扫描仪：扫描int或字符串

Question

试图使用扫描仪来获取int或字符串..下面的代码就是我所拥有的：

System.out.println("where would you like to send the following item: ");
System.out.println("1. groceries, 2. savings, 3. utilities");
System.out.print("Enter list name or list number to send: ");
String user = user_input.nextLine();
      if (user.toLowerCase().equals("groceries") || Integer.parseInt(user) == 1) {
             System.out.println("item was sent to the grocery list");

      }else if(user.toLowerCase().equals("savings") || Integer.parseInt(user) == 2) {
             System.out.println("item was sent to the savings list");

      }else if(user.toLowerCase().contains("utilities") || Integer.parseInt(user) == 3) {
             System.out.println("item was sent to the utilities list");

      }else{
             System.out.println("item is not in any category");

      }

我的问题是每当我输入一个String时，我都会收到一个NumberFormatException错误。输入int并没有给我任何问题。我认为使用String user = user_input.nextLine();获取输入并使用Integer.parseInt(user)将数字字符串转换为if语句中的int将会起作用，但它仍然给我错误消息。

任何帮助是极大的赞赏...

Answer 1

您可以尝试使用user.equals（“1”）而不是Integer.parseInt（user）== 1.或者假设异常java.lang.NumberFormatException有效并捕获该异常并以正确的方式对待它，例如：

    System.out.println("where would you like to send the following item: ");
    System.out.println("1. groceries, 2. savings, 3. utilities");
    System.out.print("Enter list name or list number to send: ");
    String user = user_input.nextLine();
    try {
          if (user.toLowerCase().equals("groceries") || Integer.parseInt(user) == 1) {
                 System.out.println("item was sent to the grocery list");

          }else if(user.toLowerCase().equals("savings") || Integer.parseInt(user) == 2) {
                 System.out.println("item was sent to the savings list");

          }else if(user.toLowerCase().contains("utilities") || Integer.parseInt(user) == 3) {
                 System.out.println("item was sent to the utilities list");

          }else{
                 System.out.println("item is not in any category");

          }
    } catch (NumberFormatException nfe) {
        System.out.println("item is not in any category");
    }

更新：

    System.out.println("where would you like to send the following item: ");
    System.out.println("1. groceries, 2. savings, 3. utilities");
    System.out.print("Enter list name or list number to send: ");
    String user = user_input.nextLine();
    if (user.toLowerCase().equals("groceries") || user.equals("1")) {
        System.out.println("item was sent to the grocery list");

    }else if(user.toLowerCase().equals("savings") || user.equals("2")) {
        System.out.println("item was sent to the savings list");

    }else if(user.toLowerCase().contains("utilities") || user.equals("3")) {
        System.out.println("item was sent to the utilities list");

    }else{
        System.out.println("item is not in any category");

    }

Answer 2

我不是100％肯定，但问题可能是当输入一个字符串时它仍然试图解析一个int，导致它失败。例如，输入“2”是有效的，因为“2”是一个有效的字符串，2是一个有效的int，可以从字符串“2”解析，而“groceries”是一个有效的字符串，但不包含一个有效的int，它可以被解析。

Answer 3

Odd822是对的。问题是当你输入“杂货”以外的任何字符串时，程序会尝试Integer.parseint（“stringwhichisnotaninteger”）。因此，它会引发异常。

这段代码应该有效：

if (user.toLowerCase().equals("groceries") || user.equals("1") ) {
        System.out.println("item was sent to the grocery list");

    }else if(user.toLowerCase().equals("savings") || user.equals("2")) {
        System.out.println("item was sent to the savings list");

    }else if(user.toLowerCase().contains("utilities") || user.equals("3")) {
        System.out.println("item was sent to the utilities list");

    }else{
        System.out.println("item is not in any category");

    }