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题解 | 牛客的课程订单分析(七)

select source, count(1) as cnt from ( select case when is_group_buy='Yes' then 'GroupBuy' else name end as source from ( select user_i...

2025-08-17

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题解 | 牛客的课程订单分析(六)

select id, is_group_buy, case when is_group_buy='No' then name else null end as client_name from ( select a.id, is_group_buy, name, count(1) o...

2025-08-17

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题解 | 牛客的课程订单分析(五)

select user_id, min(date) as first_buy_date, max(date) as second_buy_date, cnt from ( select *, count(1) over (partition by user_id) as cnt, ...

2025-08-17

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题解 | 牛客的课程订单分析(四)

select user_id, date as first_buy_date, cnt from ( select *, count(1) over (partition by user_id) as cnt, row_number() over (partition by...

2025-08-17

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题解 | 牛客的课程订单分析(三)

select id, user_id, product_name, status, client_id, date from ( select *, count(1) over (partition by user_id) as cnt from order_info whe...

2025-08-17

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题解 | 牛客的课程订单分析(二)

select user_id from order_info where date>'2025-10-15' and status='completed' and product_name in ('C++', 'Java', 'Python') group by user_id having...

2025-08-17

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题解 | 牛客的课程订单分析(一)

select id, user_id, product_name, status, client_id, date from order_info where date>'2025-10-15' and product_name in ('C++', 'Java', 'Python') and...

2025-08-17

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题解 | 考试分数(五)

select temp2.id, temp2.job, grade.score, rn as t_rank from ( select id, job, score, rn, (cnt+1)/2 as middle, case when rn=(cnt+1)/2 or ...

2025-08-17

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题解 | 考试分数(四)

with t as ( select job, count(1) as cnt from grade group by job ) select job, case when mod(cnt, 2)=1 then ceil(cnt/2) when mod(cnt, 2)=0 ...

2025-08-17

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题解 | 考试分数(三)

select id, name, score from ( select a.id, name, score, dense_rank() over (partition by language_id order by score desc) as rn from grade ...

2025-08-16

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