地铁售票系统设计思想及部分代码
设计思想:地铁售票系统的关键点在于换乘,所以首先要分为换乘和不换乘两种情况。不换乘比较简单,通过起始站名和终点站名查询他们的num,然后list打包输出到jsp就可以。换乘的话就先要找到两条线路,找到两条线路的交点也就是换乘站,然后分别输出起始站到换乘站,换乘站到终点站两段路线就完成了,这里面还涉及到一个最短路径问题,我的想法是把全部的可能线路都找到,然后比较大小就完成了。目前进度到换乘部分。
双人项目合作人:郑锦
部分源代码:
package Dao;
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.Statement;
import connection.DBUtil;
public class dao {
/**
* 通过name得到number(线路号)
* @param name
* @return
*/
public static int getNum(String name) {
String sql = "select xianluhao from aaa where name ='" + name + "'";
Connection conn = DBUtil.getConn();
Statement state = null;
ResultSet rs = null;
int number=0;
try {
state = conn.createStatement();
rs = state.executeQuery(sql);
while (rs.next()) {
number = rs.getInt("xianluhao");
}
} catch (Exception e) {
e.printStackTrace();
} finally {
DBUtil.close(rs, state, conn);
}
return number;
}
/**
* 通过name得到zhanhao(站台号)
* @param name
* @return
*/
public static int getZhanhao(String name) {
String sql = "select num from aaa where name ='" + name + "'";
Connection conn = DBUtil.getConn();
Statement state = null;
ResultSet rs = null;
int zhanhao=0;
try {
state = conn.createStatement();
rs = state.executeQuery(sql);
while (rs.next()) {
zhanhao = rs.getInt("num");
}
} catch (Exception e) {
e.printStackTrace();
} finally {
DBUtil.close(rs, state, conn);
}
return zhanhao;
}
public static String getLine1(int zhanhao1,int zhanhao2) {
String line="";
String sql = "select name from aaa where num between '"+zhanhao1+"' and '"+zhanhao2+"'order by num ASC ";//升序
Connection conn = DBUtil.getConn();
Statement state = null;
ResultSet rs = null;
try {
state = conn.createStatement();
rs = state.executeQuery(sql);
if(rs.next())
line=rs.getString("name");
while (rs.next()) {
String name=rs.getString("name");
line=line+"->"+name;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
DBUtil.close(rs, state, conn);
}
return line;
}
public static String getLine2(int zhanhao1,int zhanhao2) {
String line="";
String sql = "select name from aaa where num between '"+zhanhao1+"' and '"+zhanhao2+"' order by num DESC ";//降序
Connection conn = DBUtil.getConn();
Statement state = null;
ResultSet rs = null;
try {
state = conn.createStatement();
rs = state.executeQuery(sql);
if(rs.next())
line=rs.getString("name");
while (rs.next()) {
String name=rs.getString("name");
line=line+"->"+name;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
DBUtil.close(rs, state, conn);
}
return line;
}}
package servlet;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import Dao.dao;
@WebServlet("/servlet")
public class servlet extends HttpServlet {
private static final long serialVersionUID = 1L;
protected void service(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
req.setCharacterEncoding("utf-8");
String method = req.getParameter("method");
if ("chaxun".equals(method)) {
chaxun(req, resp);
}
}
private void chaxun(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
// TODO Auto-generated method stub
req.setCharacterEncoding("utf-8");
String qi=req.getParameter("qi");
String zhong=req.getParameter("zhong");
int zhanhao1=dao.getZhanhao(qi);
int zhanhao2=dao.getZhanhao(zhong);
int number1=dao.getNum(qi);
int number2=dao.getNum(zhong);
if(number1==number2) {
if(zhanhao1<zhanhao2)
{
String line=dao.getLine1(zhanhao1, zhanhao2);
req.setAttribute("line",line );
req.setAttribute("num",number1);
req.getRequestDispatcher("list.jsp").forward(req,resp);
}if(zhanhao1>zhanhao2)
{
String line=dao.getLine2(zhanhao2, zhanhao1);
System.out.print(line);
req.setAttribute("num",number1);
req.setAttribute("line",line );
req.getRequestDispatcher("list.jsp").forward(req,resp);
}
}}}
实验截图


项目总结分析
这个项目因为涉及到一个最短路径的问题,最开始我是想用迪杰斯特拉算法来解决这个问题。但是我尝试了很久没有成功,可能是我的水平还是太有限。所以最后用了最简单的方法来解决这个问题。我了解到有的同学是用迪杰斯特拉算法完成了这个项目,所以还是去请教一下。后期有时间的话还是想改善一下我程序的算法。
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