摘要: leetcode 1158 市場分析I select uo.user_id as buyer_id, uo.join_date, if(uo.buyer_id, uo.c, 0) as orders_in_2019 from ( select u.user_id, u.join_date, o.bu 阅读全文
posted @ 2023-05-12 17:04 Carl_ZhangJH 阅读(18) 评论(0) 推荐(0)
摘要: leetcode 1211 查詢結果的質量和佔比 select query_name,round( sum(rating / position) / count(result) , 2) as quality ,round(sum(rating < 3) / count(result) * 100, 阅读全文
posted @ 2023-05-12 15:04 Carl_ZhangJH 阅读(11) 评论(0) 推荐(0)
摘要: leetcode 1251 平均售價 select r.product_id, round(sum(r.price * r.units) / sum(r.units), 2) as average_price from ( select p.product_id, p.price , u.units 阅读全文
posted @ 2023-05-12 12:32 Carl_ZhangJH 阅读(15) 评论(0) 推荐(0)
摘要: leetcode 1280 學生們參加各科測試的次數 select s.student_id, s.student_name, st.subject_name, if(e.result, e.result, 0) as attended_exams from Students s join Subj 阅读全文
posted @ 2023-05-12 10:41 Carl_ZhangJH 阅读(13) 评论(0) 推荐(0)
摘要: leetcode 1148 文章瀏覽I select distinct author_id as id from Views where author_id = viewer_id order by author_id == 阅读全文
posted @ 2023-05-12 10:15 Carl_ZhangJH 阅读(20) 评论(0) 推荐(0)
摘要: leetcode 1141 查詢進30天活躍用戶數 select activity_date as day , count(distinct user_id) as active_users from Activity a where activity_date between DATE_SUB(' 阅读全文
posted @ 2023-05-12 09:56 Carl_ZhangJH 阅读(23) 评论(0) 推荐(0)