基于距离的查询建议

是的。由于这看起来像英国邮政编码，因此您需要另一个将邮政编码映射到 GPS 坐标(经度和纬度(的表。你可以从这里得到它。http://jamiethompson.co.uk/web/2008/07/24/full-uk-postcode-database-for-free/(其他地方也可能有，http://www.freepostcodes.org.uk/退房(

然后你需要在SQL中做一个"选择"，连接两个表，计算你可以在SQL中做的距离。"SQRT(POWER(Long1 - long2,2( + POWER(Lat1 - Lat2,2(("将为您提供大致距离。警告：由于需要为每一行计算此值，因此请将可能的结果限制为长+范围，纬度+范围。

为了获得准确的距离，您确实需要一些三角。这是我在最近的应用程序中使用的一个函数，它做类似的事情，你应该能够调整它来工作，甚至构建 int othe SQL 查询。

function CalculateDistance($Longitude1, $Latitude1, $Longitude2, $Latitude2) {
    // Not perfect for curvature of earth on the diaganol, but it's close enough ;)
    $LatDiff = abs($Latitude1 - $Latitude2);
    $LongDiff = abs($Longitude1 - $Longitude2);
    // Convert to meters - 1 deg latitude = 111.12km
    $mLatDiff = $LatDiff * 0.11112;
    $mLongDiff = $LongDiff * 0.11112 * cos(($Latitude1 + $Latitude2) / 2);
    // Work out difference.
    $Diff = sqrt( ($mLatDiff * $mLatDiff) + ($mLongDiff * $mLongDiff) );
    return $Diff;
}

编辑：添加代码示例

未经测试的代码，但从最近的项目中编辑。需要语法检查和调整，但会给出这个想法。

$UserPostcode = "AB1 2FG";
// Find the Long/Lat of the user (you don;t have to do this, but it saves a lot of DB stress)
// Postcodes is the table that links users to postcodes
$sql = 'SELECT long, lat FROM postcodes WHERE postcode="' . mysql_real_escape_string($UserPostcode) .'"';
// ... Missing a few lines of check row exists, error reporting etc
$UserLat = $row['lat'];
$UserLong = $row['long'];
// Next, come up with an acceptable range so as not to over-stress the DB:
// What are the acceptable limits?
// 1 deg latitude = 111.12km
// LATITUDE: 1km = 1/111.2 degrees of latitude = 0.0089992800575953923686105111591073
// LONGITUDE: 1km = 1/(111.2 * cos(LATITUDE))
$RangeLat = 8 * 1/111.2; // For 8km = 5 miles. Work out what you want here, the larger = more DB stress
$RangeLong = 8 * 1/(111.2 * cos(deg2rad($UserLat)); // For 8km = 5 miles. Work out what you want here, the larger = more DB stress
// Now the query you want.
// You'll need to change this to get restauants by group, but this gives you the essence of what you need:
$sql = 'SELECT r.restaurant_id, r.OtherDetails, r.postcode,
                  p.lat, p.long,
                  SQRT(POWER(p.lat-'.$UserLat.',2) + POWER(p.long-'.$UserLong.',2)) AS distance 
           FROM resturants r
             LEFT JOIN postcodes p ON r.postcode=p.postcode
           WHERE p.lat >='.($UserLat - $RangeLat).' AND p.lat <='.($UserLat + $RangeLat).' AND p.long>='.($UserLong-$RangeLong).' AND p.long>='.($UserLong+$RangeLong).'
           ORDER BY SQRT(POWER(p.lat-'.$UserLat.',2) + POWER(p.long-'.$UserLong.',2)) ASC';
if (!$result = mysql_query($sql)) { ExitError(4, $sql . '<br />' . mysql_error()); }
while ($row = mysql_fetch_assoc($result)) {
        // This is where you can count the 4 in each category, and disregard if you get more that 4.
        // If you get less than 4, offer to expand the range - or stick in a loop and do it automatically.
        // Also for efficienty, I'd suggest you remove the "distance" calculation from the SQL and use the function above
        // The function above it more accurate as it users cosine. Then simply sort by distance here instead on in SQL.
        // Also note the above function is written for the northern hemisphere. Also won't work across the date line ;)
        // But that shouldn't be a worry.
}
mysql_free_result($result);

但是 - 可以在一个查询中完成吗：当然，如果你想关闭SQL服务器

同样，没有经过测试，但会给你该怎么做的要点。不过，我强烈建议不要这样做！

$sql = 'SELECT r.restaurant_id, r.OtherDetails, r.postcode,
                  p.lat, p.long,
                  SQRT(POWER(p.lat-p2.lat,2) + POWER(p.long-p2.long,2)) AS distance 
           FROM resturants r
             LEFT JOIN postcodes p ON r.postcode=p.postcode
             LEFT JOIN postcodes p2 ON p2.postcode="' . mysql_real_escape_string($UserPostcode) .'"
           ORDER BY SQRT(POWER(p.lat-p2.lat,2) + POWER(p.long-p2.long,2)) ASC';

如果你走这条路，并且你想要超高精度，我会让你研究如何让 COS 进入那里:)。你需要的 mySQL 函数是 RADIANS(( 和 COS((