How to solve the problem of store ranking?

【 Interview questions 】

There are two tables ,“ Product list ” contain 2 A field ： product 、 product line .

“ The order sheet ” contain 7 A field ： Order id、 customer id、 Transaction date 、 product 、 The store 、 number 、 amount of money .

problem ： Query product line （category_3） The store with the second largest sales volume .

【 The problem solving steps 】

1. Their thinking

The problem is to analyze the sales of a product line . Can calculate “ sales ” Field of “ number ” stay “ The order sheet ” in ,“ product line ” stay “ Product list ” in .

therefore , It involves two tables , Think of it 《 Monkeys learn from scratch SQL》 It's used in “ Multi-table query ”

In order to retain “ The order sheet ” All the data in , With “ The order sheet ” It is the left table （ Record as sub query a1）, Make left connection .

“ Product list ” It's the right table （ Record as sub query a2）.

The association condition is field “ product ”.

select  Field 
from 
a1
left join 
a2 
on a1. product  = a2. product 

Add subquery a1、a2 Plug in ：

select a1.*,a2. product line 
from 
 The order sheet  as a1
left join 
 Product list  as a2 
on a1. product  = a2. product ;

Query results ：

Will be more than SQL Record as sub query a3.

3. Summary analysis

The problem is to check the product line （category_3） The store with the second largest sales volume , therefore ：

1） Just focus on the product line category_3 The order of ;

2） Summarize by store , To get the product line （category_3） Sales volume in each store ;

3） Sales of “ number ” Sum up , Because each order is not necessarily only 1 Products .

select a3. The store ,
       sum( number ) as category_3 sales 
from a3
where a3. product line  = 'category_3'
group by a3. The store ;

Add subquery a3 Plug in ：

select a3. The store ,
       sum( number ) as category_3 sales 
from (
select a1.*,a2. product line 
from 
 The order sheet  as a1
left join 
 Product list  as a2 
on a1. product  = a2. product 
) as a3
where a3. product line  = 'category_3'
group by a3. The store ;

Will be more than SQL Record as sub query a4.

4. Window function

Query ranking generally uses window functions , There are three common window functions about ranking ：

Prevent the situation of parallel ranking , We choose dense_rank().

select *,dense_rank() over(order by category_3 sales  desc) as  ranking 
from a4;

Add subquery a4 Plug in ：

select *,dense_rank() over(order by a4.category_3 sales  desc) as  ranking 
from (
select a3. The store ,
       sum( number ) as category_3 sales 
from (
select a1.*,a2. product line 
from 
 The order sheet  as a1
left join 
 Product list  as a2 
on a1. product  = a2. product 
) as a3
where a3. product line  = 'category_3'
group by a3. The store 
) as a4;

Will be more than SQL Record as sub query a5.

Last , The selected ranking is 2 Our shop ：

select  The store 
from a5
where  ranking  = 2;

Add subquery a5 Plug in ：

select a5. The store 
from (
select *,dense_rank() over(order by a4.category_3 sales  desc) as  ranking 
from (
select a3. The store ,
       sum( number ) as category_3 sales 
from (
select a1.*,a2. product line 
from 
 The order sheet  as a1
left join 
 Product list  as a2 
on a1. product  = a2. product 
) as a3
where a3. product line  = 'category_3'
group by a3. The store 
) as a4
) as a5
where a5. ranking  = 2;

【 The test point of this question 】

1） Check your understanding of group summary , And flexible use to solve business problems ;

2） Check your understanding of multi table connections , And flexible use to solve business problems ;

3） Check your understanding of window functions .

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