Incorporating multiple true statements into an IF ELSE structure can enhance the decision-making

Question

Incorporating multiple true statements into an IF ELSE structure can enhance the decision-making

I'm struggling with getting a true return value for numbers that are integers and have either 4 or 6 digits - no decimals or letters allowed. The issue seems to be the validation of whether it's really a number and if it has a decimal point.

Although I believe I have applied the right functions, I can't seem to properly connect them to my if statement. I want to check if three different conditions are true before returning true, but I haven't quite figured it out yet.

If possible, please only provide a hint, a link, SUDO Code, or references I can explore. I'll follow up with the answer once I've solved it myself.

JS

function validatePIN (pin) {
  //return true or false
   var result = (pin - Math.floor(pin)) !== 0; 
   if( pin.length === 4 || isNaN(pin) || result) {
     return true
   } else if ( pin.length === 6 || isNaN(pin) || result) {
     return true
   } else return false
}

Thanks

javascript

Answer 1

Answer №1

A straightforward regular expression can be implemented to verify whether a string consists of either 4 or 6 numbers.

function isValidPin (pin) {
   return /^(\d{4}|\d{6})$/.test(pin.toString());
}

console.log(isValidPin(123));
console.log(isValidPin("1234"));
console.log(isValidPin("12345"));
console.log(isValidPin("123456"));
console.log(isValidPin("1234567"));
console.log(isValidPin("12.45"));
console.log(isValidPin("12e45"));

Answer 2

A straightforward regular expression can be implemented to verify whether a string consists of either 4 or 6 numbers.

function isValidPin (pin) {
   return /^(\d{4}|\d{6})$/.test(pin.toString());
}

console.log(isValidPin(123));
console.log(isValidPin("1234"));
console.log(isValidPin("12345"));
console.log(isValidPin("123456"));
console.log(isValidPin("1234567"));
console.log(isValidPin("12.45"));
console.log(isValidPin("12e45"));

Answer 3

Answer №2

To determine if the conditions are met, you can use the AND operator (&&) in JavaScript.

function checkPasswordComplexity (password) {
  //return true or false
   var hasSpecialChar = /[!@#$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]+/.test(password); 
if( password.length >= 8 && /[0-9]/.test(password) && /[A-Z]/.test(password) && /[a-z]/.test(password) && hasSpecialChar) 
  { return true;} else { return false;}
}

Answer 4

To determine if the conditions are met, you can use the AND operator (&&) in JavaScript.

function checkPasswordComplexity (password) {
  //return true or false
   var hasSpecialChar = /[!@#$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]+/.test(password); 
if( password.length >= 8 && /[0-9]/.test(password) && /[A-Z]/.test(password) && /[a-z]/.test(password) && hasSpecialChar) 
  { return true;} else { return false;}
}

Answer 5

Answer №3

In order to pass the validation, switch either your 'and' or'veriify my PIN code'

 function checkPIN (pin) {
     //output true or false
      var outcome = (pin - Math.floor(pin)) !== 0; 
     if( pin.length === 4 && isNaN(pin) && outcome) 
     { return true} 
     else if ( pin.length === 6 && isNaN(pin) && outcome)  {
     return true
    } else return false
    }

Answer 6

In order to pass the validation, switch either your 'and' or'veriify my PIN code'

 function checkPIN (pin) {
     //output true or false
      var outcome = (pin - Math.floor(pin)) !== 0; 
     if( pin.length === 4 && isNaN(pin) && outcome) 
     { return true} 
     else if ( pin.length === 6 && isNaN(pin) && outcome)  {
     return true
    } else return false
    }

Answer 7

Answer №4

Give this a shot:

function checkPIN(pin) {
  var parsedNum = Math.parseInt(pin, 10);

  // checking if the input is not an integer
  if(pin !== parsedNum.toString()) return false;

  return pin.length === 4 || pin.length === 6;
}

Answer 8

Give this a shot:

function checkPIN(pin) {
  var parsedNum = Math.parseInt(pin, 10);

  // checking if the input is not an integer
  if(pin !== parsedNum.toString()) return false;

  return pin.length === 4 || pin.length === 6;
}

Answer 9

Answer №5

It's possible that I am mistaken, but it might be worth considering checking if the length of the code is either 4 OR 6 before proceeding with your other validations:

function validatePIN (pin) {
    //return true or false
    var result = (pin - Math.floor(pin)) !== 0; 
    if(!isNaN(pin) && (pin.length === 4 || pin.length === 6) && result) {
        return true;
    } else {
        return false;
    }
}

Additionally, I have made some modifications to your code since it didn't make sense to return true for NaN values.

Answer 10

It's possible that I am mistaken, but it might be worth considering checking if the length of the code is either 4 OR 6 before proceeding with your other validations:

function validatePIN (pin) {
    //return true or false
    var result = (pin - Math.floor(pin)) !== 0; 
    if(!isNaN(pin) && (pin.length === 4 || pin.length === 6) && result) {
        return true;
    } else {
        return false;
    }
}

Additionally, I have made some modifications to your code since it didn't make sense to return true for NaN values.

Answer 11

Answer №6

If you're looking for a solution, consider using regular expressions for ease of implementation.

console.log(445584, validatePin(445584));
console.log("445584", validatePin("445584"));
console.log("alj454", validatePin("alj454"));
console.log(4455.84, validatePin(4455.84));

function validatePin(pin){
  return /^(\d{4}|\d{6})$/.test(pin);
}

Answer 12

If you're looking for a solution, consider using regular expressions for ease of implementation.

console.log(445584, validatePin(445584));
console.log("445584", validatePin("445584"));
console.log("alj454", validatePin("alj454"));
console.log(4455.84, validatePin(4455.84));

function validatePin(pin){
  return /^(\d{4}|\d{6})$/.test(pin);
}

Answer 13

Answer №7

One solution is to utilize the isNaN() function in order to identify any non-numeric characters.

Additionally, consider converting the data to a string using .toString() so that you can easily determine its length with .length.

Answer 14

One solution is to utilize the isNaN() function in order to identify any non-numeric characters.

Additionally, consider converting the data to a string using .toString() so that you can easily determine its length with .length.

Incorporating multiple true statements into an IF ELSE structure can enhance the decision-making

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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