查询与SQL，地理坐标和半径重叠的区域

Question

我有一个由经度，纬度和值组成的MySQL数据集。我试图总结其经度和纬度坐标在其他经纬度坐标的给定半径内的值（我们称它们为“焦点”）。最棘手的是，我试图从重叠区域分离出不同的坐标 - 例如，半径1与半径2重叠的地方。

每个有半径的焦点都有多个半径的“区域”，所以对于任何给定的纬度/经度坐标，可以总结出很多东西。我设法拼凑一个查询，主要是工作，虽然这是一个有点慢：

Select 
      Sum(If(`zone`='z0_0x1_0',`value`,0)) as `z0_0x1_0`, 
      Sum(If(`zone`='z0_0x1_1',`value`,0)) as `z0_0x1_1`, 
      Sum(If(`zone`='z0_0x1_2',`value`,0)) as `z0_0x1_2`, 
      Sum(If(`zone`='z0_0x1_3',`value`,0)) as `z0_0x1_3`, 
      Sum(If(`zone`='z0_1x1_0',`value`,0)) as `z0_1x1_0`, 
      Sum(If(`zone`='z0_1x1_1',`value`,0)) as `z0_1x1_1`, 
      Sum(If(`zone`='z0_1x1_2',`value`,0)) as `z0_1x1_2`, 
      Sum(If(`zone`='z0_2x1_0',`value`,0)) as `z0_2x1_0`, 
      Sum(If(`zone`='z0_2x1_1',`value`,0)) as `z0_2x1_1`, 
      Sum(If(`zone`='z0_3x1_0',`value`,0)) as `z0_3x1_0`, 
      Sum(If(`zone`='z0_3x1_1',`value`,0)) as `z0_3x1_1`, 
      Sum(If(`zone`='z0_0',`value`,0)) as `z0_0`, 
      Sum(If(`zone`='z0_1',`value`,0)) as `z0_1`, 
      Sum(If(`zone`='z0_2',`value`,0)) as `z0_2`, 
      Sum(If(`zone`='z0_3',`value`,0)) as `z0_3`, 
      Sum(If(`zone`='z1_0',`value`,0)) as `z1_0`, 
      Sum(If(`zone`='z1_1',`value`,0)) as `z1_1`, 
      Sum(If(`zone`='z1_2',`value`,0)) as `z1_2`, 
      Sum(If(`zone`='z1_3',`value`,0)) as `z1_3` 
    From 
     (Select `lat`, `lng`, `value`, 
       Case 
          When ((`dist_0` Between 2.8723597844095 And 4.3343662110324) And (`dist_1` Between 3.6260179152491 And 5.4681062617155)) Then 'z0_0x1_0' 
          When ((`dist_0` Between 2.8723597844095 And 4.3343662110324) And (`dist_1` Between 2.1278369006061 And 3.6260179152491)) Then 'z0_0x1_1' 
          When ((`dist_0` Between 2.8723597844095 And 4.3343662110324) And (`dist_1` Between 1.3333495959677 And 2.1278369006061)) Then 'z0_0x1_2' 
          When ((`dist_0` Between 2.8723597844095 And 4.3343662110324) And (`dist_1` Between 0 And 1.3333495959677)) Then 'z0_0x1_3' 
          When ((`dist_0` Between 1.68658498678 And 2.8723597844095) And (`dist_1` Between 3.6260179152491 And 5.4681062617155)) Then 'z0_1x1_0' 
          When ((`dist_0` Between 1.68658498678 And 2.8723597844095) And (`dist_1` Between 2.1278369006061 And 3.6260179152491)) Then 'z0_1x1_1' 
          When ((`dist_0` Between 1.68658498678 And 2.8723597844095) And (`dist_1` Between 1.3333495959677 And 2.1278369006061)) Then 'z0_1x1_2' 
          When ((`dist_0` Between 1.0573158612197 And 1.68658498678) And (`dist_1` Between 3.6260179152491 And 5.4681062617155)) Then 'z0_2x1_0' 
          When ((`dist_0` Between 1.0573158612197 And 1.68658498678) And (`dist_1` Between 2.1278369006061 And 3.6260179152491)) Then 'z0_2x1_1' 
          When ((`dist_0` Between 0 And 1.0573158612197) And (`dist_1` Between 3.6260179152491 And 5.4681062617155)) Then 'z0_3x1_0' 
          When ((`dist_0` Between 0 And 1.0573158612197) And (`dist_1` Between 2.1278369006061 And 3.6260179152491)) Then 'z0_3x1_1' 
          When ((`dist_0` Between 2.8723597844095 And 4.3343662110324)) Then 'z0_0' 
          When ((`dist_0` Between 1.68658498678 And 2.8723597844095)) Then 'z0_1' 
          When ((`dist_0` Between 1.0573158612197 And 1.68658498678)) Then 'z0_2' 
          When ((`dist_0` Between 0 And 1.0573158612197)) Then 'z0_3' 
          When ((`dist_1` Between 3.6260179152491 And 5.4681062617155)) Then 'z1_0' 
          When ((`dist_1` Between 2.1278369006061 And 3.6260179152491)) Then 'z1_1' 
          When ((`dist_1` Between 1.3333495959677 And 2.1278369006061)) Then 'z1_2' 
          When ((`dist_1` Between 0 And 1.3333495959677)) Then 'z1_3' 
       End As `zone` 

