Ошибка the input in invalid - ErrorsMaster.ru - большая энциклопедия ошибок и их решений

I have written a JAVA program that takes input from the user and check if the user has entered the correct thing or not. I have taken the input from the Scanner class. If the user enters an invalid character like a String, I want to display ‘Invalid Input’.

public class Main {
    public static void main(String[] args) {
        Scanner takeInteger = new Scanner(System.in);
        System.out.println("Enter a number");
        int enteredNumber = takeInteger.nextInt();
    }
}

jpllosa

2,0661 gold badge28 silver badges28 bronze badges

asked Jul 7, 2021 at 15:37

Just ask the Scanner whether the next input is a valid int value, e.g.

Scanner takeInteger = new Scanner(System.in);
System.out.println("Enter a number");
while(!takeInteger.hasNextInt()) {
    System.out.println("Invalid Input: " + takeInteger.next());
}
int enteredNumber = takeInteger.nextInt();

This will retry the operation until the user entered a number. If you just want a single attempt, use something like

Scanner takeInteger = new Scanner(System.in);
System.out.println("Enter a number");
if(!takeInteger.hasNextInt()) {
    System.out.println("Invalid Input: " + takeInteger.next());
}
else {
    int enteredNumber = takeInteger.nextInt();
    // ... proceed with the input
}

answered Jul 7, 2021 at 16:50

HolgerHolger

283k40 gold badges429 silver badges756 bronze badges

You will get an Exception that is InputMismatchException when an invalid input is passed.(i.e except integer value),you can use a try-catch block to hold the exception and inform the user about the invalid input. Try block , Catch block

    import java.util.*; 

    Scanner takeInteger = new Scanner(System.in);
    System.out.println("Enter a number");
    try{
       int enteredNumber = takeInteger.nextInt();
    }
    catch(InputMismatchException e) {
        System.out.println("Enter a valid input");
    }

answered Jul 7, 2021 at 15:52

Ashish MishraAshish Mishra

6941 gold badge6 silver badges20 bronze badges

You can use Exception handling for the same.

public class Main {
     public static void main(String[] args) {
         Scanner takeInteger = new Scanner(System.in);
         System.out.println("Enter a number");
         try {
             int enteredNumber = takeInteger.nextInt();
         }
         catch (Exception e) {
             System.out.println("Invalid Input");
         }
     }
}

jpllosa

2,0661 gold badge28 silver badges28 bronze badges

answered Jul 7, 2021 at 15:53

Sonam GuptaSonam Gupta

3282 silver badges11 bronze badges

You can add a try-catch block in your program to check if the user’s input is a number or not.

        Scanner takeInteger = new Scanner(System.in);
        System.out.println("Enter a number");
        String input = takeInteger.next();
        int enteredNumber;
        try
        {
            enteredNumber = Integer.parseInt(input); // Input is a number
        }
        catch(NumberFormatException ex)
        {
            System.out.println("Wrong Input!"); // Invalid input
        }

answered Jul 7, 2021 at 15:51

GURU ShreyanshGURU Shreyansh

8711 gold badge6 silver badges18 bronze badges

You need to call the .nextLine method of the Scanner class and then parse to the desired type.

Example:

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a number");
        String line = sc.nextLine();

        try {
            int enteredNumber = Integer.parseInt(line);
            System.out.println("You have entered: " + enteredNumber);
        } catch (Exception e) {
            System.out.println("Invalid Input");
        }
    }
}

Result with a number:

Enter a number
12
You have entered: 12

Result a text:

Enter a number
abcd
Invalid Input

answered Jul 7, 2021 at 15:53

mamadalievmamadaliev

1332 silver badges9 bronze badges

Источник

Improve Article

Save Article

Like Article

Read

Discuss

Improve Article

Save Article

Like Article

PHP can be easily embedded in HTML files and HTML codes can also be written in a PHP file. The thing that differentiates PHP from a client-side language like HTML is, PHP codes are executed on the server whereas HTML codes are directly rendered on the browser. To display error for invalid input with HTML and PHP.

Approach:

Display error on invalid error when

input textbox is left empty.
input is wrong.

PHP code: The following is the code for “form.php” which is used in the HTML code in the later part.

To show invalid input in PHP, set the name of the input textbox which is in HTML. All the fields are first checked for empty fields and then it is validated for correctness. If all fields are correct then it shows the success message. If the input given by the user is wrong then it will show a message for ” Invalid input!!”.

PHP

<?php

$nameError = "";

$emailError = "";

$passwordError = "";

$mobileError = "";

$success = "";

function validate_input($input) {

$input = trim($input);

$input = stripslashes($input);

$input = htmlspecialchars($input);

return $input;

}

if(isset($_POST['form_submit'])) {

$name = $_POST['name'];

$password = $_POST['password'];

$email = $_POST['user_email'];

$mobile = $_POST['mobile'];

if (empty($_POST["name"])) {

$nameError = "Name is required";

} else {

$name = validate_input($_POST["name"]);

if($name == 'chetan') {

$success= "Thank you ". $name.", ";

echo $success;

}

if (empty($_POST["email"])) {

$emailError = "Email is required";

} else {

$email = validate_input($_POST["email"]);

if($email == 'test@email.com') {

$success= $email." is correct";

echo $success;

}

if (empty($_POST["password"])) {

$passwordError = "Password is required";

} else {

$password = validate_input($_POST["password"]);

if($password == 'test@123') {

$success= $password." is correct";

echo $success;

}

if (empty($_POST["mobile"])) {

$mobileError = "Mobile is required";

} else {

$mobile = validate_input($_POST["mobile"]);

if($mobile == '123456789') {

$success= $mobile." is correct";

echo $success;

}

if(empty($success))

echo "Invalid input!!!";

}

?>

HTML code: The following code uses the above PHP “form.php” code.

HTML

<!DOCTYPE html>

<html lang="en" dir="ltr">

<head>

<meta charset="utf-8">

<title>Form</title>

</head>

<body>

<form action="form.php" method="POST">

<input type="text" name="name"

placeholder="Enter name">

<input type="password" name="password"

placeholder="Password">

<input type="email" name="user_email"

placeholder="yourname@gamil.com">

<input type="tel" name="mobile"

placeholder="Mobile no">

<button type="submit" name="form_submit">

Submit

</button>

</form>

</body>

</html>

Output:

Incorrect input by user
```
Invalid input!!!
```

Correct input by user
```
Thank you chetan, 123456789 is correct
```

Last Updated :
31 Dec, 2020

Like Article

Save Article

Источник

Помогаю со студенческими работами здесь

invalid numeric input
Здравствуйте! Собственно в программе пишет вот такую ошибку. Ввожу данные из текстового файла в…

Invalid numeric input
Выходит ошибка при загрузке массива из файла. Помогите исправить. Так как проект состоит из…

Ошибка: Invalid numeric input
Эта ошибка выходит после ввода данных
program Project1;

{$APPTYPE CONSOLE}

uses
SysUtils;…

Ошибка: Invalid numeric input
unit Unit4;

interface

uses
Windows, Messages, SysUtils, Variants, Classes, Graphics,…

Искать еще темы с ответами

Или воспользуйтесь поиском по форуму:

Источник