Lad os først lave nogle testdata:
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);
Hvis du ikke havde brug for portefølje-id'et, ville det være nemt:
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 |
+-----------+----------+--------------------+
Da du har brug for portefølje-id'et, bliver tingene mere komplicerede. Lad os gøre det i trin. Først vil vi skrive en underforespørgsel, der returnerer den maksimale porteføljeværdi for hver klient:
select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id
+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
| 1 | 33.33 |
| 2 | 90.90 |
+-----------+----------+
Derefter forespørger vi porteføljetabellen, men bruger en joinforbindelse til den forrige underforespørgsel for kun at beholde de porteføljer, hvis samlede værdi er den maksimale for klienten:
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 |
+--------------+---------------+
Endelig kan vi slutte os til klienttabellen (som du gjorde) for at inkludere navnet på hver klient:
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 |
+-----------+----------+--------------+---------------+
Bemærk, at dette returnerer to rækker for John Doe, fordi han har to porteføljer med nøjagtig samme samlede værdi. For at undgå dette og vælg en vilkårlig topportefølje, tag en GROUP BY-klausul:
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 |
+-----------+----------+--------------+---------------+