How to apply conditional return statement in JavaScript

Question

I have this html input:

Rows: <input type="text" class="rows" onkeypress="return isNumber(event)"><br>

and this javascript function to validate only numbers

function isNumber(evt) {
  evt = (evt) ? evt : window.event;
  var charCode = (evt.which) ? evt.which : evt.keyCode;
  if (charCode > 31 && (charCode < 48 || charCode > 57)) {
      return false;
  }
  return true;
}

but i want to work with micro-branching to do something like this:

function isNumber(evt){
  evt = evt || window.event;
  var charCode = evt.which || evt.keyCode; 
  (charCode > 31 && (charCode < 48 || charCode > 57)) && return false;
  return true;
}

the thing is that the last 2 lines didnt work.

what you trying to do in second last line ? it seems if is missing but again what is that && return false ..is it typo ? — Indra Uprade
– Indra Uprade, Commented Jul 27, 2016 at 0:50
1. the second sample is ugly. 2. return !(charCode > 31 && (charCode < 48 || charCode > 57)); 3. The && operand must be an expression not a statement — zerkms
– zerkms, Commented Jul 27, 2016 at 0:51
Not what you're asking, but you know the code shown doesn't actually stop the user entering non-numeric data into that input, right? (Because the keyboard isn't the only option for editing...) — nnnnnn
– nnnnnn, Commented Jul 27, 2016 at 0:59
@nnnnnn i didnt think about it but you are right, im doing this: window.onload = function() { var myInput = document.querySelector('.rows'); myInput.onpaste = function(e) { e.preventDefault(); } } — imjustaguy
– imjustaguy, Commented Jul 27, 2016 at 1:13

Siguza · Accepted Answer · 2016-07-27 00:59:11Z

3

return is a statement rather than an expression, and thus cannot be used as argument to a logical operator.

In your case however, the last two lines can be rewritten into a single return statement, by simply inverting the condition to the if clause:

return !(charCode > 31 && (charCode < 48 || charCode > 57));

Or, as zerkms notes, you can lose the ! by flipping the operators (&& <=> || and < <=> >=), which, in my humble opinion, increases readability:

return charCode <= 31 || (charCode >= 48 && charCode <= 57);

edited Jul 27, 2016 at 0:59

answered Jul 27, 2016 at 0:53

Siguza

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1 Comment

zerkms Over a year ago

Btw, return charCode <= 31 || (charCode >= 48 && charCode <= 57);

Siguza · Accepted Answer · 2016-07-27 00:54:31Z

1

According to your description, it looks like you are looking for conditional check and return :

function isNumber(evt){
  evt = evt || window.event;
  var charCode = evt.which || evt.keyCode; 
  return !(charCode > 31 && (charCode < 48 || charCode > 57));
}

edited Jul 27, 2016 at 0:54

Siguza

24.7k6 gold badges58 silver badges105 bronze badges

answered Jul 27, 2016 at 0:53

Ash

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