Question

我使用过Scanner var = input.nextLine（）; 它适用于我的名字变量，但不适用于地址？这是我的代码：

Scanner input = new Scanner(System.in);

System.out.println("What is your name?");
String name = input.nextLine();

System.out.println("How old are you?");
int age = input.nextInt();

System.out.println("Where do you live?");
String address = input.nextLine();

System.out.println("Your name is " + name + ".\n You are " + age + " years old." + "\n You live at " + address + ".");

以下是它的显示内容：

What is your name?
Yassine assim
How old are you?
17
Where do you live?
Your name is Yassine assim.
 You are 17 years old.
 You live at .

Answer 1

int age = input.nextInt();会消耗您键入的int，但是，如果此消息（返回行）尚未消耗，您肯定会返回到行

所以地址的input.nextLine();会直接执行

2个选项：

int age = Integer.parseInt(input.nextLine());占据所有行并转换为int
int age = input.nextInt(); input.nextLine();接受int并使用，然后使用return返回

Java：只输出字符串的第一个字？

1 个答案: