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在SQLite中查询以获取基于Radius的记录?

sqlite android sql

Pentium10 · 技术社区 · 14 年前

我有一个在MySQL中运行良好的查询

SELECT ((ACOS(SIN(12.345 * PI() / 180) * SIN(lat * PI() / 180) +
         COS(12.345 * PI() / 180) * COS(lat * PI() / 180) * COS((67.89 - lon) * 
         PI() / 180)) * 180 / PI()) * 60 * 1.1515 * 1.609344) AS distance, poi.* 
FROM poi
WHERE lang='eng' 
HAVING distance<='30'

距离以公里为单位,输入为 lat=12.345 和 lon=67.89

SQLite是3,我不能用它运行自定义函数,因为它在Android上。我也没有acos()等。。。因为这不是标准SQLite的一部分。

3 回复 | 直到 14 年前

Community CDub 7 年前

您可以创建4个新列,作为 lat 和 lon . 自 cos(a+b) = cos a cos b - sin a sin b SIN(12.345 * PI() / 180) 可以在程序运行前计算查询,把大的“距离”表达式缩减成某种形式 P * SIN_LAT + Q * COS_LAT + ... 可以由SQLite3处理的。

顺便说一句,另见 Sqlite on Android: How to create a sqlite dist db function - to be used in the app for distance calculation using lat, long

EricLarch 13 年前

这个想法来源于KennyTM(参见公认的回答),它意味着在表中添加4列来存储纬度和经度的正弦值和余弦值。

public static void injectLocationValues(ContentValues values, double latitude, double longitude) {
    values.put(LocationColumns.LATITUDE, latitude);
    values.put(LocationColumns.LONGITUDE, longitude);
    values.put(LocationColumns.COSLAT, Math.cos(MathUtil.deg2rad(latitude)));
    values.put(LocationColumns.SINLAT, Math.sin(MathUtil.deg2rad(latitude)));
    values.put(LocationColumns.COSLNG, Math.cos(MathUtil.deg2rad(longitude)));
    values.put(LocationColumns.SINLNG, Math.sin(MathUtil.deg2rad(longitude)));
}

public static double deg2rad(double deg) {
    return (deg * Math.PI / 180.0);
}

然后可以使用以下函数构建投影:

/**
 * Build query based on distance using spherical law of cosinus
 * 
 * d = acos(sin(lat1).sin(lat2)+cos(lat1).cos(lat2).cos(long2âlong1)).R
 * where R=6371 and latitudes and longitudes expressed in radians
 * 
 * In Sqlite we do not have access to acos() sin() and lat() functions.
 * Knowing that cos(A-B) = cos(A).cos(B) + sin(A).sin(B)
 * We can determine a distance stub as:
 * d = sin(lat1).sin(lat2)+cos(lat1).cos(lat2).(cos(long2).cos(long1)+sin(long2).sin(long1))
 * 
 * First comparison point being fixed, sin(lat1) cos(lat1) sin(long1) and cos(long1)
 * can be replaced by constants.
 * 
 * Location aware table must therefore have the following columns to build the equation:
 * sinlat => sin(radians(lat))
 * coslat => cos(radians(lat))
 * coslng => cos(radians(lng))
 * sinlng => sin(radians(lng))
 *  
 * Function will return a real between -1 and 1 which can be used to order the query.
 * Distance in km is after expressed from R.acos(result) 
 *  
 * @param latitude, latitude of search
 * @param longitude, longitude of search
 * @return selection query to compute the distance
 */
public static String buildDistanceQuery(double latitude, double longitude) {
    final double coslat = Math.cos(MathUtil.deg2rad(latitude));
    final double sinlat = Math.sin(MathUtil.deg2rad(latitude));
    final double coslng = Math.cos(MathUtil.deg2rad(longitude));
    final double sinlng = Math.sin(MathUtil.deg2rad(longitude));
    //@formatter:off
    return "(" + coslat + "*" + LocationColumns.COSLAT
            + "*(" + LocationColumns.COSLNG + "*" + coslng
            + "+" + LocationColumns.SINLNG + "*" + sinlng
            + ")+" + sinlat + "*" + LocationColumns.SINLAT 
            + ")";
    //@formatter:on
}

public static double convertPartialDistanceToKm(double result) {
    return Math.acos(result) * 6371;
}

如果要使用部分距离对查询进行排序,则需要排序DESC而不是ASC。

sclaeys 10 年前

为存储在数据库中的每个项目预先计算这些值(并存储它们):

cos_lat = cos(lat * PI / 180)
sin_lat = sin(lat * PI / 180)
cos_lng = cos(lng * PI / 180)
sin_lng = sin(lng * PI / 180)

CUR_cos_lat = cos(cur_lat * PI / 180)
CUR_sin_lat = sin(cur_lat * PI / 180)
CUR_cos_lng = cos(cur_lng * PI / 180)
CUR_sin_lng = sin(cur_lng * PI / 180)
cos_allowed_distance = cos(2.0 / 6371) # This is 2km

您的SQL查询将如下所示(用刚才计算的值替换CUR\ux*)