按距参照点的距离查找全部

Question

在我的Grails 1.3.7应用程序中，我有一个具有双纬度和双经度的Building实体。我正在实现一个简单的搜索引擎，以便找到与用户所在点的给定距离(以十进制度表示的纬度和经度)的所有Building实例。我发现这个http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL很好，因为我使用的是MySQL数据库。边界框近似对我来说非常好，因为我需要执行额外的过滤和计算，我只需要一个查找器来缩小我正在过滤的实例数量。我的问题是:有人已经在Grails环境中实现了这种搜索吗?如何实现的？

Answer 1

我以前实现过类似需求的东西，并且我使用了HQL查询。那是很久以前的事了，我记得我花了相当长的时间来阅读和理解，所以希望能为你节省一些时间。

这将根据当前位置(一个简单long容器对象)和一个"name“(以开头)执行select。它选择域对象(场所)以及距离当前位置几英里的位置。它是按照几英里外的升序排列的。注意，我添加了一个“道路因子”模糊来近似道路距离。

def getVenuesInArea(venueName, location, miles, optionsMap)
{
    def max = optionsMap?.max ?: 10
    def offset = optionsMap?.offset ?: 0
    if (venueName == null) venueName = "" 
    venueName += '%'

    double roadFactor = 1.20 // add 20% for the roads, instead of as crow flies...

    def query
    def results

    def countQuery = """ select count( distinct v)

                from Venue as v
                WHERE
                v.name like :venueName AND
                ( acos
                    (
                        sin(radians(:lat))
                        * sin(radians(v.location.latitude))
                        + cos(radians(:lat))
                        * cos(radians(v.location.latitude))
                        * cos(radians(v.location.longitude) - radians(:lon))
                    ) * 3956.1676 * :roadFactor < :distance
                )

            """

    def count = Venue.executeQuery(countQuery, [venueName:venueName, lat:location.latitude, lon:location.longitude, distance:miles, roadFactor:roadFactor])[0]

    query = """ select distinct v,

                (
                    acos
                    (
                        sin(radians(:lat))
                        * sin(radians(v.location.latitude))
                        + cos(radians(:lat))
                        * cos(radians(v.location.latitude))
                        * cos(radians(v.location.longitude) - radians(:lon))
                    )
                    * 3956.1676 * :roadFactor
                ) as milesAway

                from Venue as v
                WHERE
                v.name like :venueName AND
                ( acos
                    (
                        sin(radians(:lat))
                        * sin(radians(v.location.latitude))
                        + cos(radians(:lat))
                        * cos(radians(v.location.latitude))
                        * cos(radians(v.location.longitude) - radians(:lon))
                    ) * 3956.1676 * :roadFactor < :distance
                )

                order by
                (
                    acos
                    (
                        sin(radians(:lat))
                        * sin(radians(v.location.latitude))
                        + cos(radians(:lat))
                        * cos(radians(v.location.latitude))
                        * cos(radians(v.location.longitude) - radians(:lon))
                    )
                    * 3956.1676 * :roadFactor
                )
                asc,
                v.name

            """

    results = Venue.executeQuery( query, [venueName:venueName, lat:location.latitude, lon:location.longitude, distance:miles, roadFactor:roadFactor, max:max, offset:offset])

    def venues = []
    MathContext mc = new MathContext(2)
    results.each
    { result ->
        VenueWithDetails venueDetails = new VenueWithDetails(  venue:result[0], milesFrom:new BigDecimal(result[1]).round(mc)  )
        venues.add(venueDetails)
    }
    return [venues:venues, count:count]
}

这是在grails版本1.3.4上完成的，但我非常确定它在1.3.7上应该可以很好地工作。

希望这对你有帮助克里斯。