读取字符串 next() 和 nextLine() Java [英] Reading Strings next() and nextLine() Java

问题描述

问题是当我尝试拆分 (.split" ") 每个空格时，我无法使用 next() 读取变量输入，然后数组只获取我输入的前两个单词，因此我不得不使用 keyboard.nextLine() 并且拆分过程按照它应该的方式工作，我得到了数组中的所有单词，但问题是如果我使用 nextLine() 那么我必须创建另一个键盘对象来读取第一个变量(答案)，那就是我可以让它在这里工作的唯一方法是代码

 Scanner keyboard=new Scanner(System.in);Scanner keyboard2=new Scanner(System.in);//只是为了让回答词int answer=keyboard.nextInt();//如果我在这里不使用keyboard2，那么程序将无法正常工作，但是如果我在那里使用next()而不是nextLine，这将不是问题，但是那么拆分部分是一个问题(这个变量计算程序将拥有的行数).int电流=1；int left=0,right=0,forward=0,back=0;for(int count=0;count<answer;count++,current++){字符串输入=keyboard.nextLine();字符串数组[]=input.split(" ");for (int counter=0;counter

谢谢:)

解决方案

将 keyboard.nextLine() 放在这一行之后:

int answer=keyboard.nextInt();

这是一个常见的问题，当你在Scanner类的nextInt()方法之后使用nextLine()方法时通常会发生.>

实际发生的情况是，当用户在 int answer = keyboard.nextInt(); 处输入一个整数时，扫描器将只取数字并留下换行符 .所以你需要通过调用 keyboard.nextLine(); 来做一个技巧，只是为了丢弃那个换行符，然后你可以调用 String input = keyboard.nextLine(); 没有任何问题.

The problem is I cant read the variable input with next() cause when I try to split (.split" ") every whitespace then the array just get the first two words I type so I had to use keyboard.nextLine() and the splitting process works the way it should work and I get all the words in the array but the problem is that If I use nextLine() then I have to create another keyboard object to read the first variable (answer) and that is the only way I can make it work here is the code

    Scanner keyboard=new Scanner(System.in);
    Scanner keyboard2=new Scanner(System.in);//just to make answer word

    int answer=keyboard.nextInt();//if I don't use the keyboard2 here then the program will not work as it should work, but if I use next() instead of nextLine down there this will not be a problem but then the splitting part is a problem(this variable counts number of lines the program will have).

    int current=1;
    int left=0,right=0,forward=0,back=0;

    for(int count=0;count<answer;count++,current++)
    {
        String input=keyboard.nextLine();
        String array[]=input.split(" ");
        for (int counter=0;counter<array.length;counter++)
        {
            if (array[counter].equalsIgnoreCase("left"))
            {
                left++;
            }
            else if (array[counter].equalsIgnoreCase("right"))
            {     
                right++;
            }
            else if (array[counter].equalsIgnoreCase("forward"))
            {
                forward++;   
            }
            else if (array[counter].equalsIgnoreCase("back"))
            {     
                back++;
            }
        }

   }
}

Thanks :)

解决方案

Put keyboard.nextLine() after this line:

int answer=keyboard.nextInt();

This is a common problem that usually happens when you use nextLine() method after nextInt() method of Scanner class.

What actually happens is that when the user enters an integer at int answer = keyboard.nextInt();, the scanner will take the digits only and leave the new-line character . So you need to do a trick by calling keyboard.nextLine(); just to discard that new-line character and then you can call String input = keyboard.nextLine(); without any problem.

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