adding �sql problem

Author: shreyashji

34.Practice SQLs Without Creating Tables In Your Database.sql

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+#Practice SQLs Without Creating Tables In Your Database 
+with emp as 
+(
+select 1 as emp_id,1000 as emp_salary,1 as dep_id
+union all select 2 as emp_id,2000 as emp_salary,2 as dep_id
+union all select 3 as emp_id,3000 as emp_salary,3 as dep_id
+union all select 4 as emp_id,4000 as emp_salary,4 as dep_id
+),
+dep as 
+(
+select 1 as dep_id,'d1' as dep_name
+union all select 2 as dep_id,'d2' as dep_name
+union all select 3 as dep_id,'d3' as dep_name
+union all select 4 as dep_id,'d4' as dep_name
+)
+select * from emp

35.Data Science SQL Interview Question Recommendation System.sql

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+#create table orderss (order_id int,customer_id int,product_id int);
+#insert into orderss VALUES (1, 1, 1),(1, 1, 2),(1, 1, 3),(2, 2, 1),(2, 2, 2),(2, 2, 4),(3, 1, 5)
+#create table products (id int,name varchar(10));
+#insert into products VALUES (1, 'A'),(2, 'B'),(3, 'C'),(4, 'D'),(5, 'E');
+
+#Product recommendation. Just the basic type (“customers who bought this also bought…”). 
+#That, in its simplest form, is an outcome of basket analysis. In this video we will learn how to find products which are most 
+#frequently bought together using simple SQL. Based on the history ecommerce website can recommend products to new user.
+
+select o1.order_id,o1.product_id as p1,o2.product_id as p2 from orderss o1
+inner join orderss o2 on o1.order_id=o2.order_id
+where o1.order_id=1 and o1.product_id !=o2.product_id  and o1.product_id != o2.product_id ;
+
+select o1.product_id as p1,o2.product_id as p2 ,count(1) as purchase_frequency from orderss o1
+inner join orderss o2 on o1.order_id=o2.order_id
+where o1.product_id < o2.product_id 
+group by  o1.product_id ,o2.product_id ;
+#join with product table
+select pr1.name as p1,pr2.name as p2,count(1) as purchase_frequency from orderss o1
+inner join orderss o2 on o1.order_id=o2.order_id
+inner join products pr1 on pr1.id=o1.product_id
+inner join products pr2 on pr2.id=o2.product_id
+where o1.product_id < o2.product_id 
+group by  pr1.name ,pr2.name;
+#concatenate 
+select pr1.name +' '+ pr2.name as pair,count(1) as purchase_frequency from orderss o1
+inner join orderss o2 on o1.order_id=o2.order_id
+inner join products pr1 on pr1.id=o1.product_id
+inner join products pr2 on pr2.id=o2.product_id
+where o1.product_id < o2.product_id 
+group by pr1.name ,pr2.name;
+
+select * from  products;