Leetcode SQL membership questions [Summary of vomiting blood~~] Day 4

Table of contents

1173. Instant Food Delivery I
1174. Instant Food Delivery II
1193. Monthly Transactions I
1194. Championship Winner
1204. Last person to enter the elevator
1205. Monthly Transactions II
1211. Quality and proportion of query results
1212. Query Team points
1225. Consecutive dates for reporting system status
1241. Comments per post
1251. Average sale price
1264. Page recommendations
1270. All people reporting to the CEO of the company
1280. Number of tests students took in each subject
1285. Find the starting and ending numbers of a continuous interval

1173. Instant food delivery I

• source code:

  select round(sum(case when order_date = customer_pref_delivery_date then 1 else 0 end) / count(delivery_id) * 100, 2) as immediate_percentage
  from delivery
  

• Idea: sum() + case whenUsing this method simplifies this question a lot. This is also an experience.

1174. Instant Food Delivery II

• source code:

  with first_order as (
      select delivery_id
      from delivery
      where (customer_id, order_date) in (
          select customer_id, min(order_date)
          from delivery
          group by customer_id
      )
  )
  select 
      round(
          sum(case when order_date=customer_pref_delivery_date and delivery_id in (select * from first_order) then 1 else 0 end)
          / (select count(*) from first_order) * 100
      , 2) as immediate_percentage
  from delivery
  

• Idea: It is important to understand the demand. This question is to ask for the proportion of immediate orders among first-time orders. Therefore, we can first find out which first-time orders there are, and then use subqueries to determine the number of orders that are both immediate orders and first-time orders.

1193. Monthly Transactions I

• source code:

  select
      left(trans_date, 7) as month, country,
      count(id) as trans_count, 
      count(case when state = 'approved' then 1 else null end) as approved_count,
      sum(amount) as trans_total_amount,
      sum(case when state = 'approved' then amount else 0 end) as approved_total_amount
  from transactions
  group by left(trans_date, 7), country
  

• Idea: First we need to know how to get the date in front of the string. There are two ways substr() / left(); then figure out the grouping field and then perform aggregation.

1194. Tournament Winner

• source code:

  select group_id, id player_id
  from (
      select group_id, id, row_number() over(partition by group_id order by total_score desc, id) rk
      from (
          select id, sum(score) total_score
          from (
              select match_id, first_player id, first_score score
              from matches
              union all
              select match_id, second_player id, second_score score
              from matches
          ) t
          group by id
      ) t
      join players
      on t.id = players.player_id
  ) t
  where rk = 1
  

• Idea: According to the need to accumulate the scores of the players as mentioned in the question, that is, whether they are the first player or the second player, they need to be accumulated, so we can split one table into two tables sum. In this way, we can find the cumulative score of each player, and then find the player with the highest score in the group.

1204. The last person to enter the elevator

• source code:

  select person_name
  from (
      select person_name, sum(weight) over(order by turn) sum_weight
      from queue
  ) t
  where sum_weight <= 1000
  order by sum_weight desc
  limit 1;
  

• Idea: By sumopening the window on the function, you can get the cumulative weight sum. As long as you take out the person who is less than or equal to 1000 and the largest, he will be the last person who has the right weight and enters the elevator.

1205. Monthly Transactions II **

• source code:

  with base as (
      select 'approved' tag, country, substr(trans_date, 1, 7) month, amount
      from Transactions where state = 'approved'
      union all
      select 'chargeback' tag, t.country, substr(cb.trans_date, 1, 7) month, t.amount
      from Chargebacks cb join Transactions t on cb.trans_id = t.id
  )
  select month, country, 
         sum(if(tag = 'approved', 1, 0)) approved_count,
         sum(if(tag = 'approved', amount, 0)) approved_amount,
         sum(if(tag = 'chargeback', 1, 0)) chargeback_count,
         sum(if(tag = 'chargeback', amount, 0)) chargeback_amount
  from base 
  group by month, country;
  

• Idea: First label the data, and then union allperform grouping and aggregation;
• Focus on learning the idea of ​​​​this question. I remember asking for the maximum number of continuous online people in the live broadcast room. This is also the idea.

1211. Quality and proportion of query results

• source code:

  select 
      query_name, 
      round(avg(rating / position), 2) quality, 
      round(sum(case when rating < 3 then 1 else 0 end) / count(*) * 100, 2) poor_query_percentage 
  from queries
  group by query_name
  

• Idea: The requirements are relatively simple, and there is no logic that is particularly difficult to understand.

