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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);

我有一个孩子，它使用这两个表中的外键：

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
);

有没有一种方法可以让我获取唯一的预订 ID 并参考客户几何和影院几何来计算 ST_Distance？我假设某种嵌套查询可以做到这一点，但没有运气？

干杯

更新（来自评论）

我尝试了以下代码：

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
      )
   ))

但是得到一个错误，说“用作表达式的子查询返回了不止一行”有什么想法可以解决这个问题吗？理想情况下，我希望它返回每个预订 ID 的距离

4

1 回答 1

1

使用主键/外键和小写名称更正了架构。

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
        );

用于 3 向连接的骨架：

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，使用骨架的结果作为函数参数（我对 GIS 不是很流利，这只是一个演示语法的示例）：

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
        ;
于 2013-02-17T11:42:08.123 回答