가구 판매의 비중이 높았던 날 찾기

--https://solvesql.com/problems/day-of-furniture/
--SQLite 

SELECT 
  order_date, 
  furniture, 
  ROUND((furniture * 1.0 / daily_orders )*100, 2) AS furniture_pct
FROM (
  select 
    order_date, 
    COUNT(DISTINCT CASE WHEN category = "Furniture" THEN order_id END) AS furniture, 
    COUNT(DISTINCT order_id) AS daily_orders 
  FROM records 
  GROUP BY order_date 
  HAVING COUNT(DISTINCT order_id) >= 10
      )
WHERE furniture_pct >= 40
GROUP BY order_date
ORDER BY furniture_pct DESC, order_date;