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how to find nearest Hospital location by latitude and longitude

Posted on 2013-06-05
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Last Modified: 2013-07-01
Hi Experts

I am working on a Hospital website Portal. Where i will show list of hospitals in the page. Each hospital has own Latitude and longitude. My question is that i want to show the list of hospital with the radius of 50km with the help of latitude and longitude.

e.g Hospital A has longitude and latitude. i want to display other hospital around 50km within radius of HospitalA.

Can I get a solution ?pls help me

thanks for Advance.
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Question by:krushnac
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12 Comments
 
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Expert Comment

by:Sammy
ID: 39224686
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LVL 56

Expert Comment

by:Julian Hansen
ID: 39224809
Interesting question - requires reall programming.

There are two hurdles to jump on this one.

How to efficiently narrow the search and then to find those that fit the actual requirement.

You are looking for hospitals within a circle radius of a specific point - but to pull info from a database using circle maths is not easily done.

So what you do is pull a working set - i.e find all hospitals within a square 100Km to a side with target hospital at the centre - from this you can then eliminate the ones that don't match.

Here is a possible solution

1. There are approximately .45 degrees in a 50Km distance (varies depending where you are on the globe - but lets use that for now).

2. Find all hospitals that are +.45 / -.45 degrees North / South and East / West of your target hospital

select * from hospitals where lattitude > TARGET_LATTITUDE - 0.45 and lattitude < TARGET_LATTITUDE + 0.45 and longitude > TARGET_LONGITUDE - 0.45 and longitude < TARGET_LONGITUDE + 0.45

3. Narrow your search down even further by excluding hospitals you know for definite are in the 50KM radius. You can do this by working out which fall with in the inscribed square

4. Using the fact that the side of an inscribed square is sqrt(2)r with radius 50 we have the side of an inscribed square as  70.7Km so half the distanced is 35.355Km which is approximately 0.32 degrees. So we can exclude all records that match these criteria

select * from hospitals where lattitude > TARGET_LATTITUDE - 0.32 and lattitude < TARGET_LATTITUDE + 0.32 and longitude > TARGET_LONGITUDE - 0.32 and longitude < TARGET_LONGITUDE + 0.32

This gives us our definites

5. Putting it all together we can find a working set of maybe's that we must test with this query

select * from hospitals where
lattitude > TARGET_LATTITUDE - 0.45 AND lattitude < TARGET_LATTITUDE - 0.32 AND lattitude > TARGET_LATTITUDE + 0.32 and lattitude < TARGET_LATTITUDE + 0.45 AND
longitude > TARGET_LONGITUDE - 0.45 and longitude < TARGET_LONGITUDE - 0.32 AND
longitude > TARGET_LONGITUDE + 0.32 and longitude < TARGET_LONGITUDE + 0.45

You can now apply the distance formula to the result set and add the results to the results from the query in step 4
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LVL 56

Expert Comment

by:Julian Hansen
ID: 39224821
If you are working with MySQL then you can do the full calculation in the query

http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL
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Expert Comment

by:Ray Paseur
ID: 39225469
Please see this article, where the design pattern and PHP+MySQL code demonstrates how this is done.  (You may want to update to MySQLi or PDO.)
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_4276-What-is-near-me-Proximity-calculations-using-PHP-and-MySQL.html
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Author Comment

by:krushnac
ID: 39234058
Hi there
i got the solution. thanks for the effort .
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Author Comment

by:krushnac
ID: 39234461
I've requested that this question be closed as follows:

Accepted answer: 0 points for krushnac's comment #a39234058

for the following reason:

I got the solution.
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LVL 110

Expert Comment

by:Ray Paseur
ID: 39234462
The close request cites this comment as the solution, but the comment does not contain enough information for us to understand how you solved your problem.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/Q_28149228.html#a39234058

Please post your solution and show us how you got the solution.  If any of the responses or articles here were at all helpful, it would be appropriate to award points.

Thanks, ~Ray
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LVL 110

Expert Comment

by:Ray Paseur
ID: 39235226
My recommendation is for the author to post the solution, hopefully with enough detail that we can see if there is original art in the solution.

Absent that, delete no refund since there is nothing new here.
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Accepted Solution

by:
krushnac earned 0 total points
ID: 39240503
here is the mysql query
SELECT * , ((
ACOS( SIN( ".$lat." * PI( ) /180 ) * SIN( lat_centroid * PI( ) /180 ) + COS( ".$lat." * PI( ) /180 ) * COS( lat_centroid * PI( ) /180 ) * COS( (
 ".$long." - long_centroid) * PI( ) /180 ) ) *180 / PI( )) *60 * 1.1515) AS `distance` FROM ".$this->table." HAVING distance <=100 ORDER BY distance LIMIT 0 , 10

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