Question 254111 in Javascript Question posted in
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Javascript – Add two errors to number validation

I’m using input type=numberand validation to it.
What I am trying to do:

  • if value is not number – show error only numbers allowed
  • if input is empty – show error value is required

yup.object().shape({ fieldName: yup.number().typeError("only numbers allowed").required("value required").min(0) })

But it returns always only typerror

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2

Answers


  1. 0

    You are trying to validate an input field or type = number, if an empty string is passed it bound to produce an error. The Yup validation is actively checking if the value is a valid JavaScript number, and empty values or non-numeric strings do not meet this criterion.

    1. Use yup.mixed() to allow both numbers and empty values
    2. Use the test method to define a custom validation rule

    Here’s the modified code:

    
import * as yup from 'yup';

const schema = yup.object().shape({
  fieldName: yup
    .mixed() // Use mixed() to allow both numbers and empty values
    .test('is-number-or-empty', 'Only numbers allowed or value is required', (value) => {
      // Check if the value is a number or empty
      if (!value || !isNaN(value)) {
        return true; // Validation passed
      } else {
        return false; // Validation failed
      }
    })
    .typeError('Only numbers allowed'), // Type error message
});
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  2. 0 
    function validateNumber(input) {
  if (input === "") {
    return "Error: Input cannot be empty.";
  }

  if (!/^d+$/.test(input)) {
    return "Error: Input must contain only numeric characters.";
  }

  return "Success: Input is a valid number.";
}

// Example usage:
const userInput = prompt("Enter a number:");
const validationResult = validateNumber(userInput);

console.log(validationResult);
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