在 Java 中检查用户输入是否仅为整数
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Checking for user input to be only integers in Java
提问by user2272351
I'm doing lottery game for my assignment (user inputs 6 number, i will generate 8 unique winning numbers and 2 last numbers are supplementary). I need help with user input checking if input numbers are from 1 to 45 and input must be int, when input is not integer it throws an error.
我正在为我的任务做彩票游戏（用户输入 6 个号码，我将生成 8 个唯一的中奖号码，最后 2 个号码是补充号码）。我需要帮助用户输入检查输入数字是否从 1 到 45 并且输入必须是 int，当输入不是整数时，它会引发错误。
This programming way is procedure way, how can i change it into object oriented way? I know that I must make methods in another java file and then link it back to this main. Can you suggest me how to do it?
这种编程方式是过程方式，我如何才能将其更改为面向对象的方式？我知道我必须在另一个 java 文件中创建方法，然后将它链接回这个 main.js 文件。你能建议我怎么做吗？
I have tried try and catch, if and else (for input check) but i don't know how to check user input when it's in array. Thank you for help.
我尝试过 try 和 catch、if 和 else（用于输入检查），但我不知道如何在数组中检查用户输入。谢谢你的帮助。
Here is my code:
这是我的代码：
class Lottery {
public static void main ( String[] args ) {
System.out.println("\nWelcome to the Lottery game.");
System.out.println("You can enter numbers from 1 to 45.");
// User input into an array
int[] input = new int[6];
Scanner scanner = new Scanner(System.in);
System.out.println("\nPlease enter your 6 lucky numbers: ");
for(int j = 0; j < 6; j++) {
input[j]=scanner.nextInt();
}
int check = scanner.nextInt();
if(check < 0 && check > 45) {
System.out.println("\nERROR: Please enter only numbers from 1 to 45!");
}
// Printing out unique winning numbers from random generator
System.out.println("\nWinning numbers: ");
MultiRandomGenerator mrg = new MultiRandomGenerator();
int[] set;
set = mrg.getSet();
for (int i = 0; i < set.length; i++) {
System.out.print(set[i] + " ");
}
// Loops for counting how many numbers user has guessed right
int count = 0; // for 6 numbers
int scount = 0; // for 2 last supplementary numbers
for(int i = 0; i < input.length; i++) {
for(int k = 0; k < set.length; k++) {
if (k < 6) {
if (set[k] == input[i]) {
count++;
} else {
if (set[k] == input[i]) {
scount++;
}
}
}
}
}
System.out.print("\n\nYou guessed right " + count + " winning numbers.");
System.out.print("\nYou guessed right " + scount + " suplementary numbers.");
// If statments for printing out winning prizes
if (count == 6) {
System.out.println("\nYou have won 1st price!");
} if (count == 5 && scount == 1) {
System.out.println("\nYou have won 2st price!");
} if (count == 5) {
System.out.println("\nYou have won 3st price!");
} if (count == 4) {
System.out.println("\nYou have won 4st price!");
} if (count == 3 && scount == 1) {
System.out.println("\nYou have won 5st price!");
} if (count == 1 && scount == 2) {
System.out.println("\nYou have won 6st price!");
} else {
System.out.println("\nSorry, you didn't won anything.");
}
}
}
回答by Joseph Selvaraj
Sample code to go through array and find the invalid user input.
遍历数组并查找无效用户输入的示例代码。
set = mrg.getSet();
String[] userDataStatus = new String[45];
for (int i = 0; i < set.length; i++)
{
try
{
String inputdata = set.get(i);
if(inputdata != null && inputdata.trim().length() > 0)
{
int currentNumber = Integer.parseInt(userdata);
userDataStatus[i] = "Y";//Y represent valid number
}
}
catch (NumberFormatException ex )
{
userDataStatus[i] = "N";//If it throws exception then save as 'N'
}
}
Use the above String array and display error messaage to users.
使用上面的字符串数组并向用户显示错误消息。
回答by Orn Kristjansson
You can check in your loop, something like
你可以检查你的循环，比如
int val;
try
{
input[j] = Integer.parseInt( scanner.nextString() );
}
catch (NumberFormatException ex )
{
}
