Questions with SQL - separating a column in two - sql

Question

I don't know how I can answer this question. Because the name and last name are in one column. I'm not allowed to change the columns.
"Get the average spending (per customer) of all customers who share a last name with another customer"
I thought to say in sqlite3
SELECT avg_spending
FROM customer
JOIN customer on WHERE name is name;
This is how the table is defined:
CREATE TABLE customer
(
cuid INTEGER,
name STRING,
age INTEGER,
avg_spending REAL,
PRIMARY KEY(cuid)
);
So those values are having the same last name
INSERT INTO customer VALUES (4, "Henk Krom", 65, 24);
INSERT INTO customer VALUES (9, "Bob Krom", 66, 4);

Answer 1

From the sample data you posted I guess the format of the column name is:
FirstName LastName
so you need to extract the LastName and use group by to get the average:
select
substr(name, instr(name, ' ') + 1) lastname,
avg(avg_spending) avg_spending
from customer
group by lastname
having count(*) > 1
The having clause restricts the results to those customer names that have at least 1 other customer name with the same last name.
See the demo.
For the sample data:
> cuid | name | age | avg_spending
> :--- | :-------- | :-- | :-----------
> 4 | Henk Krom | 65 | 24
> 9 | Bob Krom | 66 | 4
> 5 | Jack Doe | 66 | 4
> 7 | Jill Doe | 66 | 6
> 1 | Alice No | 66 | 44
you get results:
> lastname | avg_spending
> :------- | :-----------
> Doe | 5
> Krom | 14

Answer 2

As mentioned in the comments, the crux of this is to find a rule how to reliably extract the surname from the name. Apart from that you merely need an exists clause, because you want to select customers where another customer with the same surname exists.
("Get the average spending (per customer)" simply means get a row from the table, because each row contains exactly one customer and their average spending.)
If all names were in the format first name - blank - last name, that would be:
select *
from customer c
where exists
(
select *
from customer other
where other.cuid <> c.cuid
and substr(other.name, instr(other.name, ' ') + 1) = substr(c.name, instr(c.name, ' ') + 1)
);

Answer 3

You were correct in joining the customer table to itself but you also need to parse out the last name to compare and remove duplicates once a match was found since if nameA equals nameB then nameB has to equal nameA.
with custs AS
(
select distinct
a.name as name_1 ,
b.name as name_2
from customer a
join customer b
on substr(a.name, instr(a.name, ' ') + 1) = substr(b.name, instr(b.name, ' ') + 1)
where a.name like '%Krom%' and a.name <> b.name
)
select * from customer where name in (select name_1 from custs)
union
select * from customer where name in (select name_2 from custs)