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select count with complicated query


Hello :)
I'm working with a database of postcodes. I have an existsing query that returns the postcodes within a certain radius:

select *, acos(cos(51.496502411798 * (PI()/180)) *cos(-0.13982862499836 * (PI()/180)) *cos(lat * (PI()/180)) *cos(lng * (PI()/180))+cos(51.496502411798 * (PI()/180)) *sin(-0.13982862499836 * (PI()/180)) *cos(lat * (PI()/180)) *sin(lng * (PI()/180))+sin(51.496502411798 * (PI()/180)) *sin(lat * (PI()/180))) * 3959 as Dist from SW having Dist < 1 order by Dist

Dist 1 is the number of miles for the search radius.

I am making a script that will allow the user to enter a postcode and return a list of postcodes in the given radius. There can be over 20,000 results of which only 1000 will be displayed. Currently, I get all results into a PHP array and use the count() function to return the total, and just display upto 1000.
I would like to make the query produce only a count of relevant matches, and then make a follow up query with a limit of 1000 which will stop the unneccesary collection of data over 1000 rows.

Thanks for your time :D


Just add LIMIT 1000 after your order by.


For a moment I thought I was being REALLY stupid :D
Goedendag Pritaeious, but that only returns 1000 results even if there are 25000 matches. I want to make an initial query that will tell me the exact amount of matches and no data, and then a second query collecting the data with a limit of 1000, or one that does both if it is possible.
Thanks for your speedy consideration :D


Ah, okay. Just use the original query and select only COUNT(*) AS resultcount. That way the server does the counting, but only one record gets transferred to your script.

select count(*) as resultcount
from SW 
where (acos(cos(51.496502411798 * (PI()/180)) *cos(-0.13982862499836 * (PI()/180)) *cos(lat * (PI()/180)) *cos(lng * (PI()/180))+cos(51.496502411798 * (PI()/180)) *sin(-0.13982862499836 * (PI()/180)) *cos(lat * (PI()/180)) *sin(lng * (PI()/180))+sin(51.496502411798 * (PI()/180)) *sin(lat * (PI()/180))) * 3959) < 1

It's perfect!
Thank you Pritaeious, as usual, a fast and concise solution :D

Question Answered as of 1 Year Ago by pritaeas

If I were you I'd put them in two stored procedures. Keeps your code nice and readable.

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