?
Solved

MySQL ordering results before grouping

Posted on 2014-12-09
2
Medium Priority
?
205 Views
Last Modified: 2014-12-09
Hi,

I have the following table in a database;

CREATE TABLE IF NOT EXISTS `activity` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` int(11) NOT NULL,
  `mood_id` int(11) NOT NULL,
  `date` datetime NOT NULL,
  `lat` varchar(30) NOT NULL,
  `lon` varchar(30) NOT NULL,
  `latlng` geometry NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 ;

Open in new window


I'm running the following query on the table.

SELECT a.*,u.sex FROM activity AS a, users AS u WHERE u.id = a.user_id AND a.id > 0 AND a.lat != '' AND a.lon != ''AND MBRContains(GeomFromText('LineString(52.74128 -0.878906, 52.349471 0.152435)'), a.latlng)GROUP BY a.user_id

Open in new window


In general the query works fine in general, however there can be multiple entries in the table with the same user_id. So what I would like to do is sort the results by date so the latest entries are first, and then apply the group by a.user_id so that the query only returns the most recent entry in the table for each user.

I remember coming across this issue in the past and I believe I had to perform a sub query first to sort by date, and then group, but I can't remember.

Can someone help please?
0
Comment
Question by:SheppardDigital
2 Comments
 
LVL 143

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 2000 total points
ID: 40489298
you may want to read this article:
http://www.experts-exchange.com/Database/Miscellaneous/A_3203-DISTINCT-vs-GROUP-BY-and-why-does-it-not-work-for-my-query.html

for MySQL, the solution is a subquery :
SELECT a.*,u.sex 
FROM activity AS a, users AS u 
WHERE u.id = a.user_id AND a.id > 0 AND a.lat != '' AND a.lon != ''
AND MBRContains(GeomFromText('LineString(52.74128 -0.878906, 52.349471 0.152435)'), a.latlng)
AND a.date = ( select max(x.date) from activity as x where x.id = u.user_id                                  

Open in new window

0
 

Author Comment

by:SheppardDigital
ID: 40489669
It looks like that worked perfectly. The final query was;

SELECT a.*,u.sex 
FROM activity AS a, users AS u 
WHERE u.id = a.user_id AND a.id > 0 AND a.lat != '' AND a.lon != '' 
AND MBRContains(GeomFromText('LineString(52.74128 -0.878906, 52.349471 0.152435)'), a.latlng) 
AND a.date = (SELECT MAX(x.date) FROM activity AS x WHERE x.user_id = a.user_id)
GROUP BY a.user_id

Open in new window

0

Featured Post

Fill in the form and get your FREE NFR key NOW!

Veeam is happy to provide a FREE NFR server license to certified engineers, trainers, and bloggers.  It allows for the non‑production use of Veeam Agent for Microsoft Windows. This license is valid for five workstations and two servers.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Password hashing is better than message digests or encryption, and you should be using it instead of message digests or encryption.  Find out why and how in this article, which supplements the original article on PHP Client Registration, Login, Logo…
This article shows the steps required to install WordPress on Azure. Web Apps, Mobile Apps, API Apps, or Functions, in Azure all these run in an App Service plan. WordPress is no exception and requires an App Service Plan and Database to install
In this video, Percona Solution Engineer Dimitri Vanoverbeke discusses why you want to use at least three nodes in a database cluster. To discuss how Percona Consulting can help with your design and architecture needs for your database and infras…
In this video, Percona Solutions Engineer Barrett Chambers discusses some of the basic syntax differences between MySQL and MongoDB. To learn more check out our webinar on MongoDB administration for MySQL DBA: https://www.percona.com/resources/we…
Suggested Courses
Course of the Month15 days, 16 hours left to enroll

850 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question