2

我有一个 sqlite 数据库，其中包含一个充满地理位置的表，每个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。

这是函数，以及将度数转换为弧度的便捷宏。此功能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，余弦定律（也称为边的余弦规则）是一个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律。”

#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);
    }
}

这很好地调用了“distanceFunc”。但是，当它到达检查所有四个参数是否非空的语句时，它们都返回为空。

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是 0，对吧？

location.latitude = -37.1134; //double property
location.longitude = 145.4254; //double property:

从（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，以便我可以在 UITableView 中显示它们。

注意：使用的“distanceFunc”也可以在以下链接中找到：http ://www.thissuchiknow.co.uk/?p=71&cpage=1#comment-30834

4

1 回答 1

4

我使用了一个变体，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，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];
于 2011-01-23T20:12:05.897 回答