如何检查用户输入是否不是int值

我需要检查用户输入值是否不是int值。 我已经尝试了我所知道的不同组合，但我得到的都没有或随机错误

例如：

如果用户输入“adfadf 1324”，它将发出警告信息。

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是）我有的：

// 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); } } 

Answer 1

如果输入无效，只需抛出exception

 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"); } 

Answer 2

 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; } 

Answer 3

你有以下错误，这反过来导致你的exception，让我解释一下

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(); } 

在上面的代码中， if(!scan.hasNextInt())只有当用户输入包含字符和整数（如输入adfd 123 if(!scan.hasNextInt())才会变为true 。

但是你试图使用usrInput= sc.nextInt();在if条件中只读取整数usrInput= sc.nextInt(); 。 这是不正确的，这就是Exception in thread "main" java.util.InputMismatchException抛出Exception in thread "main" java.util.InputMismatchException的原因。

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; } 

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

请在我的第一个答案中使用代码来构建完整的逻辑。如果您需要任何解释，我知道。

Answer 4

试试这段代码[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; } } 

Answer 5

摘自相关post ：

 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; } 

Answer 6

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; } 

Answer 7

这是为了在输入为整数时继续请求输入，并查找它是否为奇数，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"); }