RE: Within-group aggregate query help please - customers and latest subscription row
2011/10/24 16:31 -0700, Daevid Vincent WHERE cs.customer_id = 7 GROUP BY customer_id Well, the latter line is now redundant. How will you make the '7' into a parameter? -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
RE: Within-group aggregate query help please - customers and latest subscription row
A kind (and shy) soul replied to me off list and suggested this solution, however, this takes 28 seconds (that's for a single customer_id, so this is not going to scale). Got any other suggestions? :-) SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs GROUP BY customer_id) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate WHERE c.customer_id = 7; There are 781,270 customers (nearly 1 million) and 1,018,092 customer_subscriptions. Our tables have many indexes on pretty much every column and for sure the ones we use here. EXPLAIN says: id select_type table typepossible_keys key key_len refrows Extra -- --- -- -- --- --- -- --- --- 1 PRIMARY c const PRIMARY PRIMARY 4 const 1 1 PRIMARY s ref date,customer_id customer_id 4 const 2 1 PRIMARY derived2 ALL (NULL)(NULL) (NULL) (NULL) 781265 Using where 2 DERIVED cs ALL (NULL)(NULL) (NULL) (NULL) 1018092 Using temporary; Using filesort -Original Message- From: Daevid Vincent [mailto:dae...@daevid.com] Sent: Monday, October 24, 2011 1:46 PM To: mysql@lists.mysql.com Subject: Within-group aggregate query help please - customers and latest subscription row I know this is a common problem, and I've been struggling with it for a full day now but I can't get it. I also tried a few sites for examples: http://www.artfulsoftware.com/infotree/queries.php#101 http://forums.devarticles.com/general-sql-development-47/select-max-datetime -problem-10210.html Anyways, pretty standard situation: CREATE TABLE `customers` ( `customer_id` int(10) unsigned NOT NULL auto_increment, `email` varchar(64) NOT NULL default '', `name` varchar(128) NOT NULL default '', `username` varchar(32) NOT NULL, ... ); CREATE TABLE `customers_subscriptions` ( `subscription_id` bigint(12) unsigned NOT NULL default '0', `customer_id` int(10) unsigned NOT NULL default '0', `date` date NOT NULL default '-00-00', ... ); I want to show a table where I list out the ID, email, username, and LAST SUBSCRIPTION. I need this data in TWO ways: The FIRST way, is with a query JOINing the two tables so that I can easily display that HTML table mentioned. That is ALL customers and the latest subscription they have. The SECOND way is when I drill into the customer, I already know the customer_id and so don't need to JOIN with that table, I just want to get the proper row from the customers_subscriptions table itself. SELECT * FROM `customers_subscriptions` WHERE customer_id = 7 ORDER BY `date` DESC; subscription_id processor customer_id date --- - --- -- 134126370 chargem 7 2005-08-04 1035167192 billme 7 2004-02-08 SELECT MAX(`date`) FROM `customers_subscriptions` WHERE customer_id = 7 GROUP BY customer_id; gives me 2005-08-04 obviously, but as you all know, mySQL completely takes a crap on your face when you try what would seem to be the right query: SELECT subscription_id, MAX(`date`) FROM `customers_subscriptions` WHERE customer_id = 7 GROUP BY customer_id; subscription_id MAX(`date`) --- --- 1035167192 2005-08-04 Notice how I have the correct DATE, but the wrong subscription_id. In the example web sites above, they seem to deal more with finding the MAX(subscription_id), which in my case will not work. I need the max DATE and the corresponding row (with matching subscription_id). Thanks, d -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
RE: Within-group aggregate query help please - customers and latest subscription row
Okay, it seems I am learning... slowly... So there needs to be a second WHERE in the sub-select... To get ONE customer's last subscription (0.038s): SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs WHERE cs.customer_id = 7 GROUP BY customer_id ) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate WHERE c.customer_id = 7; To get ALL customers and their last subscription row (1m:28s) SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs GROUP BY customer_id ) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate ORDER BY customer_id LIMIT 10; Thanks to you know who you are for pointing me in the right direction. Hopefully this helps someone else. d. -Original Message- From: Daevid Vincent [mailto:dae...@daevid.com] Sent: Monday, October 24, 2011 4:06 PM To: mysql@lists.mysql.com Subject: RE: Within-group aggregate query help please - customers and latest subscription row A kind (and shy) soul replied to me off list and suggested this solution, however, this takes 28 seconds (that's for a single customer_id, so this is not going to scale). Got any other suggestions? :-) SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs GROUP BY customer_id) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate WHERE c.customer_id = 7; There are 781,270 customers (nearly 1 million) and 1,018,092 customer_subscriptions. Our tables have many indexes on pretty much every column and for sure the ones we use here. EXPLAIN says: id select_type table typepossible_keys key key_len refrows Extra -- --- -- -- --- --- -- --- --- 1 PRIMARY c const PRIMARY PRIMARY 4 const 1 1 PRIMARY s ref date,customer_id customer_id 4 const 2 1 PRIMARY derived2 ALL (NULL)(NULL) (NULL) (NULL) 781265 Using where 2 DERIVED cs ALL (NULL)(NULL) (NULL) (NULL) 1018092 Using temporary; Using filesort -Original Message- From: Daevid Vincent [mailto:dae...@daevid.com] Sent: Monday, October 24, 2011 1:46 PM To: mysql@lists.mysql.com Subject: Within-group aggregate query help please - customers and latest subscription row I know this is a common problem, and I've been struggling with it for a full day now but I can't get it. I also tried a few sites for examples: http://www.artfulsoftware.com/infotree/queries.php#101 http://forums.devarticles.com/general-sql-development-47/select-max-datetime -problem-10210.html Anyways, pretty standard situation: CREATE TABLE `customers` ( `customer_id` int(10) unsigned NOT NULL auto_increment, `email` varchar(64) NOT NULL default '', `name` varchar(128) NOT NULL default '', `username` varchar(32) NOT NULL, ... ); CREATE TABLE `customers_subscriptions` ( `subscription_id` bigint(12) unsigned NOT NULL default '0', `customer_id` int(10) unsigned NOT NULL default '0', `date` date NOT NULL default '-00-00', ... ); I want to show a table where I list out the ID, email, username, and LAST SUBSCRIPTION. I need this data in TWO ways: The FIRST way, is with a query JOINing the two tables so that I can easily display that HTML table mentioned. That is ALL customers and the latest subscription they have. The SECOND way
