Question

在第一个循环中，它没有将第一个单词作为输入。当我没有从用户那里输入字符串长度时，它工作正常。查看未输入单词0的输出。

public class Stringinput {
        public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            System.out.println("How many words do you need to make the sentence: ");
            int n = in.nextInt();
            String[] names = new String[n];

            for (int i = 0; i < names.length; i++) {

                System.out.printf("Enter word no %d : ",i);
                names[i]= in.nextLine();
            }
            System.out.println("The sentence you made is: ");
            for (int i = 0; i < names.length; i++) {
                System.out.print(names[i]+" ");

            }
            System.out.println("");
        }
    }

这就是输出：

How many words do you need to make the sentence: 
4
Enter word no 0 : Enter word no 1 : My
Enter word no 2 : name
Enter word no 3 : is
The sentence you made is: 
 My name is

建立成功（总时间：11秒）

Answer 1

使用.next()代替.nextLine()，因为Scanner.nextInt方法不使用上一个换行符字符，因此您只需要一个单词，而不是句子。

for (int i = 0; i < names.length; i++) {

            System.out.printf("Enter word no %d : ",i);
            names[i]= in.next();
        }

输出：

How many words do you need to make the sentence: 
4
Enter word no 0 : my
Enter word no 1 : next
Enter word no 2 : r
Enter word no 3 : t
The sentence you made is: 
my next r t

Answer 2

nextInt()不会消耗在该号码后输入的换行符号，因此您的第一个nextLine()会这样，所以您需要这样做

public class Stringinput {
        public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            System.out.println("How many words do you need to make the sentence: ");
            int n = in.nextInt();
            in.nextLine(); // consume leftover new line here
            String[] names = new String[n];

            for (int i = 0; i < names.length; i++) {

                System.out.printf("Enter word no %d : ",i);
                names[i]= in.nextLine();
            }
            System.out.println("The sentence you made is: ");
            for (int i = 0; i < names.length; i++) {
                System.out.print(names[i]+" ");

            }
            System.out.println("");
        }
    }

Answer 3

Scanner.nextInt()方法不使用输入的换行符，因此它转到下一个Scanner.nextLine。如果在next（）方法

之后使用nextline，则类似

只需在循环前添加一个虚拟scanner.nextLine()并继续。

从用户获取字符串数组长度。但为什么跳过将第一个单词作为输入？

3 个答案: