Switch 语句使第一种情况跳过一行
所以我猜这个问题的解决方案会非常简单，但我不知道我在寻找什么，所以我需要一些帮助。发生的情况是，当我运行程序并选择情况 1 时，它会打印“狗的名字”和“狗的比赛”，但没有给我机会填写狗的名字。因此，当我选择案例 1 时，我一开始只填写狗的比赛、重量和年龄！这是我正在使用的代码...
do {
System.out.println("(1 - reg\n2 - tail\n3- delete\n4-exit\nEnter number: ");
// so this is where the switch stuff starts
int option=sc.nextInt();
switch (option) {
case 1: System.out.println("Dog's Name: ");
String na=sc.nextLine();
System.out.println("Dog Race: ");
String ra=sc.nextLine();
System.out.println("How heavy?");
double wey=sc.nextDouble();
System.out.println("How old?");
double ag=sc.nextDouble();
dog doggy= new dog(na, ra, wey, ag);
kennel.add(doggy);
break;
case 2: System.out.println("its a tail");
break;
case 3: System.out.println("you delete");
break;
case 4: System.out.println("QUITTING\n(Data was not saved srry.)");
play = false;
default: System.out.println("try again");
}
}while(play);
So I'm guessing that the solution to this is going to be really simple but I have no idea what I'm looking for so I'd like some help. What happens is that when I run the program, and choose case 1. It prints both "dog's name" and "dogs race" without giving me a chance to fill in the dogs name. So when I choose case 1 I start out only getting to fill in dogs race, how heavy, and how old it is! here is the code I'm using...
do {
System.out.println("(1 - reg\n2 - tail\n3- delete\n4-exit\nEnter number: ");
// so this is where the switch stuff starts
int option=sc.nextInt();
switch (option) {
case 1: System.out.println("Dog's Name: ");
String na=sc.nextLine();
System.out.println("Dog Race: ");
String ra=sc.nextLine();
System.out.println("How heavy?");
double wey=sc.nextDouble();
System.out.println("How old?");
double ag=sc.nextDouble();
dog doggy= new dog(na, ra, wey, ag);
kennel.add(doggy);
break;
case 2: System.out.println("its a tail");
break;
case 3: System.out.println("you delete");
break;
case 4: System.out.println("QUITTING\n(Data was not saved srry.)");
play = false;
default: System.out.println("try again");
}
}while(play);
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（3）
我相信您需要在调用
nextInt()之后调用nextLine()，因为这尚未将扫描器推进到下一行。I believe you need to call
nextLine()after your call tonextInt(), because that hasn't advanced the scanner to the next line yet.第一个 sc.nextInt 有换行提醒，您可以将分隔符更改为 \n 或直接调用 nextLine();阅读选项后（使用 sc.useDelimiter("\n") ）
There's a newline reminder from your first sc.nextInt, you can change the delimiter to \n or just call nextLine(); just after reading the option (Using sc.useDelimiter("\n") )
尝试：
int option=Integer.parseInt(sc.nextLine());这具有将光标前进到下一行和获取键入的数字的效果。
Try:
int option=Integer.parseInt(sc.nextLine());This has both the effect of advancing the cursor to the next line and getting the typed number.