扫描仪输入为何不起作用？

Question

因此，我是一名新Java程序员，我试图弄清楚为什么一段代码无法正常工作。我遇到的问题是：“字符串兴趣=
input.nextLine（）;”，它跳过了用户的输入并跳至下一个System.out，因此它仅在控制台中显示“您的个人资料…”在允许用户输入任何数据之前。抱歉，这是一个愚蠢的问题，我很新！

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();

其余代码可能与您上面的代码相同。

编辑：
它必须是这样的（我的意思是，不将行存储在您的任何变量中），因为nextInt()函数不会读取整行，而只是读取下一个整数。因此，当nextInt()函数读取时int，的“光标”
Scanner将位于之后的位置int。

例如，如果您在尝试读取时输入多个单词（用空格隔开）int，则在您的兴趣变量中，您将nextInt()读取之前无法读取的其余行。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，如果您将代码如下所示：

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();

您现在要拥有的值是：

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

因此，总而言之，如果您有一个nextInt()并且试图读取一个int，则可以读取它，但是光标将停留在该行的最后，您将无法读取下一行。为此，您必须阅读完整的行而不将其存储在中input.nextLine();。

希望对您有帮助！