我希望能够在表上执行 SQL SELECT (sqlite) 并按每行距任意点的距离进行排序
-
26-09-2019 - |
题
我有一个 sqlite 数据库,其中包含一个充满地理位置的表,每个位置都存储为以度为单位的纬度和经度值。我希望能够在此表上执行 SQL SELECT 并按每行与任意点的距离进行排序。我通过使用下面定义的自定义 sqlite 函数 (distanceFunc) 实现了这一点。
这是该函数,以及一个将度数转换为弧度的便捷宏。该函数基于在线距离计算器,该计算器利用余弦球面定律。
http://en.wikipedia.org/wiki/Spherical_law_of_cosines
“在球面三角学中,余弦定理(也称为边余弦定则)是与球面三角形的边和角相关的定理,类似于平面三角学中的普通余弦定理。”
#define DEG2RAD(degrees) (degrees * 0.01745327) // degrees * pi over 180
(Location.m) 中的“distanceFunc”
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);
}
我将我的方法称为“getDistanceBetweenLongLat”,它也在“Location.m”中,如下所示:
在我的(AppDelegate.m)中使用:
Location *location = [[Location alloc] init];
location.latitude = -37.1134; //double property
location.longitude = 145.4254; //double property
[location getDistanceBetweenLongLat:[self dbPath]];
我在(Location.m)中的“getDistanceBetweenLongLat”方法:
- (void) getDistanceBetweenLongLat:(NSString *)dbPath {
AppDelegate *_ad = (AppDelegate *)[[UIApplication sharedApplication] delegate];
if (sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) {
sqlite3_create_function(database, "distance", 4, SQLITE_UTF8, NULL, &distanceFunc, NULL, NULL);
const char *sql = "select * from locations order by distance(latitude, longitude, ?, ?)";
sqlite3_stmt *selectstmt;
sqlite3_bind_double(selectStmt, 1, latitude);
sqlite3_bind_double(selectStmt, 2, longitude);
if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL) == SQLITE_OK) {
while(sqlite3_step(selectstmt) == SQLITE_ROW) {
NSInteger primaryKey = sqlite3_column_int(selectstmt, 0);
Location *location = [[Location alloc] initWithPrimaryKey:primaryKey];
location.latitude = sqlite3_column_double(selectstmt, 1);
location.longitude = sqlite3_column_double(selectstmt, 2);
location.title = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 3)];
location.street1 = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 4)];
location.street2 = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 5)];
location.suburb = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 6)];
NSInteger postCodeItem = sqlite3_column_int(selectstmt, 7);
location.postcode = postCodeItem;
location.isDirty = NO;
[_ad.locations addObject:location];
[location release];
}
}
} else {
//Even though the open call failed, close the database connection to release all the memory.
sqlite3_close(database);
}
}
这调用“distanceFunc”就可以了。然而,当它到达检查所有四个参数是否为非空的语句时,它们都将返回为空。
但是,当我将上面的 sql 语句更改为以下内容时:
const char *sql = "select * from locations order by distance(latitude, longitude, '?', '?')"; //single quotes added
这四个参数不会返回为空。但是,最后两个参数值应该为 0,对吗?
location.latitude = -37.1134; //double property
location.longitude = 145.4254; //double property:
来自 (Location.m) 中的“distanceFunc”:
double lat1 = sqlite3_value_double(argv[0]); // -17.7699 from 1st result in Locations table
double lon1 = sqlite3_value_double(argv[1]); // 245.1103 from 1st result in Locations table
double lat2 = sqlite3_value_double(argv[2]); // 0
double lon2 = sqlite3_value_double(argv[3]); // 0
我使用几乎相同的方法来获取要显示的初始数据,并且该特定数组填充得很好。这次我只希望我的位置数组(在 _ad.locations 中)包含按距离排序的结果,这样我就可以在 UITableView 中显示它们。
笔记:使用的“distanceFunc”也可以在以下链接中找到: http://www.thismuchiknow.co.uk/?p=71&cpage=1#comment-30834
解决方案
我使用一种变体,其中 latVal 和 longVal 是我当前的位置:
NSString *sqlString = @"SELECT * FROM stores WHERE status = 1 AND distance(latitude, longitude, %f, %f, id) < 800";
sqlString = [NSString stringWithFormat:sqlString, latVal, longVal];
我还使用一个变量来读取 distanceFunc 内的距离值,将最后两行更改为以下内容:
float val = acos(sin(lat1rad) * sin(lat2rad) + cos(lat1rad) * cos(lat2rad) * cos(DEG2RAD(lon2) - DEG2RAD(lon1))) * 6378.1;
distanceVal = (float)val;
sqlite3_result_double(context, val);
然后,我将 distanceVal 保存到适合匹配的对象中,并对返回的项目应用排序:
NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:@"distance" ascending:YES];
[ locations sortUsingDescriptors: [ NSArray arrayWithObject: sortDescriptor ] ];
[sortDescriptor release];