Question

WITH hi AS (
  SELECT ps.id, ps.brgy_locat, ps.municipali, ps.bldg_name, fh.gridcode, ps.bldg_type
  FROM evidensapp_polystructures ps
  JOIN evidensapp_floodhazard fh ON fh.gridcode=3
                                 AND ST_Intersects(fh.geom, ps.geom)
), med AS (
  SELECT ps.id, ps.brgy_locat, ps.municipali ,ps.bldg_name, fh.gridcode, ps.bldg_type
  FROM evidensapp_polystructures ps
  JOIN evidensapp_floodhazard fh ON fh.gridcode=2
                                 AND ST_Intersects(fh.geom, ps.geom)
  EXCEPT SELECT * FROM hi
), low AS (
  SELECT ps.id, ps.brgy_locat, ps.municipali,ps.bldg_name, fh.gridcode, ps.bldg_type
  FROM evidensapp_polystructures ps
  JOIN evidensapp_floodhazard fh ON fh.gridcode=1
                                 AND ST_Intersects(fh.geom, ps.geom)
  EXCEPT SELECT * FROM hi
  EXCEPT SELECT * FROM med
)
SELECT brgy_locat, municipali, bldg_name,  bldg_type, gridcode, count( bldg_name)
FROM (SELECT brgy_locat, municipali, bldg_name, gridcode, bldg_type
      FROM hi
      GROUP BY 1, 2, 3, 4, 5) cnt_hi
FULL JOIN (SELECT brgy_locat, municipali,bldg_name, gridcode, bldg_type
      FROM med
      GROUP BY 1, 2, 3, 4, 5) cnt_med USING (brgy_locat, municipali, bldg_name,gridcode,bldg_type)
FULL JOIN (SELECT brgy_locat, municipali,bldg_name,gridcode, bldg_type
      FROM low
      GROUP BY 1, 2, 3, 4, 5) cnt_low USING (brgy_locat, municipali, bldg_name, gridcode, bldg_type)

上面的查询返回错误：

错误：列＆＃34; cnt_hi.brgy_locat＆＃34;必须出现在GROUP BY子句中   或用于聚合函数   **********错误**********

错误：列＆＃34; cnt_hi.brgy_locat＆＃34;必须出现在GROUP BY子句中   或者在聚合函数SQL状态中使用：42803

但是，如果我省略count(bldg_name)它的工作原理。但我需要根据bldg_name计算。

修改我想获得与危险值（gridcode）相交的建筑物数量：High（3），Medium（2）和Low（1）。但是，如果某个几何体已经在“高”中相交，则在其中排除“中”查询，同样与“低”相同，排除那些在“高”和“中”中相交的几何体。

PostgreSQL：9.4，PostGIS：2.1.7

表格详情：

CREATE TABLE evidensapp_floodhazard (
  id integer NOT NULL DEFAULT nextval('evidensapp_floodhazard_id_seq'::regclass),
  gridcode integer 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;

预期输出是这样但具有正确的计数：

编辑2： 尽管我尽力解释预期的输出是什么，但无论如何：

计算bldg_name而不是id，其中floodhazard中的网格代码与 EDIT 1 上面提到的条件相交。
然后将其分组为brgy_locat，brgy_municipali以及它所属的gridcode和bldg_type。

请看一下上面的图片。

Answer 1

你可能想要这个：

WITH hi AS (
   SELECT ps.brgy_locat, ps.municipali, ps.bldg_name, ps.bldg_type, fh.gridcode
        , count(*) OVER(PARTITION BY ps.bldg_name, ps.bldg_type) AS building_count
   FROM   evidensapp_polystructures ps
   JOIN   evidensapp_floodhazard    fh ON fh.gridcode = 3
                                      AND ST_Intersects(fh.geom, ps.geom)
   )
, med AS (
   SELECT ps.brgy_locat, ps.municipali, ps.bldg_name, ps.bldg_type, fh.gridcode
        , count(*) OVER(PARTITION BY ps.bldg_name, ps.bldg_type) AS building_count
   FROM   evidensapp_polystructures ps
   JOIN   evidensapp_floodhazard    fh ON fh.gridcode = 2
                                      AND ST_Intersects(fh.geom, ps.geom)
   LEFT   JOIN hi USING (bldg_name, bldg_type)
   WHERE  hi.bldg_name IS NULL
   )
TABLE hi

UNION ALL
TABLE med

UNION ALL 
   SELECT ps.brgy_locat, ps.municipali, ps.bldg_name, ps.bldg_type, fh.gridcode
        , count(*) OVER(PARTITION BY ps.bldg_name, ps.bldg_type) AS building_count
   FROM   evidensapp_polystructures ps
   JOIN   evidensapp_floodhazard    fh ON fh.gridcode = 1
                                      AND ST_Intersects(fh.geom, ps.geom)
   LEFT   JOIN hi USING (bldg_name, bldg_type)
   LEFT   JOIN med USING (bldg_name, bldg_type)
   WHERE  hi.bldg_name IS NULL
   AND    med.bldg_name IS NULL;

