这更像是一个SQL问题，而不是PostGIS问题，但我再次陷入困境:(
我在“from”和“to”列中有一个名为id的表。 我想根据邮政编码计算所有这些身份证号码之间的距离。
有一个名为doc的单独参考表，其中包含“NPI”列中的ID号和“提供商业务邮件地址邮政编码”列中的邮政编码和一个名为zctas的单独地理表，其中邮政编码列为zcta和geom列。
例如，此查询可以正常工作：
SELECT z.zcta As zip1,
z2.zcta As zip2,
ST_Distance(z.geom,z2.geom) As thedistance
FROM zctas z,
zctas z2
WHERE z2.zcta = '60611'
AND z.zcta='19611';
一个问题是“提供商商业邮寄地址邮政编码”应该=左（“提供商商业邮寄地址邮政编码”，5）。
在这一个查询中，我一直停留在参考表中加入2个邮政编码。
样本表：
参考表：
from | to | count
------------+------------+-------
1174589766 | 1538109665 | 108
1285653204 | 1982604013 | 31
期望的输出：
from | to | count | distance
------------+------------+----------------
1174589766 | 1538109665 | 108 | 53434
1285653204 | 1982604013 | 31 | 34234
\ d +
Table "public.zctas"
Column | Type | Modifiers | Storage | Stats target | Description
------------------+------------------------+-----------+----------+--------------+-------------
state | character(2) | | extended | |
zcta | character(5) | | extended | |
junk | character varying(100) | | extended | |
population_tot | bigint | | plain | |
housing_tot | bigint | | plain | |
water_area_meter | double precision | | plain | |
land_area_meter | double precision | | plain | |
water_area_mile | double precision | | plain | |
land_area_mile | double precision | | plain | |
latitude | double precision | | plain | |
longitude | double precision | | plain | |
thepoint_lonlat | geometry(Point,4269) | | main | |
thepoint_meter | geometry(Point,32661) | not null | main | |
geom | geometry(Point,32661) | | main | |
Indexes:
"idx_zctas_thepoint_lonlat" gist (thepoint_lonlat)
"idx_zctas_thepoint_meter" gist (thepoint_meter) CLUSTER
Table "public.referred"
Column | Type | Modifiers | Storage | Stats target | Description
--------+-----------------------+-----------+----------+--------------+-------------
from | character varying(25) | | extended | |
to | character varying(25) | | extended | |
count | integer | | plain | |
Has OIDs: no
Table "public.doc"
Column | Type | Modifiers | Storage | Stats target | Description
--------------------------------------------------------------+------------------------+-----------+----------+--------------+-------------
NPI | character varying(255) | | extended | |
Entity Type Code | character varying(255) | | extended | |
Replacement NPI | character varying(255) | | extended | |
Employer Identification Number (EIN) | character varying(255) | | extended | |
Provider Organization Name (Legal Business Name) | character varying(255) | | extended | |
Provider Last Name (Legal Name) | character varying(255) | | extended | |
Provider First Name | character varying(255) | | extended | |
Provider Middle Name | character varying(255) | | extended | |
Provider Name Prefix Text | character varying(255) | | extended | |
Provider Name Suffix Text | character varying(255) | | extended | |
Provider Credential Text | character varying(255) | | extended | |
Provider Other Organization Name | character varying(255) | | extended | |
Provider Other Organization Name Type Code | character varying(255) | | extended | |
Provider Other Last Name | character varying(255) | | extended | |
Provider Other First Name | character varying(255) | | extended | |
Provider Other Middle Name | character varying(255) | | extended | |
Provider Other Name Prefix Text | character varying(255) | | extended | |
Provider Other Name Suffix Text | character varying(255) | | extended | |
Provider Other Credential Text | character varying(255) | | extended | |
Provider Other Last Name Type Code | character varying(255) | | extended | |
g(255) | | extended | |
Provider Second Line Business Mailing Address | character varying(255) | | extended | |
Provider Business Mailing Address City Name | character varying(255) | | extended | |
Provider Business Mailing Address State Name | character varying(255) | | extended | |
Provider Business Mailing Address Postal Code | character varying(255) | | extended | . . . . other columns not really needed.
感谢!!!!
答案 0 :(得分：1)
这应该相对简单。
假设NPI在doc和referred中实际上具有相同的长度，您可以非常轻松地加入这些表：
SELECT ad."Provider Business Mailing Address Postal Code" as a_zip,
bd."Provider Business Mailing Address Postal Code" as b_zip,
r."count"
FROM referred r
LEFT JOIN doc ad ON r."from" = ad."NPI"
LEFT JOIN doc bd ON r."from" = bd."NPI";
显然，请根据对数据中NPI和from / to字段的仔细分析来调整此联接。如有必要，在联接中添加trim或left方法调用 - 最重要的是JOIN条件是可比数据。
现在，从这个到原始查询找到距离是微不足道的：
SELECT ad."Provider Business Mailing Address Postal Code" as a_zip,
bd."Provider Business Mailing Address Postal Code" as b_zip,
r."count",
ST_Distance(az.geom,bz.geom) As thedistance
FROM referred r
LEFT JOIN doc ad ON r."from" = ad."NPI"
LEFT JOIN doc bd ON r."from" = bd."NPI"
LEFT JOIN zctas az
ON az.zcta = left(ad."Provider Business Mailing Address Postal Code",5)
LEFT JOIN zctas bz
ON bz.zcta = left(bd."Provider Business Mailing Address Postal Code",5)
这只是一种应该有效的结构，其他许多都是可行的。这种特殊结构将确保表示referred中的每个条目，即使NPI与doc表中的条目不匹配，或者邮政编码无法匹配zctas表。
另一方面，如果NPI表中doc存在多个条目，那么提及此重复referred的任何NPI条目也将是复制。
同样，如果特定邮政编码（zctas）的zcta条目中有多个条目，您会看到重复的referred行。
这就是LEFT JOIN的工作方式，但我认为值得加入警告，因为提供商数据通常与NPI完全重复，并且邮政编码列表中的邮政编码通常是重复的邮政编码跨国界线。