Why else is not working

JavaScript
1

Hi. Why “else”, where function is calling isn’t work when I enter correct a and b? And 2nd question is how to make such action shorter?

‘use strict’;

let a = prompt(“Your name?”,‘’),
b = prompt(“yor age?”,‘’);

function user(name,age){
alert('Имя пользователя: ’ + name);
alert('Возраст: '+ age);
};

if ((a || b == null)||(a && b==null)||(a || b == ‘’)||(a && b==‘’))
{
alert(‘Not Correct!’);
} else {
user(a,b);
};

WOOOOOOOOOOOOOOOOOOOOOOW!!!!!

cat_in_bath10

2

Quotes in your code look wrongly placed…
:thinking::thinking::thinking::thinking::thinking:

3

In repl i got some issues with quotations, so that might cause additional problem for you.

Beside that… You dont need all of those checking in your if statement.

if (!a || !b) would do the job. It checks if either input of a was falsy or input of b was falsy and alerts Not correct.

There is code with console logs instead of alerts:

let a = prompt("Your name?",""),
b = prompt("yor age?",);

function user(name,age){
  console.log('Имя пользователя:' + name);
  console.log('Возраст: '+ age);
};

if (!a || !b) {
  console.log("Wrong input")
} else {
  user(a,b);
};
1 Like
5

((a || b == null)||(a && b==null)||(a || b == ‘’)||(a && b==’’))

The above is your problem. You can’t do combined conditional evaluations like that. This conditional translates to:
if

  • a is truthy OR b is equal to null
    OR
  • a is truthy AND b is equal to null
    OR
  • a is truthy AND b is an empty string
6

Thanks for answer, “!” is working, but I want to understand, what is wrong in my “if”?

7

so i need to combine (a||b)?

8

It depends on what you are trying to achieve.
(a || b) is “If a is truthy OR b is truthy.”

1 Like