I'm trying to group by multiple columns here - one on each table.
It's a scenario where I want to find the top portfolio value for each client by adding their current portfolio and cash together but a client may have more than one portfolio, so I need the top portfolio for each client.
At the moment, with the code below I'm getting the same clients multiple times for each of their top portfolios (it's not grouping by client id).
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
Yes, it is possible to use MySQL GROUP BY clause with multiple columns just as we can use MySQL DISTINCT clause. Consider the following example in which we have used DISTINCT clause in first query and GROUP BY clause in the second query, on 'fname' and 'Lname' columns of the table named 'testing'.
A GROUP BY clause can contain two or more columns—or, in other words, a grouping can consist of two or more columns.
The MySQL GROUP BY Statement The GROUP BY statement groups rows that have the same values into summary rows, like "find the number of customers in each country". The GROUP BY statement is often used with aggregate functions ( COUNT() , MAX() , MIN() , SUM() , AVG() ) to group the result-set by one or more columns.
A shortcut way to group rows or columns is to highlight the rows/columns you wish to group and use ALT+SHIFT+RIGHT ARROW to group the rows/columns, and ALT+SHIFT+LEFT ARROW to ungroup them. You can go multiple levels as well (so you could group rows 1-30, and then group rows 20-25 as a subgroup of the first).
First, let's make some test data:
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);
If you didn't need the portfolio ID, it would be easy:
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 | +-----------+----------+--------------------+
Since you need the portfolio ID, things get more complicated. Let's do it in steps. First, we'll write a subquery that returns the maximal portfolio value for each client:
select client_id, max(cash + stocks) as maxtotal from portfolio group by client_id +-----------+----------+ | client_id | maxtotal | +-----------+----------+ | 1 | 33.33 | | 2 | 90.90 | +-----------+----------+
Then we'll query the portfolio table, but use a join to the previous subquery in order to keep only those portfolios the total value of which is the maximal for the client:
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 | +--------------+---------------+
Finally, we can join to the client table (as you did) in order to include the name of each client:
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 | +-----------+----------+--------------+---------------+
Note that this returns two rows for John Doe because he has two portfolios with the exact same total value. To avoid this and pick an arbitrary top portfolio, tag on a GROUP BY clause:
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 | +-----------+----------+--------------+---------------+
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