1212. Query team points

• source code:

  select team_id, team_name, ifnull(score, 0) as num_points 
  from teams 
  left join (
      select id, sum(score) score
      from (
          -- 作为主场
          select  
              host_team id, 
              sum(case when host_goals > guest_goals then 3 when host_goals = guest_goals then 1 else 0 end) score
          from matches
          group by id
          union all 
          -- 作为客场
          select  
              guest_team id, 
              sum(case when host_goals < guest_goals then 3 when host_goals = guest_goals then 1 else 0 end) score
          from matches
          group by id
      ) t
      group by id
  ) t
  on teams.team_id = t.id
  order by num_points desc, team_id;
  

• Idea: The core logic is left jointhe logic on the right. We first calculate the home score, then calculate the away score, and finally add these two parts together.

1225. Report system status on consecutive dates **

• source code:

  select type period_state , min(dt) start_date, max(dt) end_date
  from 
  (
      select  
          type, dt,
          subdate(dt, row_number() over(partition by type order by dt)) diff
      from (
          select 'succeeded' type, success_date dt
          from succeeded
          where success_date between '2019-01-01' and '2019-12-31'
          union 
          select 'failed' type, fail_date dt
          from failed
          where fail_date between '2019-01-01' and '2019-12-31'
      ) t
  ) t
  group by type, diff
  order by start_date;
  

• Idea: 类似于求最大连续登陆天数问题, the difference is that this question asks for the start and end time of continuous login! This question requires connecting two tables, and then you need to create a field yourself (a very good idea ), and then apply the maximum continuous login problem routine date_sub + row_numberto solve it.

1241. Number of comments per post

• source code:

  select sub_id post_id, ifnull(number_of_comments, 0) number_of_comments
  from (
      select distinct sub_id
      from submissions
      where parent_id is null
  ) t1
  left join (
      select parent_id, count(distinct sub_id) number_of_comments 
      from submissions
      where parent_id is not null
      group by parent_id
  ) t2
  on t1.sub_id = t2.parent_id
  order by post_id;
  

• Idea: Just make it clear whether each piece of data is a post or a comment; then pay attention to removing duplicates!

1251. Average selling price

• source code:

  select 
      product_id, 
      round(sum(price * units) / sum(units), 2) average_price
  from 
  (
      select p.product_id, price, units 
      from prices p 
      join unitssold u 
      on p.product_id = u.product_id and purchase_date between start_date and end_date
  ) t
  group by product_id;
  

• Idea: relatively simple

1264. Page recommendation

• source code:

  select distinct page_id as recommended_page
  from (
      select page_id 
      from likes
      where user_id in (
          select case when user1_id = 1 then user2_id else user1_id end 
          from friendship 
          where user1_id = 1 or user2_id = 1
      ) 
  ) t
  where page_id not in (
      select page_id
      from likes
      where user_id = 1
  )
  ;
  

• Idea: It’s relatively simple, you just need to pay attention to some of the details.

1270. Everyone who reports to the CEO of the company *

• source code:

  select 
     e3.employee_id
  from employees e1, employees e2, employees e3 
  where 
     -- 根据题目说的要找出所有经理为1的员工id
     e1.manager_id  = 1 and 
     -- 根据上面一个条件找出直接经理为1的员工id，然后让这些员工id作为第二张表的经理，来获取所有的员工id
     e2.manager_id = e1.employee_id and  
     -- 根绝上面一个条件找出直接或者间接一次得到的所有员工id，然后让这些员工id作为第三张表的经理，来获取所有的员工id
     e3.manager_id = e2.employee_id and 
     -- 排除自己的经理就是自己的员工id
     e3.employee_id != 1
  ;
  

• Idea: It is easy to get the answer to this question without writing a program, but it is more difficult to implement it through a program. I have explained the wheremeaning of each sentence in the above source code.

1280. Number of tests students took in each subject

• source code:

  select t1.student_id, student_name, t1.subject_name, count(ex.student_id) attended_exams
  from (
     select student_id, student_name, subject_name
     from students
     join subjects
     on 1 = 1
  ) t1
  left join examinations ex
  on t1.student_id = ex.student_id and t1.subject_name = ex.subject_name
  group by t1.student_id, t1.subject_name
  order by t1.student_id, t1.subject_name;
  

• Idea: It involves a Cartesian product connection , which can be usedcross join

1285. Find the starting and ending numbers of a continuous interval

• source code:

  select 
     min(log_id) start_id,
     max(log_id) end_id
  from (
     select log_id, (log_id - row_number() over(order by log_id)) diff
     from logs
  ) t
  group by diff
  order by start_id
  

• Idea: Exactly the same as the above question, find the maximum and minimum time for continuous login