这里是一个基于C语言的自定义sqlite函数[从下面提供的链接中复制而来]。它可以在iOS应用程序中使用。它假设你有名为latitude和longitude的列，并计算它们与你提供的任何纬度/经度坐标之间的差异。非常好的写作，可直接使用。
#define DEG2RAD(degrees) (degrees * 0.01745327) // degrees * pi over 180 static void distanceFunc(sqlite3_context *context, int argc, sqlite3_value **argv) { // check that we have four arguments (lat1, lon1, lat2, lon2) assert(argc == 4); // check that all four arguments are non-null if (sqlite3_value_type(argv[0]) == SQLITE_NULL || sqlite3_value_type(argv[1]) == SQLITE_NULL || sqlite3_value_type(argv[2]) == SQLITE_NULL || sqlite3_value_type(argv[3]) == SQLITE_NULL) { sqlite3_result_null(context); return; } // get the four argument values double lat1 = sqlite3_value_double(argv[0]); double lon1 = sqlite3_value_double(argv[1]); double lat2 = sqlite3_value_double(argv[2]); double lon2 = sqlite3_value_double(argv[3]); // convert lat1 and lat2 into radians now, to avoid doing it twice below double lat1rad = DEG2RAD(lat1); double lat2rad = DEG2RAD(lat2); // apply the spherical law of cosines to our latitudes and longitudes, and set the result appropriately // 6378.1 is the approximate radius of the earth in kilometres sqlite3_result_double(context, acos(sin(lat1rad) * sin(lat2rad) + cos(lat1rad) * cos(lat2rad) * cos(DEG2RAD(lon2) - DEG2RAD(lon1))) * 6378.1); }This defines an SQL function distance(Latitude1, Longitude1, Latitude2, Longitude2), which returns the distance (in kilometres) between two points.
To use this function, add the code above ... and then add this line immediately after you call sqlite3_open:
sqlite3_create_function(sqliteDatabasePtr, "distance", 4, SQLITE_UTF8, NULL, &distanceFunc, NULL, NULL);…where sqliteDatabasePtr is the database pointer returned by your call to sqlite3_open.
Assuming you have a table called Locations, with columns called Latitude and Longitude (both of type double) containing values in degrees, you can then use this function in your SQL like this:
SELECT * FROM Locations ORDER BY distance(Latitude, Longitude, 51.503357, -0.1199)This example orders the locations in your database based on how far away they are from the London Eye, which is at 51.503357, -0.1199.
编辑：
原始链接http://www.thismuchiknow.co.uk/?p=71已经失效，因此如评论中所提到的，您可以使用这个链接：https://web.archive.org/web/20160808122817/http://www.thismuchiknow.co.uk/?p=71来获取该网页。
Haversine公式就是你需要的。使用简单的欧几里得距离公式是不够的，因为地球的曲率会影响两点之间的距离。
使用PHP、MySQL和Google Maps创建商店定位器是一篇关于如何使用MySQL实现此解决方案的文章，但SQLite不支持三角函数。
在Stack Overflow的计算大圆距离与SQLite答案中查看进一步扩展SQLite函数以解决此问题的提示。
