我正在构建一个巴士预订系统。我正在尝试根据所选行程查询巴士。 我在出发和到达表中存储时间。 我需要查询出发和到达。
下面是我的表架构
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`);
我尝试过
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
这接近我的预期答案,但我返回了 4 个结果,正如我预期的那样,因为这次行程只有两辆巴士。
在四个结果中,两个正确,两个不正确。
请您帮我正确查询此操作。
提前谢谢
**Expected Output**
BusName | tripFrom | tripTo | Departure | Arrival
Sprinter 101 104 1:00:00 2:30:00
VVIP Bus 101 104 2:30:00 3:15:00
这是我想要的输出的示例。 再次感谢
最佳答案
您的查询的问题在于您使用的是无条件的 JOIN
,因此它会创建 2 条巴士路线 x 2 条巴士路线 = 4 个结果的叉积。如果您有 3 条路线,您将得到 9 个结果。如果您在 SELECT 中包含了 t2.BusName
,您就会看到它与 t1.BusName
不同的所有情况。您需要通过添加一个条件来限制结果,以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
关于MySQL 从同一个表中进行多重选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50466288/