MySQL GROUP BY兩列

Question

我試圖在這里按多列分組 - 每個表一個。
在這種情況下，我想通過將他們當前的投資組合和現金加在一起來找到每個客戶的最高投資組合價值，但客戶可能有多個投資組合，因此我需要為每個客戶提供最高投資組合。

目前，通過下面的代碼，我為每個頂級投資組合多次獲得相同的客戶（它不是按客戶ID分組）。

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 

Answer 1

首先，讓我們制作一些測試數據：

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);

如果您不需要投資組合ID，那將很容易：

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

由於您需要投資組合ID，因此事情變得更加復雜。 讓我們分步進行。 首先，我們將編寫一個子查詢，返回每個客戶端的最大組合值：

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

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

然后我們將查詢投資組合表，但使用一個連接到前一個子查詢，以便只保留那些組合的總值是客戶端的最大值：

 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，我們可以加入客戶端表（就像你一樣），以便包含每個客戶端的名稱：

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

請注意，這會為John Doe返回兩行，因為他有兩個具有完全相同總值的投資組合。 要避免這種情況並選擇任意頂級投資組合，請在GROUP BY子句上進行標記：

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

Answer 2

在組上使用Concat將起作用

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