       From 
        (Select `lat`, `lng`, `value`, 
         (acos(0.65292272498833*sin(radians(`lat`)) + 0.75742452772129*cos(radians(`lat`))*cos(radians(`lng`)-(-1.2910922519714))) * 6371) as `dist_0`, 
         (acos(0.65251345816785*sin(radians(`lat`)) + 0.75777713538338*cos(radians(`lat`))*cos(radians(`lng`)-(-1.2916315412569))) * 6371) as `dist_1` 
        From `pop` 
        Where 
         ((`lat` Between 40.714353892125 And 40.810300107875) And (`lng` Between -74.037474145971 And -73.910799854029)) Or 
         ((`lat` Between 40.673205922895 And 40.789544077105) And (`lng` Between -74.081798776797 And -73.928273223203)) 
        ) 
       As FirstCut 
     ) 
     As Zonecut 

这里的事物的逻辑：

1. 首先，它抓住周围的最大半径边界框为每个焦点。 （这是FirstCut查询。）这将我们正在查看的数据点的数量减少了几个数量级。

2. 然后，它处理所有数据，并从焦点获取每个数据点的距离（在这种情况下，dist_0和dist_1，但可以有焦点的任意数量的 - 我已经在这个例子中使用了两个刚显示它是如何工作的）。这是大圈距离的Haversine公式。

3. 然后，Case语句启动，为每个坐标指定一个“区域”的成员，这些成员从最复杂到最不复杂的处理。区域代码仅意味着“区域X，半径Y” - 因此“z0_1”表示“区域0，半径1”。如果存在“x”，则表示它是多个区域的交集。这个“区码”只是作为一个字符串分配的。

4. 最后，通过分配区域名称和Sum（If（））语句，区域代码用于总结所有内容。 （不管出于什么原因，如果（）似乎比案件工作稍快（）在这里。）

其中输出到我的脚本（PHP）区和款项的清单。现在显然这整个过程是程序化生成的，因为你必须提前计算所有可能实际上会“命中”的区域，并且这些都是作为预处理完成的，以避免在SQL中执行。

有没有更聪明的方法来做到这一点？我给他们分配一个字符串的位，然后将该字符串过滤到字段中......它看起来很黑，并不很优雅。但我无法找到一种更好的方法，将它们分类为一个大的Case语句中的字段（它似乎比许多Case语句快得多）。

任何和所有这方面的反馈将不胜感激。 MySQL表格非常庞大（数百万行）并且被索引到所有神圣的地狱。运行上面的查询大约需要0.6秒，这并不算太糟糕，但随着更多的焦点被添加，查询开始花费更长的时间，而我只是想通过SQL逻辑思考我在这个阶段的方式。谢谢。

Answer 1

我没有彻底检查，但它看起来这可能会缩短那个大CASE一些：

CONCAT(
    (CASE 
     WHEN (dist_0 ...) THEN 'z0_0' 
     WHEN (dist_0 ...) THEN 'z0_1' 
     ... 
     ELSE ''), 
    (CASE 
     WHEN (dist_1 ...) THEN 'z1_0' 
     WHEN (dist_1 ...) THEN 'z1_1' 
     ... 
     ELSE '')) AS zone