在接受用户输入的同时打印格式

发布于 2024-12-22 15:36:27 字数 2122 浏览 1 评论 0原文

我有一项大学作业,需要打印出五金店出售的物品,获取用户的输入,对该输入执行一些计算,然后打印出发票。

我已经能够成功打印出五金店出售的物品,但在接受输入的 while 循环方面遇到问题。

该程序要求用户输入代码,然后询问相应的数量。这在循环的第一次迭代中工作正常,但在第二次迭代中,用户提示“代码:”和“数量:”出现在同一行,尽管我在提示时使用了 println用户。

我非常感谢适合编程新手的详细答复。

这是代码:

 import java.util.Scanner;

 class HardwareStore {


    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);


        System.out.println("WELCOME TO THE HARDWARE STORE!");
        System.out.println("----------------------------------------------------------------------");


        String sticky = "G22";
        String keyring = "K13";
        String screwy = "S21";
        String padlock = "I30";


        int stickyprice = 10989;
        int keyringprice = 5655;
        int screwyprice = 1099;
        int padlockprice = 4005;


        System.out.println("CODE\t\tDESCRIPTION\t\t\t\t\tPRICE");
        System.out.println("----\t\t-----------\t\t\t\t\t-----");

        System.out.println(sticky + "\t\tSTICKY Construction Glue, Heavy Duty, \n\t\t7oz, 12 Pack \t\t\t\t\t$" + stickyprice);
        System.out.println(keyring + "\t\tCAR-LO Key Ring, Quick Release, \n\t\t1 Pack\t\t\t\t\t\t$ " + keyringprice);
        System.out.println(screwy + "\t\t!GREAT DEAL! SCREW-DUP Screwy Screws, \n\t\tDry Wall Screws, 3 in. Long, 50 Pack\t\t$ " + screwyprice);
        System.out.println(padlock + "\t\tLET-IT-RAIN, Weather Proof Padlock, \n\t\tPortable, One Push Functionality\t\t$ " + padlockprice);
        System.out.println("----------------------------------------------------------------------");


        int i = 10000; 
        String [] usercode = new String[i]; 
        int [] userquantity = new int[i];


        System.out.println("PLEASE ENTER YOUR ORDER:");


        while (true) {
            System.out.println("CODE: (X to terminate)");
            usercode[i] = in.nextLine();

            if (usercode[i].equalsIgnoreCase("x")) {
                break;
            }

            System.out.println("QUANTITY: ");
            userquantity[i] = in.nextInt();
        } 
    }
}

I have a college assignment where I need to print out items sold by a hardware store, take input from a user, perform some calculations on that input, and then print out an invoice.

I have been able to successfully print out the items sold by the hardware store, but am encountering problems with the while loop that takes the input.

The program asks the user to enter a CODE and then asks for the corresponding QUANTITY. This works fine on the first iteration of the loop, but on the second iteration the user prompts for "CODE:" and "QUANTITY:" appear on the same line, despite my use of println when prompting the user.

I would greatly appreciate a detailed response appropriate for someone new in programming.

Here's the code:

 import java.util.Scanner;

 class HardwareStore {


    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);


        System.out.println("WELCOME TO THE HARDWARE STORE!");
        System.out.println("----------------------------------------------------------------------");


        String sticky = "G22";
        String keyring = "K13";
        String screwy = "S21";
        String padlock = "I30";


        int stickyprice = 10989;
        int keyringprice = 5655;
        int screwyprice = 1099;
        int padlockprice = 4005;


        System.out.println("CODE\t\tDESCRIPTION\t\t\t\t\tPRICE");
        System.out.println("----\t\t-----------\t\t\t\t\t-----");

        System.out.println(sticky + "\t\tSTICKY Construction Glue, Heavy Duty, \n\t\t7oz, 12 Pack \t\t\t\t\t$" + stickyprice);
        System.out.println(keyring + "\t\tCAR-LO Key Ring, Quick Release, \n\t\t1 Pack\t\t\t\t\t\t$ " + keyringprice);
        System.out.println(screwy + "\t\t!GREAT DEAL! SCREW-DUP Screwy Screws, \n\t\tDry Wall Screws, 3 in. Long, 50 Pack\t\t$ " + screwyprice);
        System.out.println(padlock + "\t\tLET-IT-RAIN, Weather Proof Padlock, \n\t\tPortable, One Push Functionality\t\t$ " + padlockprice);
        System.out.println("----------------------------------------------------------------------");


        int i = 10000; 
        String [] usercode = new String[i]; 
        int [] userquantity = new int[i];


        System.out.println("PLEASE ENTER YOUR ORDER:");


        while (true) {
            System.out.println("CODE: (X to terminate)");
            usercode[i] = in.nextLine();

            if (usercode[i].equalsIgnoreCase("x")) {
                break;
            }

            System.out.println("QUANTITY: ");
            userquantity[i] = in.nextInt();
        } 
    }
}

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(1

红墙和绿瓦 2024-12-29 15:36:27

当您输入 QUANTITY 时,您按下了 Enter 键。 in.nextInt(); 不使用该 newline 字符,它保留在扫描器缓冲区中,直到您滚动到 in.nextLine()< /代码> 再次。

此时,in.nextLine() 会一直读取,直到找到换行符,而该换行符恰好是缓冲区中的下一个字符。因此它再次直接跳到QUANTITY

when you enter the QUANTITY you're pressing enter. That newline character isn't used by in.nextInt();, it remains in the scanner buffer, until you roll around to in.nextLine() again.

At that point in.nextLine() reads until it finds a newline character, which just happens to be the next one in the buffer. So it skips straight to QUANTITY again.

~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文