Question

只是想知道你们是否可以帮我解决这个错误，我一直在努力修复过去一小时左右。

我在这段代码中调用带有错误的函数：

static Scanner scan = new Scanner(System.in);
static int xIn, yIn;
static String name = "";
static String input = "";

public static void main(String[] args){         
    if(args.length < 2){
        System.out.println("Not enough arguments. Temination.");
        System.exit(0);
    }
    else if(args.length > 2){
        System.out.println("Too many arguments. Termination.");
        System.exit(0);
    }
    BottyBot.Test(Integer.parseInt(args[0]), Integer.parseInt(args[1]));
    BottyBot.Greet();
}

BottyBot.Greet（）函数是：

public static void Greet(){
    name = scan.nextLine("Hello, what is your name? ");
    do{
        input = scan.nextLine("Would you like to order some boots, " + name + "? (y/n) ");
        if(input == "y"){
            System.out.println("Great! Let's get started.");
            break;
        }
        else if(input == "n"){
            System.out.println("Come back next time, " + name + ".");
            System.exit(0);
        }
        else{
            System.out.println("Invalid response. Try again.");
        }
    }
    while(true);
}

我在两个扫描仪行都遇到两个错误。具体来说，就是说

error: method nextLine in class Scanner cannot be applied to given types;
    next = scan.nextLine("Hello, what is your name? ");
               ^
required: no arguments
found: String
reason: actual and formal argument lists differ in length

如果有人能指出我做错了什么，那就太棒了！谢谢！

Answer 1

您应该这样写，因为scanner.nextLine()方法不接受任何参数

System.out.println("Hello, what is your name? ");
name = scan.nextLine();

打印部件由System.out.println()完成，扫描部件由scanner.nextLine()完成。

error: method nextLine in class Scanner cannot be applied to given types;
next = scan.nextLine("Hello, what is your name? ");
           ^
required: no arguments
found: String
reason: actual and formal argument lists differ in length

此消息为您提供了明确的提示。它声明actual and formal argument lists differ in length。所以这里的参数表示方法参数，必需提到no arguments，而是找到String，即＆＃34;你好，你叫什么名字？＆＃34;

Answer 2

我认为你会混淆Scanner所做的事情。这让我想起了QBasic中的input函数，尽管我可能不正确。

如错误所述，nextLine()方法不接受任何参数，但是您将字符串传递给它。根据我的理解，您的目的是打印到控制台并从中读取。为此，您需要将扫描仪语句拆分为两个。鉴于此：

name = scan.nextLine("Hello, what is your name? ");

把它变成：

System.out.println("Hello, what is your name?");    //Print to console.
name = scan.nextLine();                             //Read from it.

另一方也是如此。

Answer 3

Scanner.nextLine()不接受任何参数，它只是读一行。

你可以用这个：

System.out.println("Hello, what is your name?");
name = scan.nextLine();

和

System.out.println("Would you like to order some boots, " + name + "? (y/n) ");
input = scan.nextLine();

必需：没有参数，找到：String

3 个答案: