美好的一天,需要我的车辆检查数据库的帮助。您可以在下面看到结构,您也可以在http://sqlfiddle.com/#!3/4ab7e看到它。我需要的是从每个项目中提取具有至少一(1)个缺陷或违规的车辆数量。在下面的模式中,项目4的总数=两(2)辆车辆和项目9 = 1辆车。
需要的列是[Project_Name],[Vehicle_Type],[yy],[mm],[Total]
-- Vehicle Inspection Database --
-- Vehicle_Type Table
CREATE TABLE VehicleType
([VehicleTypeId] int,
[Type] varchar (36));
INSERT INTO VehicleType ([VehicleTypeId],[Type])
VALUES (1, 'Light Vehicle'),
(2, 'Tanker'),
(3, 'Goods');
-- Car Table
CREATE TABLE Vehicle
([VehicleID] varchar(36),
[PlateNo] varchar(36),
[VehicleTypeId] int,
[Project] int);
INSERT INTO Vehicle ([VehicleID], [PlateNo],[VehicleTypeId], [Project])
VALUES('A57D4151-BD49-4B44-AF10-000F1C298E05', '8112AG', 1, 4),
('C7095628-AE88-4DD0-A4FD-00363EAB767F', '60070 AD2', 2, 9),
('E714CCD7-E56C-46A8-89D5-003CA5BF6094', '68823 AD1', 3, 9);
-- Event Table
CREATE TABLE Event
([EventID] int,
[VehicleID] varchar(36),
[EventTime] smalldatetime,
[TicketStatus] varchar (10)) ;
INSERT INTO Event([EventID], [VehicleID], [EventTime], TicketStatus)
VALUES (1, 'A57D4151-BD49-4B44-AF10-000F1C298E05', '20130701', 'Open'),
(2, 'A57D4151-BD49-4B44-AF10-000F1C298E05', '20130702', 'Close'),
(3, 'A57D4151-BD49-4B44-AF10-000F1C298E05', '20130703', 'Close'),
(4, 'C7095628-AE88-4DD0-A4FD-00363EAB767F','20130705', 'Open'),
(5, 'C7095628-AE88-4DD0-A4FD-00363EAB767F','20130710', 'Open');
-- Event_Defects Table
CREATE TABLE EventDefects
([EventDefectsID] int,
[EventID] int,
[Status] varchar(15),
[DefectID] int) ;
INSERT INTO EventDefects ([EventDefectsID], [EventID], [Status], [DefectID])
VALUES
-- 1st Inspection for PlateNo. 8112AG
(1, 1, 'YES', 1),
(2, 1, 'NO', 2),
(3, 1, 'YES',3),
(4, 1, 'N/A', 4),
(5, 1, 'N/A', 5),
-- 2nd Inspection for PlateNo. 8112AG
(6, 2, 'NO', 1),
(7, 2, 'NO', 2),
(8, 2, 'NO', 3),
(9, 2, 'N/A', 4),
(10,2, 'N/A', 5),
-- 3rd Inspection for PlateNo. 8112AG
(11, 3, 'NO', 1),
(12, 3, 'NO', 2),
(13, 3, 'NO', 3),
(14, 3, 'NO', 4),
(15, 3, 'NO', 5),
-- 1st Inspection for PlateNo. 60070 AD2
(16, 3, 'NO', 1),
(17, 3, 'NO', 2),
(18, 3, 'NO', 3),
(19, 3, 'N/A', 4),
(20, 3, 'N/A', 5);
-- Defects Table
CREATE TABLE Defects
([DefectID] int,
[DefectsName] varchar (36),
[DefectClassID] int) ;
INSERT INTO Defects ([DefectID], [DefectsName], [DefectClassID])
VALUES (1, 'TYRE', 1),
(2, 'BRAKING SYSTEM', 1),
(3, 'MIRRORS AND WINDSCREEN', 2),
(4, 'OVER SPEEDING', 3),
(5, 'NOT WEARING SEATBELTS', 3);
-- Defect_Class Table
CREATE TABLE DefectClass
([Description] varchar (15),
[DefectClassID] int) ;
INSERT INTO DefectClass ([DefectClassID], [Description])
VALUES (1, 'CATEGORY A'),
(2, 'CATEGORY B'),
(3, 'CATEGORY C');
答案 0 :(得分:0)
将所有连接作为内部连接进行...并且它将消除空的记录。
你能检查一下这对你有用吗?
SELECT Vehicle.VehicleID from Vehicle
INNER JOIN Event ON Vehicle.VehicleID = Event.VehicleID
INNER JOIN EventDefects ON EventDefects.EventID = Event.EventID
INNER JOIN Defects ON EventDefects.DefectID = Defects.DefectID
GROUP BY Vehicle.VehicleID;