MySQL按两列分组

Question

我在这里尝试按多个列分组 - 每个表格一个。
我的需求是找到每个客户的最高投资组合价值，通过将其当前投资组合和现金相加，但一个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，对于每个顶级投资组合，我会多次获得相同的客户（它没有按客户ID分组）。

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 portfolios.id, clients.id
ORDER BY total DESC
LIMIT 30 

- ChrisS

Answer 1

首先，让我们创建一些测试数据：

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);

如果您不需要投资组合ID，那么这将非常简单：

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

由于您需要投资组合ID，事情会变得更加复杂。我们将分步骤进行。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价值：

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

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

然后我们将查询投资组合表，但使用连接到前一个子查询，以便仅保留客户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，我们可以像你所做的那样加入客户表，以便包括每个客户的姓名：

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，这里返回了约翰·杜的两行数据，因为他有两个投资组合具有完全相同的总价值。为避免这种情况并选择任意一个顶级投资组合，请添加一个GROUP BY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 | 
+-----------+----------+--------------+---------------+

- Vebjorn Ljosa

---

这是一个惊人的答案，一旦我在结尾处加上ORDER BY maxtotal DESC，它就完美了！;) 谢谢 - ChrisS

Answer 2

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

- Jose' Vargas

---

好的解决方案！我会在这里添加一个分隔符：CONCAT(portfolios.id, "-", clients.id)。如果没有分隔符，可能会出现相同的连接值与不同的数值对。 - TCFDS

正如maksa所说，分隔符非常重要。否则，如果你碰巧遇到这样一种情况：投资组合1属于客户11（连接：'111'），而投资组合11属于客户1（连接：也是'111'），那么它们将被分在一起。 - DaveMongoose

虽然CONCAT函数的性能成本很高，但它可能比使用子查询更有效，正如其他答案所建议的那样... - rodrigo-silveira

还有CONCAT_WS('-', field, field)可以用指定的分隔符连接字段。 - Jonathan Gruber

这是一个巧妙的解决方案！ - Aris