마케팅팀 요청 하나를 받아서,
한 사이클 전체를 라이브로

[테이블 구조]
- members: member_id PK, member_name, gender (M/F), age, joined_at (DATE)
- orders: order_id PK, member_id FK, ordered_at (DATE)
- order_items: order_id FK, product_id FK, quantity
- products: product_id PK, product_name, price, category_id FK
- categories: category_id PK, category_name

[부모-자식 관계]
- members 1:N orders
- orders 1:N order_items
- products 1:N order_items
- categories 1:N products

[샘플 데이터]
members: (1, 김철수, M, 35, '2025-01-15'), (4, 정수진, F, 31, '2025-03-05')
orders: (4, 4, '2025-03-15')
order_items: (4, 27, 1), (4, 30, 1)
products: (27, '수분 크림', 32000, 5)
categories: (5, '뷰티')

[요구사항]
30대 여성 회원이 가장 많이 산 카테고리 TOP 3.
- "많이 산" = 주문 횟수 (DISTINCT order_id)
- "30대 여성" = gender = 'F' AND age BETWEEN 30 AND 39
- 출력 컬럼: 카테고리명, 주문수
- 정렬: 주문수 많은 순
- 행 수: 3

SELECT
  c.category_name,
  COUNT(DISTINCT o.order_id) AS 주문수
FROM
  members m
JOIN orders o
  ON m.member_id = o.member_id
JOIN order_items oi
  ON o.order_id = oi.order_id
JOIN products p
  ON oi.product_id = p.product_id
JOIN categories c
  ON p.category_id = c.category_id
WHERE
  m.gender = 'F'
  AND m.age BETWEEN 30 AND 39
GROUP BY
  c.category_id,
  c.category_name
ORDER BY
  주문수 DESC
LIMIT 3;

SELECT
  *
FROM
  members
WHERE
  gender = 'F'
  AND age BETWEEN 30 AND 39;

SELECT
  m.member_name,
  COUNT(*) AS 주문수
FROM
  members m
JOIN orders o
  ON m.member_id = o.member_id
WHERE
  m.gender = 'F'
  AND m.age BETWEEN 30 AND 39
GROUP BY
  m.member_id,
  m.member_name;

[테이블 구조 + 관계 + 샘플]  (부분 A 와 동일 — 그대로 재사용)

[요구사항]
부분 A 결과의 TOP 3 카테고리 ('뷰티', '의류', '도서') 에서 —
각 카테고리에서 30대 여성이 가장 많이 산 상품 1개씩.
- "많이 산" = 판매 수량 (SUM(quantity))
- 출력 컬럼: 카테고리명, 상품명, 총수량
- 카테고리당 1행만

복잡하니까 WITH 절로 단계별로 정리해줘.

WITH female_30s AS (
  SELECT
    member_id
  FROM
    members
  WHERE
    gender = 'F'
    AND age BETWEEN 30 AND 39
),
sales_per_product AS (
  SELECT
    c.category_name,
    p.product_id,
    p.product_name,
    SUM(oi.quantity) AS total_qty
  FROM
    female_30s f
  JOIN orders o
    ON f.member_id = o.member_id
  JOIN order_items oi
    ON o.order_id = oi.order_id
  JOIN products p
    ON oi.product_id = p.product_id
  JOIN categories c
    ON p.category_id = c.category_id
  WHERE
    c.category_name IN ('뷰티', '의류', '도서')
  GROUP BY
    c.category_name,
    p.product_id,
    p.product_name
),
top_per_cat AS (
  SELECT
    category_name,
    MAX(total_qty) AS max_qty
  FROM
    sales_per_product
  GROUP BY
    category_name
)
SELECT
  spp.category_name,
  spp.product_name,
  spp.total_qty
FROM
  sales_per_product spp
JOIN top_per_cat tpc
  ON spp.category_name = tpc.category_name
  AND spp.total_qty = tpc.max_qty;

방금 만든 두 SQL 의 결과를 — 마케팅팀이 한 줄로 이해할 수 있는
요약 보고서 형태로 정리해줘.