Redshift generate rows as many as value in another column - amazon-web-services

Question

df
customer_code contract_code product num_products
C0134 AB01245 toy_1 4
B8328 EF28421 doll_4 2
I would like to transform this table based on the integer value in column num_products and generate a unique id for each row:
Expected_df
unique_id customer_code contract_code product num_products
A1 C0134 AB01245 toy_1 1
A2 C0134 AB01245 toy_1 1
A3 C0134 AB01245 toy_1 1
A4 C0134 AB01245 toy_1 1
A5 B8328 EF28421 doll_4 1
A6 B8328 EF28421 doll_4 1
unique_id can be any random characters as long as I can use a count(distinct) on it later on.
I read that generate_series(1,10000) i is available in later versions of Postgres but not in Redshift

Answer 1

You need to use a recursive CTE to generate the series of number. Then join this with you data to produce the extra rows. I used row_number() to get the unique_id in the example below.
This should meet you needs or at least give you a start:
create table df (customer_code varchar(16),
contract_code varchar(16),
product varchar(16),
num_products int);
insert into df values
('C0134', 'AB01245', 'toy_1', 4),
('B8328', 'EF28421', 'doll_4', 2);
with recursive nums (n) as
( select 1 as n
union all
select n+1 as n
from nums
where n < (select max(num_products) from df) )
select row_number() over() as unique_id, customer_code, contract_code, product, num_products
from df d
left join nums n
on d.num_products >= n.n;
SQLfiddle at http://sqlfiddle.com/#!15/d829b/12