Java 如何检查用户输入是否不是 int 值

How to check if user input is not an int value

提问by jjkk0990

I need to check if a user input value is not an int value. I've tried different combinations of what I know but I either get nothing or random errors

我需要检查用户输入值是否不是 int 值。我尝试了我所知道的不同组合，但我要么一无所获，要么出现随机错误

For example:

例如：

If the user inputs "adfadf 1324" it'll raise a warning message.

如果用户输入“adfadf 1324”，它会发出警告消息。

---

---

What I have:

我拥有的：

       // Initialize a Scanner to read input from the command line
       Scanner sc = new Scanner(System.in);
       int integer, smallest = 0, input;
       boolean error = false;

       System.out.print("Enter an integer between 1-100: ");
       range = sc.nextInt();

       if(!sc.hasNextInt()) {

          error = true;
          System.out.println("Invalid input!");
          System.out.print("How many integers shall we compare? (Enter an integer between 1-100: ");
          sc.next();
    }

       while(error) {
          for(int ii = 1; ii <= integer; ii++) {

              ...

          } // end for loop
      }
      System.out.println("The smallest number entered was: " + smallest);

      }
  }

采纳答案by Pandiyan Cool

Simply throw Exception if input is invalid

如果输入无效，只需抛出异常

Scanner sc=new Scanner(System.in);
try
{
  System.out.println("Please input an integer");
  //nextInt will throw InputMismatchException
  //if the next token does not match the Integer
  //regular expression, or is out of range
  int usrInput=sc.nextInt();
}
catch(InputMismatchException exception)
{
  //Print "This is not an integer"
  //when user put other than integer
  System.out.println("This is not an integer");
}

回答by ElliotSchmelliot

Taken from a related post:

摘自相关帖子：

public static boolean isInteger(String s) {
    try { 
        Integer.parseInt(s); 
    } catch(NumberFormatException e) { 
        return false; 
    }
    // only got here if we didn't return false
    return true;
}

回答by Rajendra_Prasad

Try this one:

    for (;;) {
        if (!sc.hasNextInt()) {
            System.out.println(" enter only integers!: ");
            sc.next(); // discard
            continue;
        }
        choose = sc.nextInt();
        if (choose >= 0) {
            System.out.print("no problem with input");

        } else {
            System.out.print("invalid inputs");

        }
    break;
  }

回答by Rajendra_Prasad

you have following errors which in turn is causing you that exception, let me explain it

您有以下错误，这反过来又导致您出现异常，让我解释一下

this is your existing code:

这是您现有的代码：

if(!scan.hasNextInt()) {
        System.out.println("Invalid input!");
        System.out.print("Enter an integer: ");
        usrInput= sc.nextInt();
    }

in the above code if(!scan.hasNextInt())will become trueonly when user input contains both characters as well as integers like your input adfd 123.

在上面的代码中，只有当用户输入既包含字符又包含整数时if(!scan.hasNextInt())才会变为true输入 adfd 123。

but you are trying to read only integers inside the if condition using usrInput= sc.nextInt();. Which is incorrect,that's what is throwing Exception in thread "main" java.util.InputMismatchException.

但是您正在尝试使用 usrInput= sc.nextInt();. 这是不正确的，这就是 throw Exception in thread "main" java.util.InputMismatchException。

so correct code should be

所以正确的代码应该是

 if(!scan.hasNextInt()) {
            System.out.println("Invalid input!");
            System.out.print("Enter an integer: ");
            sc.next(); 
            continue;
        }

in the above code sc.next()will help to read new input from user and continuewill help in executing same if condition(i.e if(!scan.hasNextInt())) again.

在上面的代码sc.next()中将有助于从用户读取新输入，并 continue有助于i.e if(!scan.hasNextInt())再次执行相同的 if 条件（）。

Please use code in my first answer to build your complete logic.let me know if you need any explanation on it.

请使用我的第一个答案中的代码来构建您的完整逻辑。如果您需要任何解释，请告诉我。

回答by Rajendra_Prasad

try this code [updated]:

试试这个代码[更新]：

Scanner scan = null;
       int range, smallest = 0, input;

     for(;;){
         boolean error=false;
        scan = new Scanner(System.in);
        System.out.print("Enter an integer between 1-100:  ");


            if(!scan.hasNextInt()) {
                System.out.println("Invalid input!");                      
                continue;
            }
         range = scan.nextInt();
            if(range < 1) {
                System.out.println("Invalid input!");
                error=true;
            }
        if(error)
        {
        //do nothing
        }
        else
        {
       break;
        }

        }
             for(int ii = 1; ii <= range; ii++) {
            scan = new Scanner(System.in);
            System.out.print("Enter value " + ii + ": ");

            if(!scan.hasNextInt()) {
                System.out.println("Invalid input!"); 
               ii--;
                continue;
            } 
        }

回答by shady

Maybe you can try this:

也许你可以试试这个：

int function(){
Scanner input = new Scanner(System.in);   
System.out.print("Enter an integer between 1-100: ");   
int range;
while(true){   
    if(input.hasNextInt()){   
    range = input.nextInt();
    if(0<=range && range <= 100)
        break;
    else
        continue;
    }
    input.nextLine();  //Comsume the garbage value
    System.out.println("Enter an integer between 1-100:");
}
return range;
}

回答by Salah Assaf

This is to keep requesting inputs while this input is integer and find whether it is odd or even else it will end.

这是为了在此输入为整数时继续请求输入，并查找它是奇数还是偶数，否则它将结束。

int counter = 1;
    System.out.println("Enter a number:");
    Scanner OddInput = new Scanner(System.in);
        while(OddInput.hasNextInt()){
            int Num = OddInput.nextInt();
            if (Num %2==0){
                System.out.println("Number " + Num + " is Even");
                System.out.println("Enter a number:");
            }
            else {
                System.out.println("Number " + Num + " is Odd");
                System.out.println("Enter a number:");
                }
            }
        System.out.println("Program Ended");
    }