issue with comparing two numbers in javascript

Question

my html code is like:

<html>
<SCRIPT type="text/javascript" language="JavaScript">   
function fun()
{
var l = document.test.low.value;
var h = document.test.high.value;
alert(l);
alert(h);
    if(l >h){
        alert("low greater than high low is -"+l+"high is -"+h);
    }
}
</SCRIPT>

<body>

<form name="test">
<input type="text" size="11" id="low" />
<input type="text" size="11" id="high" />
<input type="button" value="sss" onClick="fun()"/> 
</form>
</body>
</html>

when i compare this two values using low is 12 and high is 112 is not working. like 22 and 122, 33 and 133 etc....

I am using IE browser version 8. please help.

Sebastian Simon · Accepted Answer · 2022-10-15 06:44:21Z

32

Convert the values to numbers:

if(Number(l) > Number(h)){ /* … */ }

or alternatively, parse each string as a float:

if(Number.parseFloat(l) > Number.parseFloat(h)){ /* … */ }

or as an integer:

if(Number.parseInt(l, 10) > Number.parseInt(h, 10)){ /* … */ }

edited Oct 15, 2022 at 6:44

Sebastian Simon

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answered Feb 1, 2012 at 10:36

Pranav

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3 Comments

Sarfraz Over a year ago

It will not work if number happens to be octal format: developer.mozilla.org/en/JavaScript/Reference/Global_Objects/…

Pranav Over a year ago

yes you are right then in that case:use base value or simply use Number()

menaka Over a year ago

idk what the reason is but the Number() did not worked for me , but other stuff works

Community · Accepted Answer · 2017-05-23 12:18:05Z

11

You need to convert them to numbers:

if(+l > +h){

The unary plus operator will convert the string value to numeric value. See this question for more details:
What's the significant use of unary plus and minus operators?

Live example.

edited May 23, 2017 at 12:18

CommunityBot

11 silver badge

answered Feb 1, 2012 at 10:31

user447356

Comments

Sarfraz · Accepted Answer · 2012-02-01 10:34:28Z

6

Use parseInt with base to get correct results:

var l = parseInt(document.test.low.value, 10);
var h = parseInt(document.test.high.value, 10);

answered Feb 1, 2012 at 10:34

Sarfraz

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2 Comments

user447356 Over a year ago

Not sure parseInt is good here, for example 10a will be converted to 10 even though it's not really a number - using the unary plus will make it NaN value which is more proper IMO.

Sarfraz Over a year ago

@ShadowWizard: In that case + can be prefixed and then parseInt but using + alone won't avoid octal number problem AFAIK.

Lokesh · Accepted Answer · 2013-12-04 05:32:59Z

1

var lnum = new Number(l);
var hnmu = new Number(h);
if(lnum > hnum){
alert('if');

}else{
alert('else');
}

answered Dec 4, 2013 at 5:32

Lokesh

213 bronze badges

Comments

Pavel Vladov · Accepted Answer · 2016-10-12 07:33:44Z

0

When you compare two strings in an if statement:

if ( l > h) {}    
when l = "12" and h = "112"

It compares these two values as strings as "3" > "1". It says l > h is true

edited Oct 12, 2016 at 7:33

Pavel Vladov

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answered Feb 1, 2012 at 10:45

vireshas

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