Question

我正在建立一个公共汽车预订系统。我试图根据所选行程查询公交车。我在表格Departure and Arrival中存储时间。我需要查询出发和抵达。

下面是我的表架构

SELECT * ...

我试过

CREATE TABLE `bus_details` (
  `ID` int(11) NOT NULL,
  `Route` varchar(60) NOT NULL,
  `RouteCode` int(11) NOT NULL,
  `BusCode` int(11) NOT NULL,
  `CityCode` int(11) NOT NULL,
  `City` varchar(20) NOT NULL,
  `Departure` time DEFAULT NULL,
  `Arrival` time DEFAULT NULL,
  `FromCityCode` int(11) NOT NULL,
  `ToCityCode` int(11) NOT NULL,
  `BusName` varchar(30) NOT NULL,
  `sValid` int(11) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `bus_details` (`ID`, `Route`, `RouteCode`, `BusCode`, `CityCode`, `City`, `Departure`, `Arrival`, `FromCityCode`, `ToCityCode`, `BusName`, `sValid`) VALUES
(48, 'Accra Mall - Papaye', 10001, 1001, 101, 'Accra Mall', '01:00:00', NULL, 101, 101, 'Sprinter', 1),
(49, 'Accra Mall - Papaye', 10001, 1001, 102, 'Flower Pot', '00:30:00', '01:15:00', 101, 102, 'Sprinter', 0),
(50, 'Accra Mall - Papaye', 10001, 1001, 103, 'Palace', '02:00:00', '00:45:00', 102, 103, 'Sprinter', 0),
(51, 'Accra Mall - Papaye', 10001, 1001, 104, 'Papaye', NULL, '02:30:00', 103, 104, 'Sprinter', 1),
(52, 'Accra Mall - Papaye', 10001, 1003, 101, 'Accra Mall', '02:00:00', NULL, 101, 101, 'VVIP Bus', 1),
(53, 'Accra Mall - Papaye', 10001, 1003, 102, 'Flower Pot', '02:30:00', '02:15:00', 101, 102, 'VVIP Bus', 0),
(54, 'Accra Mall - Papaye', 10001, 1003, 103, 'Palace', '03:00:00', '02:45:00', 102, 103, 'VVIP Bus', 0),
(55, 'Accra Mall - Papaye', 10001, 1003, 104, 'Papaye', NULL, '03:15:00', 103, 104, 'VVIP Bus', 1);

ALTER TABLE `bus_details`
  ADD PRIMARY KEY (`ID`);

这与我预期的答案很接近，但我的预期结果是4，因为这次旅行只有两辆公共汽车。

在四个结果中，两个是正确的，两个不是。

请帮助我正确查询此操作。

提前谢谢

SELECT DISTINCT(t1.BusCode), t1.BusName, t1.CityCode, t1.FromCityCode, t2.ToCityCode, t1.Departure, t2.Arrival
FROM
    (SELECT BusName, BusCode, CityCode, FromCityCode, ToCityCode, Departure From bus_details Where CityCode IN(101) AND FromCityCode IN(101) Group By BusCode) As t1,
    (SELECT BusName, BusCode, CityCode, FromCityCode, ToCityCode, Arrival From bus_details Where CityCode IN(104) AND ToCityCode IN(104) Group By BusCode) As t2

这是我想要输出的样本。再次感谢

Answer 1

您的查询的问题在于您使用的JOIN没有条件，因此它创建了2条公交线路的交叉乘积x 2条公交线路= 4条结果。如果您有3条路线，您将获得9条结果。如果您在SELECT中添加了t2.BusName，则会看到与t1.BusName不同的所有情况。您需要通过添加条件来限制结果，该条件确保两条路线上的总线相同，即t1.BusCode = t2.BusCode（或t1.BusName = t2.BusName）

SELECT t1.BusName, t1.FromCityCode AS tripFrom, t2.ToCityCode AS tripTo, t1.Departure, t2.Arrival
FROM
    (SELECT BusName, BusCode, CityCode, FromCityCode, ToCityCode, Departure 
     FROM bus_details 
     WHERE CityCode IN(101) AND FromCityCode IN(101) 
     GROUP BY BusCode) As t1
JOIN
    (SELECT BusName, BusCode, CityCode, FromCityCode, ToCityCode, Arrival 
     FROM bus_details 
     WHERE CityCode IN(104) AND ToCityCode IN(104)
     GROUP BY BusCode) As t2
ON t1.BusCode = t2.BusCode

输出（Demo）：

BusName     tripFrom    tripTo  Departure   Arrival
Sprinter    101         104     01:00:00    02:30:00
VVIP Bus    101         104     02:00:00    03:15:00

MySQL从同一个表中选择多个

1 个答案: