Optimizes SQL queries, designs database schemas, and troubleshoots performance issues. Use when a user asks why their query is slow, needs help writing complex
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EXPLAIN ANALYZE and confirm no sequential scans on large tables; if query does not meet sub-100ms target, iterate on index selection or query rewrite before proceedingLoad detailed guidance based on context:
| Topic | Reference | Load When |
|---|---|---|
| Query Patterns | references/query-patterns.md | JOINs, CTEs, subqueries, recursive queries |
| Window Functions | references/window-functions.md | ROW_NUMBER, RANK, LAG/LEAD, analytics |
| Optimization | references/optimization.md | EXPLAIN plans, indexes, statistics, tuning |
| Database Design | references/database-design.md | Normalization, keys, constraints, schemas |
| Dialect Differences | references/dialect-differences.md | PostgreSQL vs MySQL vs SQL Server specifics |
-- Isolate expensive subquery logic for reuse and readability
WITH ranked_orders AS (
SELECT
customer_id,
order_id,
total_amount,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date DESC) AS rn
FROM orders
WHERE status = 'completed' -- filter early, before the join
)
SELECT customer_id, order_id, total_amount
FROM ranked_orders
WHERE rn = 1; -- latest completed order per customer
-- Running total and rank within partition — no self-join required
SELECT
department_id,
employee_id,
salary,
SUM(salary) OVER (PARTITION BY department_id ORDER BY hire_date) AS running_payroll,
RANK() OVER (PARTITION BY department_id ORDER BY salary DESC) AS salary_rank
FROM employees;
-- PostgreSQL: always use ANALYZE to see actual row counts vs. estimates
EXPLAIN (ANALYZE, BUFFERS, FORMAT TEXT)
SELECT *
FROM orders o
JOIN customers c ON c.id = o.customer_id
WHERE o.created_at > NOW() - INTERVAL '30 days';
Key things to check in the output:
ANALYZE <table> to refresh statisticsread count signals missing cache / index-- BEFORE: correlated subquery, one execution per row (slow)
SELECT order_id,
(SELECT SUM(quantity) FROM order_items oi WHERE oi.order_id = o.id) AS item_count
FROM orders o;
-- AFTER: single aggregation join (fast)
SELECT o.order_id, COALESCE(agg.item_count, 0) AS item_count
FROM orders o
LEFT JOIN (
SELECT order_id, SUM(quantity) AS item_count
FROM order_items
GROUP BY order_id
) agg ON agg.order_id = o.id;
-- Supporting covering index (includes all columns touched by the query)
CREATE INDEX idx_order_items_order_qty
ON order_items (order_id)
INCLUDE (quantity);
When implementing SQL solutions, provide: