[RESOLVED] ideas needed... distance / radius searching.

stevesweetland

i have a list of uk postcodes, and these include long / lat references and x / y metre references.

based on a user-entered postcode, and a user-selected distance (likely in miles) i want to query the database on what postcodes are within the selected distance.

im just a bit stuck as to the syntax of how id go about doing this query, im not a beginner, but its got me a bit baffed (probably cause im knackered).

im assuming if, i got a grid ref (x + y) for the entered postcode, i could just do it in sql...

...IF x + distance <= x OR x - distance <= x...

does this make sense? or am i going about it all wrong?

its the idea of radius that has me scared...

stevesweetland

getting scared now. the only bit of help iv currently found, from where i actually got the list... is that i can calculate distances using.... pythagoras theorem.

its been a while > has anyone a scooby how id do this?

Installer

What exactly do the postcode records look like? What do the "x / y metre references" refer to?

Edit: 50 pounds annually just for Postzon data? You should complain to your MP. 🙂

stevesweetland

Installer wrote:
What exactly do the postcode records look like? What do the "x / y metre references" refer to?

Edit: 50 pounds annually just for Postzon data? You should complain to your MP. 🙂

lol > who said i paid for it? yes i do need to... lol... know much about it?

the x/y metre references go from the bottom left tip of the uk... lands end i believe.

the records are like

postcode (first half) | x | y | lat | long

Installer

Assuming that x is the E-W axis and y the N-S axis (ore vice-versa) then, yes, it would be a simple application of the Pythagorean Theorem to find the distance from the reference point:

Straight-line distance = square root of (x squared + y squared)

$distance = sqrt(pow($x, 2) + pow($y, 2));

Finding the distance of one point from another would take a little more figuring, but it would still be basic trigonometry:

$distance = sqrt(pow($x_1 - $x_2, 2) + pow($y_1 - $y_2, 2));

(Of course, this all assumes that the earth is flat. In case it's not, and you want more accurate results, you could apply the Haversine formula to the lat/long.)

stevesweetland

Installer wrote:
Assuming that x is the E-W axis and y the N-S axis (ore vice-versa) then, yes, it would be a simple application of the Pythagorean Theorem to find the distance from the reference point:

Straight-line distance = square root of (x squared + y squared)
$distance = sqrt(pow($x, 2) + pow($y, 2));
Finding the distance of one point from another would take a little more figuring, but it would still be basic trigonometry:
$distance = sqrt(pow($x_1 - $x_2, 2) + pow($y_1 - $y_2, 2));
(Of course, this all assumes that the earth is flat. In case it's not, and you want more accurate results, you could apply the Haversine formula to the lat/long.)

excellent. thankyou sir