使用st_distance父母几何

Question

我有两个父表如下

CREATE TABLE GISD.CUSTOMERS 
(CUSTOMER_ID INTEGER NOT NULL, 
FIRST_NAME VARCHAR (15) NOT NULL, 
SURNAME VARCHAR (20) NOT NULL, 
DATE_OF_BIRTH DATE NOT NULL, 
HOUSE_NUMBER VARCHAR (5) NOT NULL      
POST_CODE VARCHAR(8) NOT NULL, 
STREET VARCHAR (25) NOT NULL, 
TOWN VARCHAR (25) NOT NULL 
); 

SELECT ADDGEOMETRYCOLUMN('gisd','customers', 'customers_geom', '27700','POINT',2); 

和

CREATE TABLE GISD.CINEMAS 
(CINEMA_ID INTEGER NOT NULL, 
CINEMA_NAME VARCHAR(25) NOT NULL, 
ADDRESS_NUMBER INTEGER NOT NULL, 
POST_CODE VARCHAR(8) NOT NULL, 
STREET VARCHAR (25) NOT NULL, 
TOWN VARCHAR (25) NOT NULL, 
OPENING_TIME TIME NOT NULL, 
CLOSING_TIME TIME NOT NULL 
); 

SELECT ADDGEOMETRYCOLUMN('gisd','cinemas', 'cinemas_geom', '27700','POLYGON',2); 
SELECT ADDGEOMETRYCOLUMN('gisd','cinemas', 'centroid', '27700','POINT',2); 

我有一个使用外键从这两个表作为这样的孩子：

CREATE TABLE GISD.BOOKING 
(BOOKING_ID INTEGER NOT NULL, 
CUSTOMER_ID INTEGER NOT NULL, 
CINEMA_ID INTEGER NOT NULL, 
TIME TIME NOT NULL, 
DATE DATE NOT NULL, 
FILM VARCHAR(50) NOT NULL, 
BOOKING_METHOD VARCHAR (15) NOT NULL, 
BOOKING_FEE NUMERIC NOT NULL -- Numeric is suggested by postgresql.org for currency 
TICKET_PRICE NUMERIC NOT NULL 
); 

是否有一种方法可以让我获取唯一的预订ID并参考客户几何图形和电影地理metry来计算ST_Distance？我假设某种嵌套查询可以做到这一点，但没有运气？

欢呼

UPDATE（从评论）

我曾尝试下面的代码：

SELECT (ST_DISTANCE(
    (SELECT centroid 
     FROM GISD.CINEMAS 
     INNER JOIN GISD.BOOKING ON CINEMAS.CINEMA_ID=BOOKING.CINEMA_ID 
    ),(
     SELECT customers_geom 
     FROM GISD.CUSTOMERS 
     INNER JOIN GISD.BOOKING ON CUSTOMERS.CUSTOMER_ID=BOOKING.CUSTOMER_ID 
    ) 
    )) 

却得到一个错误说“由子查询返回多行作为一种表达'任何想法如何绕过这个？理想情况下，我希望它返回每个预订ID的距离。

Answer 1

更正了具有主键/外键和LOWERCASED名称的模式。

DROP SCHEMA tmp CASCADE; 
CREATE SCHEMA tmp; 
SET search_path=tmp; 

CREATE TABLE tmp.customers 
     (customer_id INTEGER NOT NULL PRIMARY KEY 
     , first_name VARCHAR (15) NOT NULL 
     , surname VARCHAR (20) NOT NULL 
     , date_of_birth DATE NOT NULL 
     , house_number VARCHAR (5) NOT NULL 
     , post_code VARCHAR(8) NOT NULL 
     , street VARCHAR (25) NOT NULL 
     , town VARCHAR (25) NOT NULL 
     ); 

SELECT addgeometrycolumn('tmp','customers', 'customers_geom', '27700','POINT',2); 

-- and 

CREATE TABLE tmp.cinemas 
     (cinema_id INTEGER NOT NULL PRIMARY KEY 
     , cinema_name VARCHAR(25) NOT NULL 
     , address_number INTEGER NOT NULL 
     , post_code VARCHAR(8) NOT NULL 
     , street VARCHAR (25) NOT NULL 
     , town VARCHAR (25) NOT NULL 
     , opening_time TIME NOT NULL 
     , closing_time TIME NOT NULL 
     ); 

SELECT addgeometrycolumn('tmp','cinemas', 'cinemas_geom', '27700','POLYGON',2); 
SELECT addgeometrycolumn('tmp','cinemas', 'centroid', '27700','POINT',2); 

-- I have a junction table between these tables as such: 

CREATE TABLE tmp.BOOKING 
     (booking_id INTEGER NOT NULL PRIMARY KEY 
     , customer_id INTEGER NOT NULL REFERENCES tmp.customers (customet_id) 
     , cinema_id INTEGER NOT NULL REFERENCES tmp.cinemas (cinema_id) 
     , zdatetime timestamp NOT NULL 
     , film VARCHAR(50) NOT NULL 
     , booking_method VARCHAR (15) NOT NULL 
     , booking_fee NUMERIC NOT NULL 
     , ticket_price NUMERIC NOT NULL 
     ); 

骨架3路连接：

SELECT bo.booking_id, bo.zdatetime, bo.film 
     , ci.cinemas_geom 
     , ci.centroid 
     , cu.customers_geom 
FROM booking bo 
JOIN customers cu ON cu.customer_id = bo.customer_id 
JOIN cinemas ci ON ci.cinema_id = bo.cinema_id 
     ; 

现在，只需添加函数调用，使用从skeletton作为函数参数的结果（我不是在GIS流利，这只是一个示例来演示语法）：

SELECT bo.booking_id, bo.zdatetime, bo.film 
     , st_distance(ci.centroid , cu.customers_geom) AS the_distance 
FROM booking bo 
JOIN customers cu ON cu.customer_id = bo.customer_id 
JOIN cinemas ci ON ci.cinema_id = bo.cinema_id 
     ;