公交车路线查询系统后台数据库设计--引入步行路线
在《查询算法》和《关联地名和站点》两篇文章中,已经实现了通过地名或站点进行路线查询的算法,但是在现实中,从起点到终点不一定全程都是乘车,例如,有以下3条路线:
R1: S1->S2->S3->S4->S5
R2: S6->S7->S2->S8
R3: S8->S9->S10
假如现在要从站点S1到S7,如果用Inquiry查询路线,显然没有合适的乘车方案。但是S2和S7相距仅仅一个站的距离,可以用步行代替,因此可以先从S1乘坐R1到S2再步行到S7。
为了实现在乘车路线中插入步行路线,在数据库使用WalkRoute(StartStop, EndStop, Distance, Remark)(StartStop-起始站点,EndStop-目的站点,Distance-距离,Remark-备注)储存距离较近的两个站点。
加入表WalkRoute后,查询算法也要作相应的修改,其实WalkRoute和RouteT0很相似,因此只需把WalkRoute看成是特殊的直达线路即可,修改后的InqueryT1如下:
/* 查询站点@StartStops到站点@EndStops之间的一次换乘乘车路线,多个站点用'/'分开,如: exec InquiryT1 '站点1/站点2','站点3/站点4' */ CREATE proc InquiryT1(@StartStops varchar(32),@EndStops varchar(32)) as begin declare @ss_tab table(name varchar(32)) declare @es_tab table(name varchar(32)) insert @ss_tab select Value from dbo.SplitString(@StartStops,'/') insert @es_tab select Value from dbo.SplitString(@EndStops,'/') if(exists(select * from @ss_tab sst,@es_tab est where sst.name=est.name)) begin raiserror ('起点集和终点集中含有相同的站点',16,1) return end declare @stops table(name varchar(32)) insert @stops select name from @ss_tab insert @stops select name from @es_tab declare @result table( StartStop varchar(32), Route1 varchar(256), TransStop varchar(32), Route2 varchar(256), EndStop varchar(32), StopCount int ) declare @count int set @count=0 --查询"步行-乘车"路线 insert @result select sst.name as StartStop, '从'+r1.StartStop+'步行到'+r1.EndStop as Route1, r1.EndStop as TransStop, r2.Route as Route2, est.name as EndStop, r2.StopCount as StopCount from @ss_tab sst, @es_tab est, (select * from WalkRoute where EndStop not in (select name from @stops)) r1, RouteT0 r2 where sst.name=r1.StartStop and r1.EndStop=r2.StartStop and r2.EndStop=est.name order by r2.StopCount set @count=@@rowcount --查询"乘车-步行"路线 insert @result select sst.name as StartStop, r1.Route as Route1, r1.EndStop as TransStop, '从'+r2.StartStop+'步行到'+r2.EndStop as Route2, est.name as EndStop, r1.StopCount as StopCount from @ss_tab sst, @es_tab est, RouteT0 r1, (select * from WalkRoute where StartStop not in (select name from @stops)) r2 where sst.name=r1.StartStop and r1.EndStop=r2.StartStop and r2.EndStop=est.name order by r1.StopCount set @count=@count+@@rowcount if(@count=0) begin --查询"乘车-乘车"路线 insert @result select sst.name as StartStop, r1.Route as Route1, r1.EndStop as TransStop, r2.Route as Route2, est.name as EndStop, r1.StopCount+r2.StopCount as StopCount from @ss_tab sst, @es_tab est, (select * from RouteT0 where EndStop not in (select name from @stops)) r1, RouteT0 r2 where sst.name=r1.StartStop and r1.EndStop=r2.StartStop and r2.EndStop=est.name and r1.Route<>r2.Route order by r1.StopCount+r2.StopCount end select StartStop as 起始站点, Route1 as 路线1, TransStop as 中转站点, Route2 as 路线2, EndStop as 目的站点, StopCount as 总站点数 from @result end
作者:卢春城
E-mail:mrlucc@126.com
出处:http://lucc.cnblogs.com/
本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接。
E-mail:mrlucc@126.com
出处:http://lucc.cnblogs.com/
本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接。