问题描述
注意:尽管我使用的是带有荷兰邮政编码的邮政编码数据库,但这个问题与国家/地区无关.
我有一个数据库,其中包含荷兰的每个邮政编码及其x和y坐标(经/纬度).
I have a database with every zipcode in the Netherlands + its x and y coordinate (lat/long).
例如,我的邮政编码:$baseZipCode = 1044;
具有以下坐标:
I have for example zipcode: $baseZipCode = 1044;
with the following coordinates:
x coordinate = 4,808855
y coordinate = 52,406332
现在,我想从$baseZipCode
中找到所有其他带有$range
的邮政编码.
Now, I want to find all other zipcodes with $range
from $baseZipCode
.
例如:
SELECT
zipcode
FROM
zipcodes
WHERE
????? // Need help here
问题在于地球并不完全是圆形的.我发现很多有关from a to b
计算的教程,但这不是我所需要的.
The problem is that the earth is not completely round. I find a lot of tutorials with from a to b
calculations but that's not what I need.
有人有什么主意吗?
更新多亏了Captaintokyo,我发现了这一点:
UPDATEThanks to Captaintokyo I found this:
是否要查找所有邮政编码和距另一个邮政编码或某个点的特定英里/公里半径内的对应距离?这个问题需要经纬度坐标来解决.对地址进行地理编码可为您提供地址的纬度/经度坐标.
Want to find all zipcodes and corresponding distances within a certain mile/kilometer radius from another zipcode or point? This problems require latitude and longitude coordinates to solve. Geocoding the address gives you latitude/longitude coordinates from an address.
首先,您需要一个包含所有邮政编码及其对应的纬度和经度坐标的数据库:
First you will need a database of all zipcodes and their corresponding latitude and longitude coordinates:
CREATE TABLE `zipcodes` (
`zipcode` varchar(5) NOT NULL DEFAULT '',
`city` varchar(100) NOT NULL DEFAULT '',
`state` char(2) NOT NULL DEFAULT '',
`latitude` varchar(20) NOT NULL DEFAULT '',
`longitude` varchar(20) NOT NULL DEFAULT '',
KEY `zipcode` (`zipcode`),
KEY `state` (`state`)
)
因此,一旦有了数据库,您便要查找中心点一定英里半径内的所有邮政编码.如果中心点是另一个邮政编码,只需在数据库中查询该邮政编码的纬度和经度坐标即可.然后代码如下:
So once you have the database you want to find all zipcodes within a certain mile radius of a central point. If the central point is another zipcode, simply query the database for the latitude and longitude coordinates of that zipcode. Then the code is as follows:
// ITITIAL POINT
$coords = array('latitude' => "32.8", 'longitude' => "-117.17");
//RADIUS
$radius = 30;
// SQL FOR KILOMETERS
$sql = "SELECT zipcode, ( 6371 * acos( cos( radians( {$coords['latitude']} ) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians( {$coords['longitude']} ) ) + sin( radians( {$coords['latitude']} ) ) * sin( radians( latitude ) ) ) ) AS distance FROM zipcodes HAVING distance <= {$radius} ORDER BY distance";
// SQL FOR MILES
$sql = "SELECT zipcode, ( 3959 * acos( cos( radians( {$coords['latitude']} ) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians( {$coords['longitude']} ) ) + sin( radians( {$coords['latitude']} ) ) * sin( radians( latitude ) ) ) ) AS distance FROM zipcodes HAVING distance <= {$radius} ORDER BY distance";
// OUTPUT THE ZIPCODES AND DISTANCES
$query = mysql_query($sql);
while($row = mysql_fetch_assoc($query)){
echo "{$row['zipcode']} ({$row['distance']})<br>\n";
}
(Yahoo和Google都提供免费的地理编码服务.)
(Both Yahoo and Google offer free geocoding services.)
推荐答案
您要执行以下操作:
SELECT zipcode FROM zipcodes WHERE DistanceFormula(lat, long, 4.808855, 52.406332) < $range
如果您的邮政编码表很大,可能会很慢.您可能还想查看MySQL的地理空间扩展.
It may be slow if your table of zip codes is large. You may also want to check out the geospatial extensions for MySQL.
这篇关于mySQL在y范围内的x公里/英里内选择邮政编码的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!