关于你的问题,我的一些想法:
我需要使用函数ST_MakeLine将其转换为linestring并存储,是吗?
是的,要将多个点合并成LINESTRING
,可以使用ST_MakeLine
函数:
SELECT ST_AsText(
ST_MakeLine(ARRAY[ST_MakePoint(-122.21,37.77),
ST_MakePoint(-157.82,21.29),
ST_MakePoint(178.43,-18.14),
ST_MakePoint(153.02,-27.46)]));
st_astext
---------------------------------------------------------------------
LINESTRING(-122.21 37.77,-157.82 21.29,178.43 -18.14,153.02 -27.46)
(1 Zeile)
如何根据我上面提到的要求进行查询
在您的LINESTRING
每个点上创建一个缓冲区,并检查其他几何图形是否在其内部。
首先,您必须使用ST_DumpPoints
将您的路由LINESTRING
分割成POINT
。
db=
st_astext
----------------------
POINT(-122.21 37.77)
POINT(-157.82 21.29)
POINT(178.43 -18.14)
POINT(153.02 -27.46)
(4 Zeilen)
然后使用ST_Buffer
创建每个点周围的缓冲区。ST_Buffer返回一个几何图形,表示点(或任何其他几何类型)周围的区域。例如,如果我没有交换x和y,则将路线的第一个点(位于旧金山某个地方)POINT(-122.21 37.76)
:
![enter image description here](https://istack.dev59.com/xY8XP.webp)
db=# SELECT ST_AsText(
ST_Buffer('POINT(-122.21 37.76)'::GEOMETRY,0.0001, 'quad_segs=16'));
返回此几何图形:
POLYGON((-122.2099 37.76,-122.209900481527 37.759990198286,-122.209901921472 37.7599804909678,-122.209904305966 37.7599709715323,-122.209907612047 37.7599617316568,-122.209911807874 37.7599528603263,-122.209916853039 37.7599444429767,-122.209922698955 37.7599365606716,-122.209929289322 37.7599292893219,-122.209936560672 37.7599226989547,-122.209944442977 37.7599168530388,-122.209952860326 37.7599118078736,-122.209961731657 37.7599076120467,-122.209970971532 37.7599043059664,-122.209980490968 37.759901921472,-122.209990198286 37.7599004815273,-122.21 37.7599,-122.210009801714 37.7599004815273,-122.210019509032 37.759901921472,-122.210029028468 37.7599043059664,-122.210038268343 37.7599076120467,-122.210047139674 37.7599118078736,-122.210055557023 37.7599168530388,-122.210063439328 37.7599226989547,-122.210070710678 37.7599292893219,-122.210077301045 37.7599365606716,-122.210083146961 37.7599444429767,-122.210088192126 37.7599528603263,-122.210092387953 37.7599617316568,-122.210095694034 37.7599709715323,-122.210098078528 37.7599804909678,-122.210099518473 37.759990198286,-122.2101 37.76,-122.210099518473 37.760009801714,-122.210098078528 37.7600195090322,-122.210095694034 37.7600290284677,-122.210092387953 37.7600382683432,-122.210088192126 37.7600471396737,-122.210083146961 37.7600555570233,-122.210077301045 37.7600634393284,-122.210070710678 37.7600707106781,-122.210063439328 37.7600773010453,-122.210055557023 37.7600831469612,-122.210047139674 37.7600881921264,-122.210038268343 37.7600923879533,-122.210029028468 37.7600956940336,-122.210019509032 37.760098078528,-122.210009801714 37.7600995184727,-122.21 37.7601,-122.209990198286 37.7600995184727,-122.209980490968 37.760098078528,-122.209970971532 37.7600956940336,-122.209961731657 37.7600923879533,-122.209952860326 37.7600881921264,-122.209944442977 37.7600831469612,-122.209936560672 37.7600773010453,-122.209929289322 37.7600707106781,-122.209922698955 37.7600634393284,-122.209916853039 37.7600555570233,-122.209911807874 37.7600471396737,-122.209907612047 37.7600382683432,-122.209904305966 37.7600290284677,-122.209901921472 37.7600195090322,-122.209900481527 37.760009801714,-122.2099 37.76))
![enter image description here](https://istack.dev59.com/j4sdA.webp)
如果你想了解这个缓冲区的形状,请阅读答案。
使用此查询,您可以检查另一个几何对象是否在此缓冲区内(ST_Within
):
db=# SELECT
ST_Within('POINT(-122.21 37.76)'::GEOMETRY,
ST_Buffer('POINT(-122.21 37.76)'::GEOMETRY,0.0001, 'quad_segs=16'));
st_within
t
(1 Zeile)
为了将所有内容整合在一起,您可以使用
CTE
(也称为 WITH 子句),并编写以下内容:
WITH j AS (
SELECT
(ST_DumpPoints('LINESTRING(-122.21 37.77,-157.82 21.29,178.43 -18.14,153.02 -27.46)'::GEOMETRY)).geom AS g)
SELECT id, name, address, geom
FROM Seattle_Starbucks
WHERE ST_Within(geom,ST_Buffer(j.g,0.0001, 'quad_segs=16'))