Search by zipcode

[Newsgroup_User]

Anyone do a 'search by zip code within a certain distance' with a variable
of distance (i.e. 25 miles)?
I downloaded a zip code database with a ton of zip codes in it. I can get
the longitude and latitude of each zip code when querying it with the
entered zip code.

How do I then query the database to search within 25 miles of the longitude
and latitude? I would like the function to query the database and grab a
bunch of zip codes within the distance and then query that against the
possible zip codes in the rescues database to match only those within the 25
miles.

Any tutorials? Anyone have an example finished (code)? Anyone know the
correct way to do this? I am only guessing on how it is actually done.

Thanks!
--
Wally Kolcz
MyNextPet.org
Founder / Developer
586.871.4126

[Newsgroup_User]

Any tutorials? Anyone have an example finished (code)? Anyone know the
correct way to do this? I am only guessing on how it is actually done.

Thanks!

I do not have the exact code, but Google does.

When you go searching you are going to get formulas that you use to
determine the min and max long and lat that contain your 25 mile radius
area. Which one you use depends on how you want to balance performance
with accuracy and whether you OK with looking at the world as a flat
grid or need to take the curvature of the global sphere into account.

Once you have chosen your algorithm, you just plug in the numbers and
you will get your min and max long and lat numbers. You then uses these
in your query to get all zip codes between these four values.

[philh]

Wally, look up the "great circle" calculation, and anything to do with geocoding.

Here's a link to a neat VB version:

http://groups.google.com/group/microsoft.public.vb.general.discussion/browse_frm/thread/3ae913a047c0...

Enjoy!

[Newsgroup_User]

Thanks. I looked using google, but I dont know which to find.
Most show me 2 longitudes and 2 latitudes and gives me the difference in
miles.

(a^2) + (b^2) = (c^2)
If a = latitudinal distance, b = longitudinal distance, then c = distance
between the two points.
For each ZIP code, you'll have to do
a = x1 - x2
b = y1 - y2
c = Sqrt((a^2) + (b^2))
Where x1 is latitude of the "from" ZIP, x2 is lat. of the "to" ZIP,
y1 is long. of the "from" ZIP and y2 is long. of the "to" ZIP.

Someone supplied me with this CF function, but it also appears to do the
same:

Here is the code from cflib.org that calculates the distance.


<cfscript>
function LatLonDist(lat1,lon1,lat2,lon2,units)
{
// Check to make sure latitutdes and longitudes are valid
if(lat1 GT 90 OR lat1 LT -90 OR
lon1 GT 180 OR lon1 LT -180 OR
lat2 GT 90 OR lat2 LT -90 OR
lon2 GT 280 OR lon2 LT -280) {
Return ("Incorrect parameters");
}

lat1 = lat1 * pi()/180;
lon1 = lon1 * pi()/180;
lat2 = lat2 * pi()/180;
lon2 = lon2 * pi()/180;
UnitConverter = 1.150779448; //standard is statute miles
if(units eq 'nm') {
UnitConverter = 1.0;
}

if(units eq 'km') {
UnitConverter = 1.852;
}

distance = 2*asin(sqr((sin((lat1-lat2)/2))^2 +
cos(lat1)*cos(lat2)*(sin((lon1-lon2)/2))^2)); //radians

if(units neq 'radians'){
distance = UnitConverter * 60 * distance * 180/pi();
}

Return (distance) ;
}
</cfscript>


I need one that shows the range of longitude and latitudes to use in my
query to find all the zips that fall in the range.

Any ideas on what I should be looking for.

Sorry.

[Newsgroup_User]

Ok, I am retarded. Each I find asks for 2 longitudes and 2 latitudes. What
am I missing. I can grab the longtitude and latitude from the user's zip
code from the database of zipcodes I have. I can ask for the max distance
(25 miles, 50 miles, etc.) But then I need to plug those values into a
formula to generate a range of zipcodes that fall within, right?

I found this while searching for the formula:

What is my second long and lat when I am supplying the first long and lat
and distance from the user's information.

<!---UDF distance Formula||--->
<CFSCRIPT>
//calculate the Deltas
function distance(lat1,lat2,lon1,lon2)
{ km=6367;
deltaLon=twoRadians(lon1)-twoRadians(lon2);
deltaLat=twoRadians(lat1)-twoRadians(lat2);
//intermediate values
intMedVal = sin2(deltaLat/2) + cos(twoRadians(lat1)) *
cos(twoRadians(lat2)) * sin2(deltaLon/2);
//the great circle distance in radians gcd = 2 *
arcSin(getMin(1,Sqr(intMedVal)));
//multiply the the radians by the radius to get the distance in specified
units d = km * gcd;
//this only applies to us (ANDRONICS.CO.UK)
//it relevant to the vendor that is displaying the map data
//the map is coming back in rectangular format they use the length as the
zoom distance
//we multiply it by 2 two get an equal zoom distance from top to bottom
d = round(d * 2); }
//Convert to Radians
function twoRadians(currDegree){
(currDegree * (Pi()/180));
}
function sin2(currVal){
(1 - cos(2*currVal))/2;
} function arcSin(x){
atn(x/Sqr(-x * x + 1));
} function getMin(y,z)
{ if(y lte z){ getMin=y;
}else{
getMin=z;
}
}
</CFSCRIPT>
<CFSET
ATTRIBUTES.D=distance(maxLatitude,minLatitude,maxLongitude,minLongitude)>

I appreciate the help and the patients. This one really confuses me... 🙂

[Newsgroup_User]

Would the second longitute and latitude be the base plus the milage
difference?

Say the base long. was 25, would the second be 50? And the same for the
latitude?

[philh]

The basic idea is to query the table for the ZIP codes whose lat/longs, when plugged into the formula, fall within the desired range of the "home" ZIP code. Hint: have the database, not ColdFusion, do the math.

[Newsgroup_User]

Thanks. I got the thing all finished and it works great!