Question

我需要一些帮助加入两张桌子。我有：

my_type_table，其中包含以下列：

type（VARCHAR）
纬度（十进制）
经度（十进制）

...和neighborhood_shapes，其中包含以下列：

邻里（VARCHAR）
neighborhood_polygons（geometry）

我有一个名为myWithin的函数，它检查纬度和经度，看它们是否在邻域多边形中。它以lat long和邻域形状为参数。如果该点不在多边形中，则myWithin函数返回0;如果该点在多边形内，则返回1。

我可以按如下方式制作一个选择语句：

SELECT type, latitude, longitude, 'Newport' AS neighborhood
  FROM my_type_table
WHERE myWithin(POINTFROMTEXT( CONCAT( 'POINT(', latitude, ' ', longitude, ')' ) ) , 
               (SELECT neighborhood_shapes.neighborhood_polygons
                  FROM neighborhood_shapes
                 WHERE neighborhood_shapes.neighborhoods = 'Newport')) = 1

此选择的结果例如：

type  |  latitude | longitude | neighborhood
---------------------------------------------
small |  30.3030  | -70.7070  | Newport

我的问题是我有很多社区。我不想每次都输入邻居。有没有办法删除“纽波特”？基本上，我希望函数在每个点上运行并给出类型，纬度，经度以及该点所在的邻域？

我可以复制上面的选择并使用UNION ALL加入select语句，但是在每个邻居的名字中键入一个噩梦。必须有更好的方法。

有什么建议吗？

Answer 1

我当然没有对此进行测试......但似乎它可以像这样进行重组：

SELECT t.type, t.latitude, t.longitude, s.neighborhood 
  FROM my_type_table t, neighborhood_shapes s
WHERE myWithin(POINTFROMTEXT( CONCAT( 'POINT(', t.latitude, ' ', t.longitude, ')' ) ) ,               s.neighborhood_shapes.neighborhood_polygons ) = 1

Answer 2

你试过......

SELECT DISTINCT type,
                latitude,
                longitude,
                ns.neighborhoods AS neighborhood
    FROM my_type_table,
         neighborhood_shapes ns
    WHERE myWithin(POINTFROMTEXT( CONCAT( 'POINT(', latitude, ' ', longitude, ')' ) ) , 
                   (SELECT neighborhood_shapes.neighborhood_polygons
                        FROM  neighborhood_shapes
                        WHERE neighborhood_shapes.neighborhoods = ns.neighborhoods)) = 1

Answer 3

function getNeighborhood ($neighborhood) {


$sql = "SELECT type, latitude, longitude, 'Newport' AS neighborhood
  FROM my_type_table
WHERE myWithin(POINTFROMTEXT( CONCAT( 'POINT(', latitude, ' ', longitude, ')' ) ) , 
               (SELECT neighborhood_shapes.neighborhood_polygons
                  FROM neighborhood_shapes
                 WHERE neighborhood_shapes.neighborhoods = $neighborhood)) = 1";

$result = /**  Do Query Here**/

return $result;
}

如何使用函数连接两个MySQL表

3 个答案: