import java.util.Scanner;
public class day08zy {
public static void main(String[] args) {
int size = 50;
String[] orderNameArray = new String[size];//菜品名称
String[] orderNumberArray = new String[size];//菜品数量
String[] orderPriceArray = new String[size];//菜品价格
String[] orderTimeArray = new String[size];//送餐时间
String[] orderWayArray = new String[size];//送餐地址
String[] orderDianzanArray = new String[size];//点赞数
String[] orderPhoneArray = new String[size];//送餐电话
String[] orderManArray = new String[size];
orderNameArray[0] = “惠林顿牛排”;
orderNumberArray[0] = “2”;
orderPriceArray[0] = “500”;
orderTimeArray[0] = “2021年7月19日”;
orderWayArray[0] = “汤臣一品”;
orderDianzanArray[0] = “100”;
orderPhoneArray[0] = “15002711507”;
orderManArray[0] = “川哥”;
boolean isExit = false;
while (!isExit) {
Scanner sc = new Scanner(System.in);
System.out.println("********【欢迎来到在线点餐系统】**********");
System.out.println("1.我要订餐");
System.out.println("2.查看餐袋");
System.out.println("3.签收订单");
System.out.println("4.删除订单");
System.out.println("5.我要点赞");
System.out.println("6.删除订单");
String choice = sc.nextLine();
switch (choice) {
case "1":
int tempIndex = -1;
//循环遍历数组,查看应该存在的位置,第一次 null 值出现的位置
for (int i = 0; i < orderNameArray.length; i++) {
//如果为null,当前的下标位置就是 i 的值
if (orderNameArray[i] == null) {
tempIndex = i;
//跳出for循环
break;
}
}
//判断一下
if (tempIndex == -1) {
System.out.println("订单已装满...");
//跳出switch语句
break;
}
//键盘录入数据啦
System.out.println("请输入订单的菜品名称:");
orderNameArray[tempIndex] = sc.nextLine();
System.out.println("请输入订单的数量:");
orderNumberArray[tempIndex] = sc.nextLine();
System.out.println("请输入订单的价格:");
orderPriceArray[tempIndex] = sc.nextLine();
System.out.println("请输入订单的时间:");
orderTimeArray[tempIndex] = sc.nextLine();
System.out.println("请输入送餐的地址:");
orderTimeArray[tempIndex] = sc.nextLine();
System.out.println("请输入姓名:");
orderManArray[tempIndex] = sc.nextLine();
System.out.println("请输入电话:");
orderPhoneArray[tempIndex] = sc.nextLine();
System.out.println("下单成功!");
break;
case "2":
//判断内容是否全部是 null 值,如果全部是 null 值,则表示没有商品,如果没有商品,后续的代码不再执行
int count = 0;
for (int i = 0; i < orderNameArray.length; i++) {
if (orderNameArray[i] == null) {
count++;
}
}
//最后当 count == size 也就是20相等的话,表示全部是null
if (count == size) {
System.out.println("当前没有添加订单,请先添加订单");
break;
}
System.out.println("编号\t菜品名称\t菜品数量\t菜品价格\t订餐时间\t订餐人地址\t订餐人姓名\t订餐人联系电话");
//定义总价格
double sum = 0;
for (int i = 0; i < orderNameArray.length; i++) {
String name = orderNameArray[i];
if (name == null) {
continue;
}
String number = orderNumberArray[i];
String price = orderPriceArray[i];
String time = orderTimeArray[i];
String Way = orderWayArray[i];
String Man = orderManArray[i];
String Phone = orderPhoneArray[i];
String index = (i + 1) + "";
Double s1 =Double.parseDouble(orderNumberArray[i]);
Double s2 =Double.parseDouble(orderPriceArray[i]);
sum += s1*s2;
System.out.println(index + "\t\t" + name + "\t\t" + number + "\t\t" + price + "\t" + time + "\t" + Way + "\t" + Man + "\t\t" + Phone);
}
System.out.println("商品总价格:¥" + sum + "元");
break;
case "3":
System.out.println("已完成");
break;
case "4":
System.out.println("请输入要删除的订单号:");
Scanner scanner = new Scanner(System.in);
int b = scanner.nextInt();
orderNameArray[b - 1] = null;
orderNumberArray[b - 1] = null;
orderPriceArray[b - 1] = null;
orderTimeArray[b - 1] = null;
orderWayArray[b - 1] = null;
orderDianzanArray[b - 1] = null;
orderPhoneArray[b - 1] = null;
orderManArray[b - 1] = null;
System.out.println("订单删除成功");
break;
case "5":
//判断内容是否全部是 null 值,如果全部是 null 值,则表示没有商品,如果没有商品,后续的代码不再执行
int count1 = 0;
for (int i = 0; i < orderNameArray.length; i++) {
if (orderNameArray[i] == null) {
count1++;
}
}
//最后当 count == size 也就是20相等的话,表示全部是null
if (count1 == size) {
System.out.println("当前没有添加订单,请先添加订单");
break;
}
System.out.println("请输入要点赞的菜品序号:");
System.out.println("编号\t菜品名称\t菜品价格");
Scanner sca = new Scanner(System.in);
int d = sca.nextInt();
int one = Integer.parseInt( orderDianzanArray[d-1]);
orderDianzanArray[d-1]+=1;
System.out.println("点赞成功");
break;
case "6":
default:
System.out.println("程序退出");
isExit = true;
break;
}
}
}
}
运行结果:版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
本文链接:https://blog.csdn.net/Zyw907155124/article/details/118913451
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