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So, in short,

  1. What should be the data type of latitude and longitude?
  2. What SQL command I should call to get the first 100 nearest restaurants for example?


I have 100k biz record each with lattitude and longitude. I see that MySQL actually support a data type called point. Should I use that instead?

Does MySQL support KDTree storage system

Is it best to use point data type rather than regular float data type to store latitutude and longitude?

Eventually I want to find things like the first 100 restaurants closest to points 105,6 for example and my databases contains a lot of biz and points. Obviously computing the distance one by one for every records and for every points would be O(n) and hence sucks.

Notice that I am aware of a simpler solution described in How do Application Like Yelp Retrieve distance information from data base efficiently and will implement that my self too for a start. That's a good answer.

However, I think there is one creme of the crop answer that should outperform that right? In fact, storing location based on latitude and longitude and finding stuffs nearest to it is a very common problem I expect mysql to have a special design pattern for that. Does it have that?

Where can I learn more about it? Thanks.

share|improve this question
Have you seen this SO question? – Jack Douglas Aug 1 '11 at 10:30
It looks like the solution in here… is the best solution. I mean there is this thing called MYSQL spatial. However you can't pull out things like where (distance (x)<20). It's not implemented yet. – Jim Thio Aug 1 '11 at 11:27
up vote 8 down vote accepted

As far as design patterns, the Yelp question is pretty standard stuff.

For a more complex answer, you will probably need the geospatial distance. Here is a fascinating powerpoint about that topic (and here is a pdf version of that as well). However, the math involved is quite ugly.

From their slide:

set @orig_lat=122.4058; set @orig_lon=37.7907;
set @dist=10;

SELECT *, 3956 * 2 * ASIN(SQRT(
POWER(SIN((@orig_lat - abs( * pi()/180 / 2), 2) +  COS(@orig_lat * pi()/180 ) * COS(abs( * pi()/180) *  POWER(SIN((@orig_lon – dest.lon) * pi()/180 / 2), 2) )) as  distance
FROM hotels dest 
having distance < @dist
ORDER BY distance limit 10

There's a longer, more in-depth answer about geospatial distance on Stack Overflow.

But you still want to limit the results by latitude and longitude.

Ultimately, I would avoid the POINT datatype and go with latitude/longitude. There's currently no way to determine the distance between two POINTs, so you're going to have to store latitude/longitude for that calculation anyways.

One last link: you may also want to check out this SO thread regarding speeding up the queries using spatial indexes.

share|improve this answer
[ERROR in query 4] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '– dest.lon) * pi()/180 / 2), 2) )) as distance FROM network_pos dest having d' at line 2 – Felipe Micaroni Lalli Jan 23 '13 at 19:53
Hi, the @dist is on milles? thank – Olaf Erlandsen Nov 10 '15 at 23:52

Point datatypes are OK; you can just invoke X(coord)/Y(coord) to get the Lat/Lon values.

For example:

    * acos(
        * cos(radians(Y(coord)))
        * cos(radians(X(coord)) - radians(-122)) 
        + sin(radians(37))
        * sin(radians(Y(coord)))
) AS distance 
FROM markers HAVING distance < 25 
ORDER BY distance LIMIT 20;
share|improve this answer
37 is lat and -122 is lon? And 25 is meters or km? – Felipe Micaroni Lalli Jan 23 '13 at 20:03

Find the 100 restaurants nearest to some coordinate: See efficient code in It includes a stored function for computing "great cirle" distance.

share|improve this answer
Great find! Thanks. – Nirmal Nov 28 '13 at 17:14

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