BigQuery Nested Challenge Involving Joins and Having (or Where) Clauses - sql

Question

I have been given a challenge that is a bit out of my scope, so I'm just going to jump right in.
I have a sample dataset in BigQuery you can find here for testing purposes: https://bigquery.cloud.google.com/table/robotic-charmer-726:bl_test_data.complex_problem
I need to figure out the SQL code to query my table and do the following:
By aggregating using the following rules (I'll start with just one email address, and add the other in at the end):
As a general note up front, everything is to be made lowercase such that Ben=ben when aggregating.
Email is the broadest aggregation, and is aggregated by the lowercase version.
The amounts for all of those lowercase emails are summed, as is pictured below in blue.
First and last names are considered next, and they are selected based on the sum amount of the lowercase of the first AND last name.
Note, first or last names are NOT considered separately. See below where Ben has a sum amount of 160 and Kathleen only has a sum amount of 150, but Kathleen is still selected because her full name has a sum amount higher than any other full name.
Next the lowercase full address of the SELECTED NAME is chosen based on the highest sum amount.
Similar to the names, the full address considers all columns together.
Now I'll add in another email address, and we'll do the same thing.
Each lowercase email address is considered separately. I'm now realizing that I should have made that more clear with my pictures, but I don't want to do it all again... too much work. So I hope I have made it clear enough.
I hope you find this to be a very fun challenge!

Answer 1

There are probably cleaner ways of doing this, but this will give you the answer you need:
select email, first_name, last_name, address, city, state, zip, total_amount amount
from (
select d.email email, d.first_name first_name, d.last_name last_name, d.amount amount, d.total_amount total_amount, e.address address, e.city city, e.state state, e.zip zip, row_number() over (partition by e.email order by e.amount desc) ord
from (
select a.email email, a.first_name first_name, a.last_name last_name, b.amount amount, c.amount total_amount
from (
SELECT
lower(email) email, lower(first_name) first_name, lower(last_name) last_name, lower(concat(first_name, last_name)) as name_group, lower(address) address, lower(city) city, lower(state) state, lower(concat(address,city,state)) as location_group, zip, sum(amount) amount
FROM [robotic-charmer-726:bl_test_data.complex_problem]
group by 1,2,3,4,5,6,7,8,9
) a
inner join (
select email, first_name, last_name, name_group, amount
from (
select email, first_name, last_name, name_group, amount, row_number() over (partition by email order by amount desc) as ord
from (
select lower(email) email , lower(first_name) first_name, lower(last_name) last_name, lower(concat(first_name,last_name)) as name_group, sum(amount) amount,
from [robotic-charmer-726:bl_test_data.complex_problem]
group by 1, 2, 3, 4
)
)
where ord = 1
) b
on a.name_group = b.name_group
inner join (
select lower(email) email, sum(amount) amount
from [robotic-charmer-726:bl_test_data.complex_problem]
group by 1
) c
on a.email = c.email
group by 1,2,3,4,5
) d
inner join (
select lower(email) email, lower(first_name) first_name, lower(last_name) last_name, lower(address) address, lower(city) city, lower(state) state, zip,lower(concat(lower(address),lower(city), lower(state), zip)) as location_group, sum(amount) amount
from [robotic-charmer-726:bl_test_data.complex_problem]
group by 1,2,3,4,5,6,7,8
) e
on d.email = e.email and d.first_name = e.first_name and d.last_name = e.last_name
)
where ord = 1