SQL join: selecting the last records in a one-to-m

问题:

Suppose I have a table of customers and a table of purchases. Each purchase belongs to one customer. I want to get a list of all customers along with their last purchase in one SELECT statement. What is the best practice? Any advice on building indexes?

Please use these table/column names in your answer:

• customer: id, name
• purchase: id, customer_id, item_id, date

And in more complicated situations, would it be (performance-wise) beneficial to denormalize the database by putting the last purchase into the customer table?

If the (purchase) id is guaranteed to be sorted by date, can the statements be simplified by using something like LIMIT 1?

回答1:

This is an example of the greatest-n-per-group problem that has appeared regularly on StackOverflow.

Here\'s how I usually recommend solving it:

SELECT c.*, p1.*
FROM customer c
JOIN purchase p1 ON (c.id = p1.customer_id)
LEFT OUTER JOIN purchase p2 ON (c.id = p2.customer_id AND 
    (p1.date < p2.date OR p1.date = p2.date AND p1.id < p2.id))
WHERE p2.id IS NULL;

Explanation: given a row p1, there should be no row p2 with the same customer and a later date (or in the case of ties, a later id). When we find that to be true, then p1 is the most recent purchase for that customer.

Regarding indexes, I\'d create a compound index in purchase over the columns (customer_id, date, id). That may allow the outer join to be done using a covering index. Be sure to test on your platform, because optimization is implementation-dependent. Use the features of your RDBMS to analyze the optimization plan. E.g. EXPLAIN on MySQL.

---

Some people use subqueries instead of the solution I show above, but I find my solution makes it easier to resolve ties.

回答2:

You could also try doing this using a sub select

SELECT  c.*, p.*
FROM    customer c INNER JOIN
        (
            SELECT  customer_id,
                    MAX(date) MaxDate
            FROM    purchase
            GROUP BY customer_id
        ) MaxDates ON c.id = MaxDates.customer_id INNER JOIN
        purchase p ON   MaxDates.customer_id = p.customer_id
                    AND MaxDates.MaxDate = p.date

The select should join on all customers and their Last purchase date.

回答3:

You haven\'t specified the database. If it is one that allows analytical functions it may be faster to use this approach than the GROUP BY one(definitely faster in Oracle, most likely faster in the late SQL Server editions, don\'t know about others).

Syntax in SQL Server would be:

SELECT c.*, p.*
FROM customer c INNER JOIN 
     (SELECT RANK() OVER (PARTITION BY customer_id ORDER BY date DESC) r, *
             FROM purchase) p
ON (c.id = p.customer_id)
WHERE p.r = 1

回答4:

Another approach would be to use a NOT EXISTS condition in your join condition to test for later purchases:

SELECT *
FROM customer c
LEFT JOIN purchase p ON (
       c.id = p.customer_id
   AND NOT EXISTS (
     SELECT 1 FROM purchase p1
     WHERE p1.customer_id = c.id
     AND p1.id > p.id
   )
)

回答5:

I found this thread as a solution to my problem.

But when I tried them the performance was low. Bellow is my suggestion for better performance.

With MaxDates as (
SELECT  customer_id,
                MAX(date) MaxDate
        FROM    purchase
        GROUP BY customer_id
)

SELECT  c.*, M.*
FROM    customer c INNER JOIN
        MaxDates as M ON c.id = M.customer_id 

Hope this will be helpful.

回答6:

Try this, It will help.

I have used this in my project.

SELECT 
*
from
customer c
OUTER APPLY(SELECT top 1 * FROM purchase pi 
WHERE pi.Id = p.Id order by pi.Id desc) AS [LastPurchasePrice]

回答7:

Tested on SQLite:

SELECT c.*, p.*, max(p.date)
FROM customer c
LEFT OUTER JOIN purchase p
ON c.id = p.customer_id
GROUP BY c.id

The max() aggregate function will make sure that the latest purchase is selected from each group (but assumes that the date column is in a format whereby max() gives the latest - which is normally the case). If you want to handle purchases with the same date then you can use max(p.date, p.id).

In terms of indexes, I would use an index on purchase with (customer_id, date, [any other purchase columns you want to return in your select]).

The LEFT OUTER JOIN (as opposed to INNER JOIN) will make sure that customers that have never made a purchase are also included.

回答8:

Please try this,

SELECT 
c.Id,
c.name,
(SELECT pi.price FROM purchase pi WHERE pi.Id = MAX(p.Id)) AS [LastPurchasePrice]
FROM customer c INNER JOIN purchase p 
ON c.Id = p.customerId 
GROUP BY c.Id,c.name;

回答9:

If you\'re using PostgreSQL you can use DISTINCT ON to find the first row in a group.

SELECT customer.*, purchase.*
FROM customer
JOIN (
   SELECT DISTINCT ON (customer_id) *
   FROM purchase
   ORDER BY customer_id, date DESC
) purchase ON purchase.customer_id = customer.id

PostgreSQL Docs - Distinct On

Note that the DISTINCT ON field(s) -- here customer_id -- must match the left most field(s) in the ORDER BY clause.

Caveat: This is a nonstandard clause.