为什么我不能在函数内部保存来自Scanner的用户输入？

huangapple
go评论158阅读模式
英文:

Why can't I save user input from Scanner inside a function?

问题

我有一个Phone类，只有一个字段：Contact类的实例的ArrayList。Contact类只有两个字段：名称和号码。
当你运行程序时，

  • 它会显示编号的指令
  • 你输入一个数字
  • 它使用switch语句调用执行相应操作的函数，如下所示：
  1. switch (num) {
  2.     case 0:
  3.         printInstructions();
  4.         break;
  5.     case 1:
  6.         addContact();
  7.         break;
  8. }

switch语句位于一个while循环中。
我已经将提示用户输入信息（例如名称）的代码放在了函数内部：

  1. public static void addContact() {
  2.     System.out.print("Name: ");
  3.     String name = scanner.nextLine();
  4.     scanner.nextLine();
  5.     
  6.     System.out.print("Number: ");
  7.     String number = scanner.nextLine();
  8.     phone.addContact(name, number);
  9. }

phone.addContact()：

  1. public void addContact(String name, String number) {
  2.     // 确保联系人不存在的检查
  3.     contacts.add(new Contact(name, number));
  4. }

我的问题是：我无法将我输入的名称保存下来。无论我输入什么，name的值始终是空字符串。
更奇怪的是，number保存正确。为什么会发生这种情况呢？

编辑：而且，如果我去掉循环、switch语句和函数，并在main()中直接运行代码，它就能正常工作。

（我还需要分享其他代码吗？）

有人要求我包含Contact类，所以——

  1. public class Contact {
  2.     private String name;
  3.     private String number;
  4.     
  5.     public Contact(String name, String number) {
  6.         this.name = name;
  7.         this.number = number;
  8.         System.out.println("Contact created with name " + name + " and number " + number);
  9.     }
  10.     
  11.     public Contact(String name) {
  12.         this(name, null);
  13.     }
  14.     
  15.     public Contact() {
  16.         this("J. Doe", null);
  17.         System.out.println("ERROR: blank contact somehow generated");
  18.     }
  19.     
  20.     public String getName() {
  21.         return name;
  22.     }
  23.     
  24.     public String getNumber() {
  25.         return number;
  26.     }
  27.     
  28.     @Override
  29.     public String toString() {
  30.         return "Contact { \n" +
  31.                 "    name = \"" + name + "\",\n" +
  32.                 "    number = \"" + number + "\"" +
  33.                 "\n}";
  34.     }
  35. }

以及while循环：

  1. while (true) {
  2.     System.out.print("Enter an instruction number: ");
  3.     num = phoneScnr.nextInt();
  4.     switch (num) {
  5.         case 0:
  6.             printInstructions();
  7.             break;
  8.         case 1:
  9.             addContact();
  10.             break;
  11.         case 2:
  12.             editContact();
  13.             break;
  14.         case 3:
  15.             removeContact();
  16.             break;
  17.         case 4:
  18.             findContact();
  19.             break;
  20.         case 5:
  21.             listContacts();
  22.             break;
  23.         case -1:
  24.             quit = true;
  25.             break;
  26.         default:
  27.             System.out.println("Sorry, that character is not recognized.");
  28.             break;
  29.     }
  30. }
英文:

I have a Phone class that has only one field: an ArrayList of instances of the Contact class. The Contact class has only two fields: name and number.
<br> When you run the program,

  • It displays numbered instructions
  • You enter a number
  • It calls the function the performs the corresponding action using a switch statement, like so:
  2.     switch (num) {
  3.         case 0:
  4.             printInstructions();
  5.             break;
  6.         case 1:
  7.             addContact();
  8.             break;
  9.     }

The switch statement is in a while loop.
I've put the code to prompt the user for information (like a name) inside the functions:

  2.     public static void addContact() {
  3.         System.out.print(&quot;Name: &quot;);
  4.         String name = scanner.nextLine();
  5.         scanner.nextLine();
  6.     
  7.         System.out.print(&quot;Number: &quot;);
  8.         String number = scanner.nextLine();
  9.         phone.addContact(name, number);
  10.     }

phone.addContact():

  2.     public void addContact(String name, String number) {
  3.         //there is a check to make sure the contact doesn&#39;t already exist
  4.         contacts.add(new Contact(name, number));
  5.     }

My problem: I can't get it to save the name I enter. The value of name stays an empty string no matter what I enter.
What's stranger is that the number is saved correctly. Why is this happening?

<br>Edit: Also, It works fine if I get rid of the loop and the switch statement and the function and run the code directly in main().

(Is there any other code I should share?)

I was asked to include the Contact class, so —

  2.     public class Contact {
  3.         private String name;
  4.         private String number;
  5.     
  6.         public Contact(String name, String number) {
  7.             this.name = name;
  8.             this.number = number;
  9.             System.out.println(&quot;Contact created with name &quot; + name + &quot; and number &quot; + number);
  10.         }
  11.     
  12.         public Contact(String name) {
  13.             this(name, null);
  14.         }
  15.     
  16.         public Contact() {
  17.             this(&quot;J. Doe&quot;, null);
  18.             System.out.println(&quot;ERROR: blank contact somehow generated&quot;);
  19.         }
  20.     
  21.         public String getName() {
  22.             return name;
  23.         }
  24.     
  25.         public String getNumber() {
  26.             return number;
  27.         }
  28.     
  29.         @Override
  30.         public String toString() {
  31.             return &quot;Contact { \n&quot; +
  32.                     &quot;    name = \&quot;&quot; + name + &quot;\&quot;,\n&quot; +
  33.                     &quot;    number = \&quot;&quot; + number + &quot;\&quot;&quot; +
  34.                     &quot;\n}&quot;;
  35.         }
  36.     }

And the while loop:

  2.     while (true) {
  3.                 System.out.print(&quot;Enter an instruction number: &quot;);
  4.                 num = phoneScnr.nextInt();
  5.                 switch (num) {
  6.                     case 0:
  7.                         printInstructions();
  8.                         break;
  9.                     case 1:
  10.                         addContact();
  11.                         break;
  12.                     case 2:
  13.                         editContact();
  14.                         break;
  15.                     case 3:
  16.                         removeContact();
  17.                         break;
  18.                     case 4:
  19.                         findContact();
  20.                         break;
  21.                     case 5:
  22.                         listContacts();
  23.                         break;
  24.                     case -1:
  25.                         quit = true;
  26.                         break;
  27.                     default:
  28.                         System.out.println(&quot;Sorry, that character is not recognized.&quot;);
  29.                         break;
  30.                 }
  31.             }

答案1

得分: 4

请不要在读取姓名后再次读取扫描仪的行：

  1. public static void addContact() {
  2.     System.out.print("姓名：");
  3.     String name = scanner.nextLine();
  4.     //scanner.nextLine(); <-- 这里是原因
  6.     System.out.print("号码：");
  7.     String number = scanner.nextLine();
  8.     phone.addContact(name, number);
  9. }
英文:

try not reading the scanner's Line again after reading the Name :

  1. public static void addContact() {
  2.     System.out.print(&quot;Name: &quot;);
  3.     String name = scanner.nextLine();
  4.     //scanner.nextLine(); &lt;-- here is the reason
  6.     System.out.print(&quot;Number: &quot;);
  7.     String number = scanner.nextLine();
  8.     phone.addContact(name, number);
  9. }

huangapple
  • 本文由 发表于 2020年8月23日 19:28:55
  • 转载请务必保留本文链接：https://go.coder-hub.com/63546395.html
  • java
  • java.util.scanner
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定