In the simple query... the city field showed the result 'Los Angeles' in every row the distance field showed incorrect results to :(
City | Distance Los Angeles 18 Los Angeles 5 Los Angeles 7 ...
On Apr 1, 2005, at 1:59 PM, Peter Brawley wrote:
What was wrong with Graham's simpler query?
PB
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Graham Anderson wrote:
I upgraded my local mysql version to 4.1.10a and the below query finally works :)
How can I now amend the query so it works on my remote server running mysql 3.23.58 ? From one headache to another ;)
SELECT (
SELECT City FROM Cities WHERE CityID = N.CityID ), N.Distance FROM Cities C JOIN Nbc N ON C.CityID = N.PrimaryCityID WHERE C.City = 'Los Angeles' AND N.Distance <20
many many thanks to all those that replied :)
g
On Mar 31, 2005, at 11:49 PM, Philip M. Gollucci wrote:
Graham Anderson wrote:
What is the proper way to say this ?
SELECT C.City, N.Distance
FROM Cities C
JOIN Nearbycities N ON C.CityId =ci N.PrimaryCityId
WHERE N.CityId =
(SELECT Cities.CityId FROM Cities WHERE Cities.city = 'Los Angeles')
AND N.distance < 20
I am trying to enter in a city and get all the nearby cites with 20 miles
Somehow, I need to join NearbyCities.PrimaryCityId, Cities.CityId, and Cities.city
learning :)
Unless I missed something... Why did you make it so hard ? SQL is meant to be easy :)
SELECT c.city, n.distance FROM Cities c, Nearbycities n WHERE c.cityid = n.primarycityid AND c.city = 'Los Angeles' AND n.distance < 20
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