MySQL由两列组成的组

我试图在这里按多个列进行分组—每个表上有一列。
在这种情况下，我希望通过将客户的当前投资组合和现金相加，找到每个客户的最高投资组合价值，但客户可能有多个投资组合，因此我需要每个客户的最高投资组合

目前，使用下面的代码，我将为每个顶级投资组合多次获得相同的客户（不是按客户id分组）

---

首先，让我们做一些测试数据：

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);

如果您不需要公文包ID，那么很容易：

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 | 
+-----------+----------+--------------------+

因为您需要公文包ID，所以事情变得更加复杂。让我们分步做吧。首先，我们将编写一个子查询，返回每个客户的最大投资组合值：

select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id

+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
|         1 |    33.33 | 
|         2 |    90.90 | 
+-----------+----------+

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 | 
+-----------+----------+--------------+---------------+

然后我们将查询portfolio表，但使用前一个子查询的联接，以便只保留那些对客户端来说总价值最大的投资组合：

 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 | 
+--------------+---------------+

最后，我们可以连接到客户机表（正如您所做的那样），以便包含每个客户机的名称：

select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id

+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
|         1 |    33.33 | 
|         2 |    90.90 | 
+-----------+----------+

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 | 
+-----------+----------+--------------+---------------+

请注意，这将为John Doe返回两行，因为他有两个总价值完全相同的投资组合。要避免这种情况并选择任意的顶级投资组合，请标记GROUPBY子句：

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 | 
+-----------+----------+--------------+---------------+

---

在group by上使用Concat将起作用

SELECT clients.id, clients.name, portfolios.id, SUM ( portfolios.portfolio + portfolios.cash ) AS total
FROM clients, portfolios
WHERE clients.id = portfolios.client_id
GROUP BY CONCAT(portfolios.id, "-", clients.id)
ORDER BY total DESC
LIMIT 30

---

这是一个令人惊讶的答案，一旦我把maxtotal DESC的订单放在最后，它就完美了！；）我想说，谢谢你，聪明。对我来说也是很好的解决方案！我会在那里添加一个delimeter:CONCAT（portfolions.id，“-”，clients.id）。如果没有delimeter，可能会有相同的连接值和不同的值对。正如maksa所说，如果您碰巧遇到投资组合1属于客户机11（concat:'111'）而投资组合11属于客户机1（concat:'111'）的情况，则分隔符非常重要然后它们将被分组在一起。虽然

CONCAT

确实有很高的性能成本，但正如另一个答案所建议的那样，它可能比使用子查询性能更好……还有

CONCAT_WS（'-'，field，field）

到具有指定分隔符的CONCAT字段。