读取字符串next()和nextLine()Java

Question

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),这是唯一的方法我可以让它在这里工作是代码

    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 :)

Answer 1

放在keyboard.nextLine()这一行之后:

int answer=keyboard.nextInt();

这是当你使用通常发生在一个共同的问题nextLine()之后方法nextInt()的方法Scanner类.

实际发生的情况是,当用户输入整数时int answer = keyboard.nextInt();,扫描仪将仅采用数字并保留换行符\n.所以你需要通过调用keyboard.nextLine();只是为了丢弃新行字符然后你可以String input = keyboard.nextLine();毫无问题地调用.