英文:
Try-Catch gets in a loop for my Scanner - Java
问题
在我的Java程序中，我正在询问玩家的年龄，如果用户在问题提出时输入了一个字符串，我需要显示错误信息checkError(4)，然后再次询问正确的玩家年龄，直到用户输入一个数字为止。
问题在于，当年龄是字符串时，代码陷入循环，然后始终打印字符串“玩家年龄是多少？”我在一些网站上看到，如果在while语句中使用in.hasNextInt()，可以解决这个问题，但在这种情况下，我只有在"check"为true时才运行while循环。
如何解决这个问题？谢谢
check = true;
while (check){
try {
System.out.println("玩家年龄是多少？");
if(in.hasNextInt()){
edad = in.nextInt();
if (edad > 6 && edad < 100){
check = false;
} else {
System.out.println(checkError(3));
}
} else { }
} catch(Exception e) { // 年龄是字符串
System.out.println(checkError(4));
}
}
英文:
In my java program, I am asking which is the age of the player, if the user writes a string when it was made the question I need to show the error checkError(4) and ask again which is the correct age of the player until the user writes a number.
The problem here is that the code gets in a loop when the age is a string and then always prints the string "Age of the player?". I saw in some websites that it can be solved the problem if in the while statement I use in.hasNextInt() but in this case, I am running the while if "check" is true
How can I solve this issue? Thanks
check = true;
while (check){
try {
System.out.println("Age of the player?");
if(in.hasNextInt()){
edad = in.nextInt();
if (edad > 6 && edad < 100){
check = false;
} else {
System.out.println(checkError(3));
}
} else { }
} catch(Exception e) { // Edad is a string
System.out.println(checkError(4));
}
}
答案1
得分: 1
尝试像这样进行：
public static void main(String[] args) {
int edad;
boolean check = true;
Scanner in = new Scanner(System.in);
while (check) {
try {
System.out.println("玩家的年龄是多少？");
if (in.hasNext()) {
String s = in.next(); // 以字符串形式读取
edad = Integer.parseInt(s, 10); // 转换为整数，如果是NaN则抛出异常
if (edad > 6 && edad < 100) {
check = false;
}
} else {
}
} catch (NumberFormatException e) { // edad 是一个字符串
System.out.println(e.toString());
}
}
}
英文:
try it like this :
public static void main(String[] args) {
int edad;
boolean check = true;
Scanner in = new Scanner(System.in);
while (check) {
try {
System.out.println("Age of the player?");
if (in.hasNext()) {
String s = in.next(); //read as string
edad = Integer.parseInt(s, 10); // Convet to int and throw exception if NaN
if (edad > 6 && edad < 100) {
check = false;
}
} else {
}
} catch (NumberFormatException e) { // Edad is a string
System.out.println(e.toString());
}
}
}
答案2
得分: 0
只是更容易（我认为）根本不要使用**Try/Catch**：
public static void main(String[] args) {
int edad = 0;
boolean check = true;
Scanner in = new Scanner(System.in);
while (check) {
System.out.println("玩家的年龄（从6到100岁）？");
String s = in.nextLine(); // 获取用户输入
// 用户提供了整数数值吗？
if (!s.matches("\\d+")) {
// 没有...
System.err.println("无效输入！（" + s + "）。请重试...");
}
// 有...
else {
edad = Integer.parseInt(s, 10); // 转换为整数
if (edad > 6 && edad < 100) { // 检查范围
check = false; // 在范围内。 :)
}
// 超出范围...
else {
System.err.println("年龄（" + edad + "）超出范围！请重试...");
}
}
}
System.out.println("玩家年龄为： --> " + edad);
}
英文:
It's just easier (I think) to not even play with the Try/Catch:
public static void main(String[] args) {
int edad = 0;
boolean check = true;
Scanner in = new Scanner(System.in);
while (check) {
System.out.println("Age of the player (6 to 100 inclusive)?");
String s = in.nextLine(); // Get User input
// Did the User provide an integer numerical value?
if (!s.matches("\\d+")) {
// Nope...
System.err.println("Invalid Entry! (" + s + "). Try Again...");
}
// Yup...
else {
edad = Integer.parseInt(s, 10); // Convert it to integer
if (edad > 6 && edad < 100) { // Check Range
check = false; // is in range. :)
}
// Is out of range...
else {
System.err.println("Age (" + edad + ") is out of range! Try Again...");
}
}
}
System.out.println("Player's Age Is: --> " + edad);
}
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