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N022-M4 Tier 2 · Core SQL · medium hr · Helix Systems

Return the employee name, department name, and salary amount for every salary record belonging to an Engineering employee

Part of Joining Multiple Tables in SQL

The problem

An Engineering compensation reviewer at Helix Systems needs a list of every Engineering employee alongside their salary records.

Write a query to return the employee name, department name, and salary amount for every salary record belonging to an Engineering employee.

Assumptions:

  • The chain reaches: employees → departments, employees → salaries.
  • The Engineering department is identified by departments.name = 'Engineering'; the condition can equivalently use department_id = 1.
  • An Engineering employee with three salary records contributes three rows to the result.

Output:

  • One row per qualifying salary record, with columns employee_name, department_name, and amount.
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,
  s.amount
FROM
  employees e
  JOIN departments d ON e.department_id = d.id
  JOIN salaries s ON e.id = s.employee_id
WHERE
  d.name = 'Engineering'

The shape

The chain reaches from employees sideways to departments and forward to salaries. A WHERE filter on departments.name = 'Engineering' narrows the result to salary records belonging to Engineering employees only. The row count comes from the salaries side — one row per qualifying salary record, not per Engineering employee.

Clause by clause

  • SELECT e.name AS employee_name, d.name AS department_name, s.amount pulls one column from each of the three tables. Both employees and departments have a name column, so the aliases are load-bearing.
  • FROM employees e anchors the chain.
  • JOIN departments d ON e.department_id = d.id attaches the department to each employee. One-to-one on the employee side, no row multiplication.
  • JOIN salaries s ON e.id = s.employee_id attaches every salary record for that employee. An Engineering employee with two salary records contributes two rows — Sarah Chen's appearance twice in the result is exactly this. The chain produces one row per salary record before any filter runs.
  • WHERE d.name = 'Engineering' filters the joined result down to rows whose employee belongs to the Engineering department.

Why this and not WHERE e.department_id = 1

Both predicates return the same rows, because the prompt notes the equivalence — Engineering's department ID is 1. The textual filter is preferred for the same reason the prompt mentions it: d.name = 'Engineering' is a self-documenting expression, readable to anyone who didn't write the schema. e.department_id = 1 requires the reader to know that 1 means Engineering. For a one-off ad-hoc query against a familiar schema, the integer form is fine. For a query that lands in a report someone else has to read, the textual form pays for itself.

You practiced filtering a multi-table chain by an attribute on a joined dimension (departments.name). The same pattern from M3 — WHERE on a joined table — works regardless of whether the filtering column is on a central table or a leaf.

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