MySQL GROUP BY两列
我试图在这里按多列进行分组 – 每个桌上有一列。
在这种情况下，我想通过将当前的投资组合和现金加在一起来为每个客户find最高的投资组合价值，但客户可能拥有多个投资组合，所以我需要为每个客户提供最高的投资组合。
目前，在下面的代码中，我为每个顶级投资组合多次获取相同的客户（这不是按客户ID进行分组）。
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 首先，我们来做一些testing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，那很简单：
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 | +-----------+----------+--------------------+
既然你需要组合ID，事情变得更加复杂。 让我们一步一步来做。 首先，我们将编写一个子查询，返回每个客户端的最大投资组合值：
select client_id, max(cash + stocks) as maxtotal from portfolio group by client_id +-----------+----------+ | client_id | maxtotal | +-----------+----------+ | 1 | 33.33 | | 2 | 90.90 | +-----------+----------+
然后，我们将查询投资组合表，但是对前一个子查询使用连接，以便只保留客户端的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join到客户端表（如您所做的那样），以包含每个客户端的名称：
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，因为他有两个总计值完全相同的投资组合。 为了避免这种情况，并select一个任意的顶级投资组合，在GROUP BY子句上标记：
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 | +-----------+----------+--------------+---------------+
在群组上使用Concat将会起作用
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
