Question

我有通过Hibernate调用Native SQL（Oracle）的Java方法：

public List<Location> getLocationsAround(double latitude, double longitude, double radius, long retailerId) {
    List<Location> locationList = (List<Location>) SessionManager.getSession().createSQLQuery(
        "SELECT loc.*, distance(ci.coord1, ci.coord2, :latitude, :longitude) as dist " +
        "FROM location loc " +
        "join rl_retailer_location rrl on rrl.location_id = loc.location_id " +
        "join contact_info ci on ci.contact_info_id=loc.contact_info_id " +
        "WHERE rrl.retailer_id=:retailerId " +
        "and NVL(distance(ci.coord1, ci.coord2, :latitude, :longitude), :limit) <= :radius " + 
        "ORDER BY dist ASC"
    ).addEntity("loc", DAO.getInstance().getMappedClass(Location.class))
    .setLong("retailerId", retailerId)
    .setDouble("latitude", latitude)
    .setDouble("longitude", longitude)
    .setDouble("radius", radius)
    .setDouble("limit", radius + 1.)
    .list();
    return locationList;
}

对于dist计算，使用FUNCTION（存储过程）distance，它有4个参数（数字）：latitude1，longitude1，latitude2，longitude2并返回NUMBER（18,6）或null（如果参数）无效）。在ORDER BY中使用dist进行结果集排序。

此版本按预期工作。

问题：如何在WHERE中重写部分dist的查询？

目标：消除2x计算distance(ci.coord1, ci.coord2, :latitude, :longitude)

Answer 1

在Oracle中，您可以在FROM子句中使用查询作为表格，这样您就可以执行以下操作：

SELECT *
  FROM (
    SELECT loc.*
         , distance(ci.coord1, ci.coord2, :latitude, :longitude) as dist
      FROM location loc
      JOIN rl_retailer_location rrl on rrl.location_id = loc.location_id
      JOIN contact_info ci on ci.contact_info_id=loc.contact_info_id
     WHERE rrl.retailer_id=:retailerId
     ) loc
 WHERE dist <= :radius
 ORDER BY dist ASC

如何重用查询的一部分

1 个答案: