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，我无法读取带有next（）的变量输入，然后数组只得到我输入的前两个单词，所以我不得不使用keyboard.nextLine（ ）和分裂过程的工作方式应该工作，我得到数组中的所有单词但问题是如果我使用nextLine（）然后我必须创建另一个键盘对象来读取第一个变量（答案），这是我能在这里工作的唯一方法是代码

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++;
            }
        }

   }
}

谢谢：）

推荐答案

放 keyboard.nextLine（）在这一行之后：

int answer=keyboard.nextInt();

当您使用 nextLine（）时，这是一个常见问题/ code> nextInt（） 扫描程序类的方法之后的方法。

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

实际发生的情况是，当用户在 int answer = keyboard.nextInt（）; 中输入一个整数时，扫描仪将仅采用数字和保留换行符 \ n 。所以你需要通过调用 keyboard.nextLine（）; 来放弃那个新行字符，然后你可以调用 String input = keyboard.nextLine（）; 没有任何问题。

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 \n. 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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