Question

我正在开发一个公交车票预订系统。在这里，我已经为所有路线制作了表格，在表格中我添加了诸如bus_number，票价等字段。现在我们的公共汽车每隔一天往返，反之亦然，即如果公共汽车在21-12-2011从X-> Y行驶，则相同的公共汽车将在下一个日期行驶Y-> X.那么如何获得公交车的方向呢？如果我为每个公交服务提供商制作表格并添加公交车号码，请为日期添加“to”标识符，我认为可以知道所有下一天的状态。我不知道这是个好主意，所以请帮助我。

Answer 1

如果不了解现有表格的精确细节，提供最终解决方案有点困难。无论如何，这里有一个建议，你可以如何保持公共汽车以及他们的站点和票价：

CREATE TABLE `bus` (
  `id` int unsigned not null primary key auto_increment,
  `bus_number` varchar(55) not null,
   UNIQUE KEY `busUidx1` (`bus_number`)
) ENGINE=InnoDB;

CREATE TABLE `bus_stop` (
  `id` int unsigned not null primary key auto_increment,
  `stop_description` varchar(250) not null,
  UNIQUE KEY `bus_stopUidx1` (`stop_description`)
) ENGINE=InnoDB;

CREATE TABLE `bus_route` (
  `id` int unsigned not null primary key auto_increment,
  `bus_id` int unsigned not null,
  `route_date` date not null,
  `bus_start_stop_id` int unsigned not null,
  `bus_end_stop_id` int unsigned not null,
  `fare` decimal (10,2) not null,
   UNIQUE KEY `bus_stopUidx1` (`bus_id`,`route_date`),
CONSTRAINT `fk_bus_route_bus_fk1` FOREIGN KEY (`bus_id`) REFERENCES `bus` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `fk_bus_route_stop_fk1` FOREIGN KEY (`bus_start_stop_id`) REFERENCES `bus_stop` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `fk_bus_route_stop_fk2` FOREIGN KEY (`bus_end_stop_id`) REFERENCES `bus_stop` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB;

使用此模型，您应该能够存储总线列表（在bus表中），所有可能的停靠点列表（在bus_stop表中）和给定日期的总线路由。它还可以让您灵活地打破“X＆gt;的公共汽车旅行”。 Y然后总是从Y＆gt;返回X'规则，如果我过去旅行过的公共汽车都可以使用，可能会证明是有用的; - ）

修改

所以这里有一些示例数据可以尝试进一步说明我的答案：

insert into bus (bus_number) values ('Red Bus 1');
insert into bus (bus_number) values ('Red Bus 2');
insert into bus (bus_number) values ('Yellow Bus 1');
insert into bus (bus_number) values ('Yellow Bus 2');

insert into bus_stop (stop_description) values ('Stop 1');
insert into bus_stop (stop_description) values ('Stop 2');
insert into bus_stop (stop_description) values ('Stop 3');
insert into bus_stop (stop_description) values ('Stop 4');

insert into bus_route (bus_id,route_date,bus_start_stop_id,bus_end_stop_id,fare)
values (
(select id from bus where bus_number = 'Red Bus 1'),
'2011-12-11',
(select id from bus_stop where stop_description = 'Stop 1'),
(select id from bus_stop where stop_description = 'Stop 2'),
3.45);

insert into bus_route (bus_id,route_date,bus_start_stop_id,bus_end_stop_id,fare)
values (
(select id from bus where bus_number = 'Red Bus 1'),
'2011-12-12',
(select id from bus_stop where stop_description = 'Stop 2'),
(select id from bus_stop where stop_description = 'Stop 1'),
3.45);

insert into bus_route (bus_id,route_date,bus_start_stop_id,bus_end_stop_id,fare)
values (
(select id from bus where bus_number = 'Yellow Bus 1'),
'2011-12-11',
(select id from bus_stop where stop_description = 'Stop 3'),
(select id from bus_stop where stop_description = 'Stop 4'),
1.95);

insert into bus_route (bus_id,route_date,bus_start_stop_id,bus_end_stop_id,fare)
values (
(select id from bus where bus_number = 'Yellow Bus 1'),
'2011-12-12',
(select id from bus_stop where stop_description = 'Stop 4'),
(select id from bus_stop where stop_description = 'Stop 3'),
1.95);

最后是一个将表连接在一起的查询：

select b.bus_number,
       br.route_date,
       bs.stop_description as start,
       be.stop_description as end,
       br.fare
from bus_route br
inner join bus b on b.id = br.bus_id
inner join bus_stop bs on bs.id = br.bus_start_stop_id
inner join bus_stop be on be.id = br.bus_end_stop_id;

选择每隔一天选择巴士行程方向

1 个答案: