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N017-M3 Tier 2 · Core SQL · medium hr · Helix Systems

Return each employee's name and department name for every employee assigned to that department

Part of INNER JOIN in SQL

The problem

An engineering manager at Helix Systems needs a staff list for department 3.

Write a query to return each employee's name and department name for every employee assigned to that department.

Assumptions:

  • The employees table contains every active and former employee at Helix Systems.
  • The departments table contains every department; department 3 is identified by departments.id = 3.
  • The department condition can be expressed against either the employee's department_id or the department's id — both produce the same result because they are equal across the assembled rows.

Output:

  • One row per qualifying employee, with columns employee_name and department_name.
Schema · hr 4 tables
departments
id integer
name text
location text
budget numeric
salaries
id integer
employee_id integer
amount numeric
effective_date date
end_date? date
employees
id integer
name text
email text
department_id integer
manager_id? integer
hire_date date
title text
is_active boolean
job_history
id integer
employee_id integer
title text
department_id integer
start_date date
end_date? date
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Solution query
SELECT
  e.name AS employee_name,
  d.name AS department_name
FROM
  employees e
  JOIN departments d ON e.department_id = d.id
WHERE
  d.id = 3

The shape

The join pairs each employee with their department on e.department_id = d.id, and the WHERE narrows to the rows where the department side is 3. Because the ON condition forces those two columns to hold the same value on every assembled row, filtering on either side produces the same eleven-employee Sales roster.

Clause by clause

  • FROM employees e is the source: every active and former employee at Helix Systems.
  • JOIN departments d ON e.department_id = d.id pairs each employee with the department row identified by their department_id. For every employee, PostgreSQL finds the matching department row and combines both into a single output row.
  • WHERE d.id = 3 filters the assembled rows to the ones whose department side is department 3 — Sales. The eleven employees assigned to that department come through; everyone else drops.
  • SELECT e.name AS employee_name, d.name AS department_name returns the two name columns from each side. The qualifiers e. and d. are mandatory because both tables have a name column; the output aliases then label each column by its role.

Why this and not WHERE e.department_id = 3

Both conditions produce the exact same result. The join's ON e.department_id = d.id guarantees that on every row in the assembled result, e.department_id and d.id hold the same value. They're equated by definition. So WHERE d.id = 3 and WHERE e.department_id = 3 are equivalent filters.

Reaching through the dimension table (d.id = 3) is the more general pattern. When the filter is on an attribute the fact table doesn't carry — d.name = 'Sales', d.region = 'EMEA' — the condition has to ride on the dimension side. Writing the filter on the dimension keeps the shape consistent across cases where the criterion lives on attributes the employee row doesn't store.

You practiced applying a condition on the matched-key value. Once two tables are paired on e.department_id = d.id, those two columns hold the same value for every row in the result — the condition on either side gives the same answer.

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