我正在尝试创建一个输入验证，它不会让你输入同一个数字两次，也不会让你在一个数字范围内，除非它是整数，否则什么都不能输入。我目前正在创建一个彩票计划，我不确定如何做到这一点。任何帮助都将不胜感激。我的数字范围验证有效，但其他两个验证无效。我尝试了不重复的数字验证，但我不确定如何进行仅数字验证。有人能告诉我如何构造这个吗。
这个方法在我的播放器类中
public void choose() {
int temp = 0;
for (int i = 0; i<6; i++) {
System.out.println("Enter enter a number between 1 & 59");
temp = keyboard.nextInt();
keyboard.nextLine();
while ((temp<1) || (temp>59)) {
System.out.println("You entered an invalid number, please enter a number between 1 and 59");
temp = keyboard.nextInt();
keyboard.nextLine();
}
if (i > 0) {
while(temp == numbers[i-1]) {
System.out.println("Please enter a different number as you have already entered this");
temp = keyboard.nextInt();
keyboard.nextLine();
}
}
numbers[i] = temp;
}
}
按如下操作：
import java.util.Arrays;
import java.util.Scanner;
public class Main {
static int[] numbers = new int[6];
static Scanner keyboard = new Scanner(System.in);
public static void main(String args[]) {
// Test
choose();
System.out.println(Arrays.toString(numbers));
}
static void choose() {
int temp;
boolean valid;
for (int i = 0; i < 6; i++) {
// Check if the integer is in the range of 1 to 59
do {
valid = true;
System.out.print("Enter in an integer (from 1 to 59): ");
temp = keyboard.nextInt();
if (temp < 1 || temp > 59) {
System.out.println("Error: Invalid integer.");
valid = false;
}
for (int j = 0; j < i; j++) {
if (numbers[j] == temp) {
System.out.println("Please enter a different number as you have already entered this");
valid = false;
break;
}
}
numbers[i] = temp;
} while (!valid); // Loop back if the integer is not in the range of 1 to 100
}
}
}
样本运行：
Enter in an integer (from 1 to 59): 100
Error: Invalid integer.
Enter in an integer (from 1 to 59): -1
Error: Invalid integer.
Enter in an integer (from 1 to 59): 20
Enter in an integer (from 1 to 59): 0
Error: Invalid integer.
Enter in an integer (from 1 to 59): 4
Enter in an integer (from 1 to 59): 5
Enter in an integer (from 1 to 59): 20
Please enter a different number as you have already entered this
Enter in an integer (from 1 to 59): 25
Enter in an integer (from 1 to 59): 6
Enter in an integer (from 1 to 59): 23
[20, 4, 5, 25, 6, 23]
对于测试数字数组中存在的值，请使用Arrays.asList(numbers).contains(temp)
如果您使用ArrayList来存储数字，可能会更好。
我会递归重写该方法以避免多个循环。
如果您不熟悉递归方法，那么它基本上是一个在方法内部调用自己的方法。通过使用巧妙的参数，可以将递归方法用作循环。例如
void loop(int index) {
if(index == 10) {
return; //End loop
}
System.out.println(index);
loop(index++);
}
通过呼叫CCD_ 2，将打印数字1至9。
在您的情况下，递归方法可能看起来像
public void choose(int nbrOfchoices, List<Integer> taken) {
if(nbrOfChoices < 0) {
return; //Terminate the recursively loop
}
System.out.println("Enter enter a number between 1 and 59");
try {
int temp = keyboard.nextInt(); //Scanner.nextInt throws InputMismatchException if the next token does not matches the Integer regular expression
} catch(InputMismatchException e) {
System.out.println("You need to enter an integer");
choose(nbrOfChoices, taken);
return;
}
if (value < 1 || value >= 59) { //Number not in interval
System.out.println("The number " + temp + " is not between 1 and 59.");
choose(nbrOfChoices, taken);
return;
}
if (taken.contains(temp)) { //Number already taken
System.out.println("The number " + temp + " has already been entered.");
choose(nbrOfChoices, taken);
return;
}
taken.add(temp);
choose(nbrOfChoices--, taken);
}
现在，您可以通过调用select(您的NumberOfchoices，您的ArrayList take(来递归启动该方法。如果你想改变你的数字间隔，你也可以很容易地添加两个附加参数。
您想要做的是使用递归，这样您就可以要求他们再次提供输入。您可以定义选项而不是6。您可以将maxExclusive定义为59(在本例中为60(。由于集只能包含唯一的非null值，因此可以跟踪选定的整数值集。在每次select调用结束时，我们再次调用select，只剩下1个选项，而不是for循环。在每个方法调用开始时，我们检查选项是否<0，如果是，我们将阻止执行。
public void choose(Scanner keyboard, int choices, int maxExclusive, Set<Integer> chosen) {
if (choices <= 0) {
return;
}
System.out.println("Enter enter a number between 1 & " + (maxExclusive - 1));
int value = keyboard.nextInt();
keyboard.nextLine();
if (value < 1 || value >= maxExclusive) {
System.out.println("You entered an invalid number.");
choose(keyboard, choices, maxExclusive, chosen);
return;
}
if (chosen.contains(value)) {
System.out.println("You already entered this number.");
choose(keyboard, choices, maxExclusive, chosen);
return;
}
chosen.add(value);
choose(keyboard, --choices, maxExclusive, chosen);
}
choose(new Scanner(System.in), 6, 60, new HashSet<>());
我希望这会有所帮助，如果是，请投票支持
import java.util.ArrayList;
import java.util.Scanner;
public class Test {
private ArrayList<String> choose() {
Scanner scanner = new Scanner(System.in);
ArrayList<String> alreadyEntered = new ArrayList<>(6); // using six because your loop indicated me that you're taking six digits
for(int i = 0 ; i < 6 ; ++i){ // ++i is more efficient than i++
System.out.println("Enter a number between 1 & 59");
String digit;
digit = scanner.nextLine().trim();
if(digit.matches("[1-5][0-9]|[0-9]" && !alreadyEntered.contains(digit))// it checks if it is a number as well as if it is in range as well if it is not already entered, to understand this learn about regular expressions
alreadyEntered.add(digit);
else {
System.out.println("Invalid input, try again");
--i;
}
}
return alreadyEntered // return or do whatever with the numbers as i am using return type in method definition i am returning
}
scanner.close();
}
使用字符串的arraylist只是为了让事情变得简单，否则我将不得不从整数到字符串和字符串到整数进行一些解析