Check if number is in range

Question

I am checking if my number coming from a input field is in range

function timeCheck(){

        var time = $.trim($('#enterTime').value());
            Number.prototype.between = function(min,max){
                return this > min && this < max;
            };
            if ((time).between(1,9)){
                    alert("test");
                }
            }

But somehow it does not work .. the alert is never triggered

Thanks for help and fast answer

`if((time).between(1,9)) /* <- parenthesis here*/ {` – xlecoustillier Sep 02 '15 at 13:34 — xlecoustillier, Sep 02 '15 at 13:34
@X.L.Ant hmm still not triggered – TamoDaleko Sep 02 '15 at 13:34 — TamoDaleko, Sep 02 '15 at 13:34
Appear missing closing `)` at `if` condition ? – guest271314 Sep 02 '15 at 13:37 — guest271314, Sep 02 '15 at 13:37
@palaѕн no that all wrong it still does not work .. – TamoDaleko Sep 02 '15 at 13:41 — TamoDaleko, Sep 02 '15 at 13:41

score 2 · Answer 1 · answered Sep 02 '15 at 13:35

2

Not a closing bracket but a missing parenthesis:

if((time).between(1,9){

Should be:

if ((time).between(1,9)){

or

if (time.between(1,9)){

answered Sep 02 '15 at 13:35

Daniel Clements

21
2

score 1 · Answer 2 · answered Sep 02 '15 at 13:45

FIDDLE Try the following code

function timeCheck(){
    var time = parseInt($.trim($('#enterTime').val()),10);
        Number.prototype.between = function(min,max){
            var num = parseInt(this,10);

            return num >= min && num <= max; 
        }
            if((time).between(1,9)){
                alert("test");
            }
}

The issue was with type conversion object time was not of type Number.

Hope it helps .....

LostMyGlasses · Accepted Answer · 2015-09-02T13:50:15.913

Extending @Daniel's answer, there are two other errors: first, $('#enterTime').value() is not a valid jQuery function, should be $('#enterTime').val(). Second, you need to convert your value to type Number. Otherwise, you will be trying to access the between property of a string, which doesn't exist. The final code would be:

function timeCheck(){
    var time = new Number($.trim($('#enterTime').val()));
    Number.prototype.between = function(min,max){
        return this > min && this < max;
    };
    if(time.between(1,9)){
        alert("test");
    }
}

score 0 · Answer 4 · edited May 23 '17 at 12:29

0

This question was already asked: here

Try the function I propose (in the mentioned post it was for PHP) that is valid for increasing or decreasing ranges and not limited to integers:

function between(n, a, b)
{
    var chk= false;
   if ((a==n)&&(b==n)){
      chk = true;
    }else{
      chk = (n-a)*(n-b)<0;
    }
    return chk;
}

The function TRUE if n is between a and b

edited May 23 '17 at 12:29

Community

1
1

answered Oct 04 '15 at 17:54

Luis Rosety

396
4
10

Check if number is in range

4 Answers4

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