The Query:
SELECT purchases.*
FROM purchases
LEFT JOIN purchases as p
ON
p.customer = purchases.customer
AND
purchases.total < p.total
WHERE p.total IS NULL
HOW DOES THAT WORK! (I've been there)
We want to make sure that we only have the highest total for each purchase.
Some Theoretical Stuff (skip this part if you only want to understand the query)
Let Total be a function T(customer,id) where it returns a value given the name and id To prove that the given total (T(customer,id)) is the highest we have to prove that We want to prove either
OR
The first approach will need us to get all the records for that name which I do not really like.
The second one will need a smart way to say there can be no record higher than this one.
Back to SQL
If we left joins the table on the name and total being less than the joined table:
LEFT JOIN purchases as p
ON
p.customer = purchases.customer
AND
purchases.total < p.total
we make sure that all records that have another record with the higher total for the same user to be joined:
+--------------+---------------------+-----------------+------+------------+---------+
| purchases.id | purchases.customer | purchases.total | p.id | p.customer | p.total |
+--------------+---------------------+-----------------+------+------------+---------+
| 1 | Tom | 200 | 2 | Tom | 300 |
| 2 | Tom | 300 | | | |
| 3 | Bob | 400 | 4 | Bob | 500 |
| 4 | Bob | 500 | | | |
| 5 | Alice | 600 | 6 | Alice | 700 |
| 6 | Alice | 700 | | | |
+--------------+---------------------+-----------------+------+------------+---------+
That will help us filter for the highest total for each purchase with no grouping needed:
WHERE p.total IS NULL
+--------------+----------------+-----------------+------+--------+---------+
| purchases.id | purchases.name | purchases.total | p.id | p.name | p.total |
+--------------+----------------+-----------------+------+--------+---------+
| 2 | Tom | 300 | | | |
| 4 | Bob | 500 | | | |
| 6 | Alice | 700 | | | |
+--------------+----------------+-----------------+------+--------+---------+
And that's the answer we need.