根据您对问题和聊天的评论，现在按 (bldg_name, bldg_type) 计算 - 排除已在较高级别相交的建筑物 - 再次基于(bldg_name, bldg_type)。

所有其他列与计数（id，geom，...）不同（brgy_locat，municipali）或功能相关的噪音。 如果不是，请在PARTITION BY子句中添加更多列以消除建筑物的歧义。并将相同的列添加到JOIN条件的USING子句中。

如果某个建筑物与evidensapp_floodhazard中的多个行相交且相同 gridcode，那么它会被计算多次。见另一击。

由于您实际上并不想聚合行而只是依靠分区，因此关键功能是将count()用作window function，而不是像原始版本那样使用聚合函数。基本解释：

Best way to get result count before LIMIT was applied

count(*)在这方面做得更好：

使用LEFT JOIN / IS NULL代替EXCEPT。详细说明：

Select rows which are not present in other table

我没有在外部查询中看到FULL JOIN的目的。改为使用UNION ALL。

替代查询

这会计算构建一次，无论它在同一网格代码级别与evidensapp_floodhazard相交多少次

此外，此变体（与第一个不同！）假定同一(bldg_name, bldg_type)的所有行在同一网格代码级别上匹配，这可能是也可能不是这样：

SELECT brgy_locat, municipali, bldg_name, bldg_type, 3 AS gridcode
     , count(*) OVER(PARTITION BY bldg_name, bldg_type) AS building_count
FROM   evidensapp_polystructures ps
WHERE  EXISTS (
   SELECT 1 FROM evidensapp_floodhazard fh
   WHERE  fh.gridcode = 3 AND ST_Intersects(fh.geom, ps.geom)
   )

UNION ALL
SELECT brgy_locat, municipali, bldg_name, bldg_type, 2 AS gridcode
     , count(*) OVER(PARTITION BY bldg_name, bldg_type) AS building_count
FROM   evidensapp_polystructures ps
WHERE  EXISTS (
   SELECT 1 FROM evidensapp_floodhazard fh
   WHERE  fh.gridcode = 2 AND ST_Intersects(fh.geom, ps.geom)
   )
AND    NOT EXISTS (
   SELECT 1 FROM evidensapp_floodhazard fh
   WHERE  fh.gridcode > 2  -- exclude matches on **all** higher gridcodes
   AND    ST_Intersects(fh.geom, ps.geom)
   )

UNION ALL 
SELECT brgy_locat, municipali, bldg_name, bldg_type, 1 AS gridcode
     , count(*) OVER(PARTITION BY bldg_name, bldg_type) AS building_count
FROM   evidensapp_polystructures ps
WHERE  EXISTS (
   SELECT 1 FROM evidensapp_floodhazard fh
   WHERE  fh.gridcode = 1 AND ST_Intersects(fh.geom, ps.geom)
   )
AND    NOT EXISTS (
   SELECT 1 FROM evidensapp_floodhazard fh
   WHERE  fh.gridcode > 1 AND ST_Intersects(fh.geom, ps.geom)
   );

还演示了没有CTE的变体，根据数据分布，这些变体可能会或可能不会表现得更好。

索引

将gridcode添加到索引可能会提高性能。（未使用PostGis进行测试）：

您需要首先安装附加模块btree_gist。详细说明：

Equivalent to exclusion constraint composed of integer and range

CREATE INDEX evidensapp_floodhazard_geom_id
  ON evidensapp_floodhazard USING gist (gridcode, geom);

Answer 2

错误是要求您在<Directory /path/to/home*> Order deny,allow Deny from all </Directory> <Directory /path/to/home/subdirectory-for-dev*> Order deny,allow Deny from all allow from 111.222.333.444 #developer 1's ip allow from 222.222.333.444 #developer 2's ip #etc.. for other developers </Directory>子句中包含选择列表列;你可以这样做

GROUP BY

Answer 3

我不知道这对你有用，因为我对postgresql知之甚少。也不确定这是否会给你你想要的东西。但是，试一试。您只需要在using子句中包含building_count。

SELECT brgy_locat, municipali, bldg_name,  bldg_type, 
gridcode, building_count
FROM (SELECT brgy_locat, municipali, bldg_name, gridcode, bldg_type,
      count( bldg_name) as building_count
      FROM hi
      GROUP BY 1, 2, 3, 4, 5) cnt_hi
FULL JOIN (SELECT brgy_locat, municipali,bldg_name, gridcode, bldg_type,
      count(bldg_name) as building_count
      FROM med
      GROUP BY 1, 2, 3, 4, 5) cnt_med 
USING (brgy_locat, municipali, bldg_name,gridcode,bldg_type, building_count)
FULL JOIN (SELECT brgy_locat, municipali,bldg_name,gridcode, bldg_type,
      count(bldg_name) as building_count
      FROM low
      GROUP BY 1, 2, 3, 4, 5) cnt_low 
USING (brgy_locat, municipali, bldg_name, gridcode, bldg_type, building_count);

我不是在追求声誉。我刚刚更新了Rahul的回答。希望能帮助到你。干杯! ：）

PostgreSQL：查询不使用count工作

3 个答案:

替代查询

索引