Primero, hagamos algunos datos de prueba:
create table client (client_id integer not null primary key auto_increment,
name varchar(64));
create table portfolio (portfolio_id integer not null primary key auto_increment,
client_id integer references client.id,
cash decimal(10,2),
stocks decimal(10,2));
insert into client (name) values ('John Doe'), ('Jane Doe');
insert into portfolio (client_id, cash, stocks) values (1, 11.11, 22.22),
(1, 10.11, 23.22),
(2, 30.30, 40.40),
(2, 40.40, 50.50);
Si no necesitara la identificación de la cartera, sería fácil:
select client_id, name, max(cash + stocks)
from client join portfolio using (client_id)
group by client_id
+-----------+----------+--------------------+
| client_id | name | max(cash + stocks) |
+-----------+----------+--------------------+
| 1 | John Doe | 33.33 |
| 2 | Jane Doe | 90.90 |
+-----------+----------+--------------------+
Dado que necesita la identificación de la cartera, las cosas se complican más. Hagámoslo por pasos. Primero, escribiremos una subconsulta que devuelva el valor máximo de la cartera para cada cliente:
select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id
+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
| 1 | 33.33 |
| 2 | 90.90 |
+-----------+----------+
Luego, consultaremos la tabla de carteras, pero usaremos una combinación con la subconsulta anterior para mantener solo aquellas carteras cuyo valor total sea el máximo para el cliente:
select portfolio_id, cash + stocks from portfolio
join (select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal
+--------------+---------------+
| portfolio_id | cash + stocks |
+--------------+---------------+
| 5 | 33.33 |
| 6 | 33.33 |
| 8 | 90.90 |
+--------------+---------------+
Finalmente, podemos unirnos a la tabla de clientes (como hiciste tú) para incluir el nombre de cada cliente:
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal
+-----------+----------+--------------+---------------+
| client_id | name | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
| 1 | John Doe | 5 | 33.33 |
| 1 | John Doe | 6 | 33.33 |
| 2 | Jane Doe | 8 | 90.90 |
+-----------+----------+--------------+---------------+
Tenga en cuenta que esto devuelve dos filas para John Doe porque tiene dos carteras con exactamente el mismo valor total. Para evitar esto y elegir una cartera superior arbitraria, etiquete una cláusula GROUP BY:
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal
group by client_id, cash + stocks
+-----------+----------+--------------+---------------+
| client_id | name | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
| 1 | John Doe | 5 | 33.33 |
| 2 | Jane Doe | 8 | 90.90 |
+-----------+----------+--------------+---------------+
