为什么扫描仪输入不工作？

Question

因此，我是一个新的Java程序员，我正试图弄清楚为什么一段代码不能工作。我遇到的问题是“String interests=input.nextLine（）；”这一行，它跳过了用户的输入并跳转到下一个System.out，所以它只显示“Your Profile...”。在允许用户输入任何数据之前。抱歉，如果这是一个愚蠢的问题，我是很新的！

System.out.println("Hello, " + name + "! What is your gender?");
String gender = input.nextLine();
System.out.println("Okay, " + name + ", you are a " + gender + ". Now, tell me, what is your age?");
int age = input.nextInt();
System.out.println("Great! We're almost done. What are three interests you have?");
String interests = input.nextLine();

System.out.println("...Your profile...");
System.out.println("Name: " + name);
System.out.println("Gender: " + gender);

Answer 1

这样说：

int age = input.nextInt();
input.nextLine();
System.out.println("Great! We're almost done. What are three interests you have?");
String interests = input.nextLine();

其余的代码可以等于上面的代码。

编辑：它必须是这样的（我的意思是，不将行存储在任何变量中），因为nextInt()函数不读取整行，只读取下一个整数。因此，当nextInt()函数读取int时，扫描器的“光标”将位于int之后的位置。

例如，如果您在尝试读取int时放了多个单词（用空格隔开），则您将在interests变量中读取nextint()之前无法读取的行的其余部分。因此，如果您有：

System.out.println("Okay, " + name + ", you are a " + gender + ". Now, tell me, what is your age?");
//Here you enter --> 18 This is a prove
int age = input.nextInt();
System.out.println("Great! We're almost done. What are three interests you have?");
//Here you won't be able to put anything
String interests = input.nextLine();

现在您将存储：

age = 18;
interests = "This is a prove";

但是如果你这样放代码：

System.out.println("Okay, " + name + ", you are a " + gender + ". Now, tell me, what is your age?");
//Here you enter --> 18 This is a prove
int age = input.nextInt();
input.nextLine();
//Now the Scanner go to a new line (because of the nextLine) but without storing the value
System.out.println("Great! We're almost done. What are three interests you have?");
//Now you are in a new line and you are going to be able to write. For example --> I am reading this.
String interests = input.nextLine();

您现在将要获得的值是：

age = 18;
interests = "I am reading this.";

希望对你有帮助！