PostgreSQL:如果下一个查询与上一个查询相交,则在下一个查询中排除几何体
我用这个查询来获取在另一个几何体中相交的几何体的数量:PostgreSQL:如果下一个查询与上一个查询相交,则在下一个查询中排除几何体,postgresql,postgis,Postgresql,Postgis,我用这个查询来获取在另一个几何体中相交的几何体的数量: SELECT count(evidensapp_polystructures.brgy_locat) AS high, evidensapp_polystructures.brgy_locat AS barangay, evidensapp_polystructures.municipali AS municipality FROM evidensapp_floodhazard INNER JOIN evi
SELECT count(evidensapp_polystructures.brgy_locat) AS high,
evidensapp_polystructures.brgy_locat AS barangay,
evidensapp_polystructures.municipali AS municipality
FROM evidensapp_floodhazard
INNER JOIN evidensapp_polystructures
ON st_intersects(evidensapp_floodhazard.geom, evidensapp_polystructures.geom)
AND evidensapp_floodhazard.hazard= 'High'
GROUP BY evidensapp_polystructures.brgy_locat, evidensapp_polystructures.municipali;
正如您所注意到的,其危险
等于高
我还想得到与危险
值相交的几何体的数量:中等
和低
。但是,如果某个几何体已经在High
中相交,则在Medium
查询中排除,同样的情况也发生在Low
中,排除那些在High
和Medium
中相交的几何体
我有这个想法,可能使用或我需要获得几何体的id
,然后尝试不在查询中,但不知道如何做。可能是因为我对PostgreSQL或任何数据库工作都是新手
以下是上述查询的示例结果:
预期结果应如下所示:
表格详情:
CREATE TABLE evidensapp_floodhazard (
id serial NOT NULL,
hazard character varying(6) NOT NULL,
date_field character varying(60),
geom geometry(MultiPolygon,32651),
CONSTRAINT evidensapp_floodhazard_pkey PRIMARY KEY (id)
);
CREATE INDEX evidensapp_floodhazard_geom_id
ON evidensapp_floodhazard USING gist (geom);
ALTER TABLE evidensapp_floodhazard CLUSTER ON evidensapp_floodhazard_geom_id;
CREATE TABLE evidensapp_polystructures (
id serial NOT NULL,
bldg_name character varying(100) NOT NULL,
bldg_type character varying(50) NOT NULL,
brgy_locat character varying(50) NOT NULL,
municipali character varying(50) NOT NULL,
province character varying(50) NOT NULL,
geom geometry(MultiPolygon,32651),
CONSTRAINT evidensapp_polystructures_pkey PRIMARY KEY (id)
);
CREATE INDEX evidensapp_polystructures_geom_id
ON evidensapp_polystructures USING gist (geom);
ALTER TABLE evidensapp_polystructures CLUSTER ON evidensapp_polystructures_geom_id;
由于在比较方面对字符串“High”、“Medium”和“Low”几乎无能为力,因此必须使用子查询。带有某些CTE的解决方案可能是最干净的:
WITH hi AS (
SELECT ps.id, ps.brgy_locat, ps.municipali
FROM evidensapp_polystructures ps
JOIN evidensapp_floodhazard fh ON fh.hazard = 'High'
AND ST_Intersects(fh.geom, ps.geom)
), med AS (
SELECT ps.id, ps.brgy_locat, ps.municipali
FROM evidensapp_polystructures ps
JOIN evidensapp_floodhazard fh ON fh.hazard = 'Medium'
AND ST_Intersects(fh.geom, ps.geom)
EXCEPT SELECT * FROM hi
), low AS (
SELECT ps.id, ps.brgy_locat, ps.municipali
FROM evidensapp_polystructures ps
JOIN evidensapp_floodhazard fh ON fh.hazard = 'Low'
AND ST_Intersects(fh.geom, ps.geom)
EXCEPT SELECT * FROM hi
EXCEPT SELECT * FROM med
)
SELECT brgy_locat AS barangay, municipali AS municipality, high, medium, low
FROM (SELECT brgy_locat, municipali, count(*) AS high
FROM hi
GROUP BY 1, 2) cnt_hi
FULL JOIN (SELECT brgy_locat, municipali, count(*) AS medium
FROM med
GROUP BY 1, 2) cnt_med USING (brgy_locat, municipali)
FULL JOIN (SELECT brgy_locat, municipali, count(*) AS low
FROM low
GROUP BY 1, 2) cnt_low USING (brgy_locat, municipali);
在三个CTE中,您首先识别属于“高”危险类别的行,然后识别属于“中等”危险类别但除外的行,然后识别属于“低”危险类别的行,除了列为“高”或“中等”的行。然后,在主查询中,将3个CTE与每个CTE的子查询中计算的每个barangay和市政计数合并。使用完全连接
,这样,没有“高”危险等级结构的barangays和市政当局也会出现在结果中。感谢这一点,我如何在一行中得出结果?如果是同一个barangay,则是同一行。您使用的是哪一版本的PG?它是9.3。基于预期输出,结果是正确的,但即使barangay相同,结果也不在一行中。顺便说一句,谢谢你的解释。请参阅更新的答案:使用联接而不是联合。在9.4中,您可以将FILTER
子句与count()
聚合一起使用,这就是我的问题。在提问时,你应该始终指出你的PG版本。好的..谢谢..我需要大约15分钟(874043毫秒)才能得到结果。但无论如何,我会研究更多如何优化它。你的回答将作为我的依据。