我正在尝试制作一个 if 语句来捕获程序用户是否为程序末尾提出的问题“继续?(y/n):”输入除 y 或 n 之外的值。但无论我输入“y”、“n”或无效的值是什么,我都会从控制台收到消息“无效输入,请重试”。只有当选择不是 y 或 n 时才会发生这种情况有人知道为什么无论我输入什么它都会发生吗?
import java.util.Scanner;
public class ProductApp
{
public static void main(String[] args)
{
//display a welcome message
System.out.println("Welcome to the Product Selector ");
System.out.println();
// perform 1 or more selections
Scanner sc = new Scanner(System.in);
String choice = "y";
while (choice.equalsIgnoreCase("y"))
{
System.out.println("Enter Product Code: ");
String productCode = sc.next(); //read the product code
sc.nextLine() ; //discard any other data entered on the line
//make sure the case the user enters for a product code doesn't matter
productCode = productCode.toLowerCase();
// get the Product object
Product p = ProductDB.getProduct(productCode) ;
// display the output
System.out.println();
if (p != null)
System.out.println(p);
else
System.out.println("No product matches this product code. \n");
System.out.println("Product count: " + Product.getCount() + "\n" );
// see if the user wants to continue
System.out.println("Continue? (y/n): ");
choice = sc.nextLine() ;
System.out.println();
if( !choice.equalsIgnoreCase("y") && !choice.equalsIgnoreCase("n") );
{
System.out.println("Invalid input try again");
continue;
}
}
}
}
此外,无论我收到消息“无效输入,请重试”,程序都会要求一次新输入,然后继续前进,无论它是否有效。如果是“y”,它会再次运行,如果是其他任何内容,它会关闭,而不是再次询问有效输入。
最佳答案
您的条件不起作用,因为 if 语句后有 ;。
if( !choice.equalsIgnoreCase("y") && !choice.equalsIgnoreCase("n") );
^^
改变
if( !choice.equalsIgnoreCase("y") && !choice.equalsIgnoreCase("n") );
{
System.out.println("Invalid input try again");
continue;
}
致
if(!(choice.equalsIgnoreCase("y") || choice.equalsIgnoreCase("n")))
{
System.out.println("Invalid input try again");
continue;
}
关于java - 验证 if 语句不能正常工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15425862/