Have you learned the common SQL interview questions on the short video platform?

【 subject 】

There are three tables in the database of a short video company , List of user video information 、 Anchor start schedule 、 Detailed list of user information in the live broadcasting room .

The list of user video information includes users id, Release video id, Video likes , Video release date , Video category and number of fans .

The anchor start schedule contains the anchor id, The host opens the live broadcast room id Number , And the time of broadcasting .

The user information list of the live broadcasting room includes the audience who enter the live broadcasting room to watch id, Enter the live studio id And the time the audience entered .

Business needs ：

1. Find the video with the highest number of likes per user , When the number of likes is the same, follow the video id The biggest record .

2. Find out the number of fans in 6 month 2 No. 1 has been promoted the most 20 Users id ( contrast 6 month 1 Number ).

3. Find out the number of the live broadcast room that no one has entered within three minutes of broadcasting .

【 Their thinking 】

1. Find the video with the highest number of likes per user , When the number of likes is the same, follow the video id The biggest record .

Let's first translate this business requirement into Chinese ：

1） The fields required for query results are users id、 video id、 Number of likes

2） According to the user id The group is then sorted according to the number of likes of each user's video , If the likes are the same, follow the video id Sort

3） Select the video with the highest number of likes per user

Each video uploaded by each user is required to be displayed , We know group by The number of rows in the table is changed after grouping , There is only one category in a row . Using the window function does not reduce the number of rows in the original table .

By user id grouping (partiotion by user id)、 And click the most like number 、 video id Descending order (order by Number of likes , video id ), Descending order desc Use the syntax of nested window functions , Get the following SQL sentence ：

select
 user id , video id , Number of likes  ,
row_number()over(partition by  user id order by  Number of likes  desc,  video id desc) as  ranking 
from  List of user video information ;

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Query results ：

Rank the bottom according to the video likes of each user , We ranked first by screening , That is, the video with the most likes .SQL It is written as follows ：

select  user id , video id , Number of likes 
from
     (select  user id , video id , Number of likes  ,row_number()over(partition by  user id
order by  Number of likes  desc,  video id desc) as  ranking 
     from  List of user video information  )t
where  ranking =1;

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Query results ：

2. Find out the number of fans in 6 month 2 No. 1 has been promoted the most 3 Users id ( contrast 6 month 1 Number ).

Let's split this business requirement first ：

1） Get that each user is 6 month 2 The number of rising fans of No

2） Before finding out 3 Users with the most fans id

1） Get that each user is 6 month 2 The number of rising fans of No

Take a look at the list of user video information , To find out 6 month 2 No. 1 is the fastest growing user , Will know 6 month 1 What is the number of fans of user No , hold 6 month 2 The number of fans on the number minus 6 month 1 The number of fans per user can be calculated from the number of fans per user .

We will limit the release time to 6 month 1 Number -2 Between , Use if Functions and sum Function to calculate the number of rising powder , If the release date is 6 month 2 Number , Show fields ” Accumulated fans of users ”, If not, it shows ”（ negative ）- Accumulated fans of users ”, Finally, sum to get the number of rising powder .

SQL It is written as follows ：

select  user id ,sum(if ( Release date  ="2022/6/2", Accumulated fans of users ,- Accumulated fans of users )) as " Rising powder number "
from  List of user video information 
where  Release date  in ("2022/6/2","2022/6/1")
group by  user id;

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Query results ：

2） Before finding out 3 Users with the most fans id

Get that each user is 6 month 2 As a temporary table t, use order by Sort the number of users in descending order （desc） after , use limit 3 Get the top three users with the most fans id.

SQL It is written as follows ：

select  user id, Rising powder number 
from
     (select  user id ,sum(if ( Release date  ="2022/6/2", Accumulated fans of users ,- Accumulated fans of users )) as " Rising powder number "
      from  List of user video information 
      where  Release date  in ("2022/6/2","2022/6/1")
      group by  user id )t
order by  Rising powder number  desc
limit 3;

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Query results ：

3. Find out the number of the live broadcast room that no one has entered within three minutes of broadcasting .

Observe the anchor start schedule and the live room user information schedule , We can know the starting time of each anchor and the time when the audience enters the live room . Use the studio id Connect the two tables to get the user information of the live broadcast room .

SQL It is written as follows ：

select a. The host id, a. studio id, The audience id,a. Start time ,b. Entry time 
from  Anchor start schedule  a
left join  Detailed list of user information in the live broadcasting room  b 
on a. studio id =b. Enter the live studio id;

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Query results ：

It is obvious from the query results that R004 There is no audience in this studio , We can use the audience id Whether it is empty is used to judge whether there is an audience in the live broadcasting room （ The audience id is null）,R005 The audience of this live studio will enter the live studio three minutes later .

Business requirements: we find out the live broadcast that has no audience within three minutes after the anchor starts broadcasting id, use date_add Function to calculate the start time and the audience's entry time to calculate the time difference .date_add Function usage is as follows ：

SQL It is written as follows ：

b. Entry time  > date_add(a. Start time ,interval +3 minute)

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Substitute into the whole SQL in

select a. The host id, a. studio id, The audience id,a. Start time ,b. Entry time 
from  Anchor start schedule  a
left join  Detailed list of user information in the live broadcasting room  b 
on a. studio id =b. Enter the live studio id  
and b. Entry time  > date_add(a. Start time ,interval +3 minute) 
where b. The audience id is null;

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Query results ：

【 The test point of this question 】

1. Be familiar with the usage of window functions , Most of them are used to sort similar business requirements for each category under each user .

2. Encounter complex business requirements , Try to disassemble the multi-dimensional analysis method into several simple problems .

3. Apply to multi table information , First, think about multi table join , Then the connection type is obtained according to the specific business scenario .