE id = department_id) AS dept_name FROM employees;

AEmployee names with NULL for all departments

BEmployee names with their department names

COnly employees from the first department

DSyntax error due to subquery

Step-by-Step Solution

Solution:

Step 1: Understand the scalar subquery usage
The subquery fetches the department name matching each employee's department_id.
Step 2: Analyze the output
Each employee's name is paired with their department name from the departments table.
Final Answer:
Employee names with their department names -> Option B
Quick Check:
Scalar subquery returns matching department name [OK]

Quick Trick: Scalar subqueries can fetch related single values per row [OK]

Common Mistakes:

Assuming subquery returns multiple rows causing error
Confusing department_id with department name
Expecting NULLs when matching rows exist

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Given the tables:

Step 1: Understand the scalar subquery usage

Step 2: Analyze the output

Final Answer:

Quick Check:

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