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Truncating without Rounding a number using javascript or xsl?

Truncating without Rounding a number using javascript or xsl?

i shall pass the "number of digits to truncate".

examples:

number digitsTruncate Output

123.456 (3) 123.456
123.456 (2) 123.45
123.456789 (7) 123.4567890
123 (2) 123.00
123 (3) 123.000
123.4 (2) 123.40
123.4 (4) 123.4000

Thank you,

Regards
vh

Zvonko

Use toFixed() method of Number objects.
Like this:

<script>
alert((123.456).toFixed(3)+
"\n"+(123.456).toFixed(2)+
"\n"+(123.456789).toFixed(7)+
"\n"+(123).toFixed(2)+
"\n"+(123).toFixed(3)+
"\n"+(123.4).toFixed(2)+
"\n"+(123.4).toFixed(4));
</script>

ASKER CERTIFIED SOLUTION

Batalf

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Batalf

Zvonko,

I think toFixed() is rounding the number.

i.e.

var number = 123.456;
alert(number.toFixed(2));

yields 123.46 and not 123.45

Batalf

GwynforWeb

toFixed() rounds (as said)

<script>
function truncDigits(inputNumber,digits){
fact= Math.pow(10,digits)
return Math.floor(inputNumber*fact)/fact
}

number = 123.456789
alert(truncDigits(number,4))
alert(truncDigits(number,3))
alert(truncDigits(number,0))
</script>

Zvonko

That was also my first idea Gwyn, but then I tested with nums < 0.

Here my prototype:

<script>

Number.prototype.fixTo = fixToPlaces;

alert((123.456).fixTo(3)+
"\n"+(123.456).fixTo(2)+
"\n"+(123.456789).fixTo(7)+
"\n"+(123).fixTo(2)+
"\n"+(123).fixTo(3)+
"\n"+(123.4).fixTo(2)+
"\n"+(123.4).fixTo(4));

function fixToPlaces(p){
for(var s="",i=0;i<p;i++) s+="0";
var x = (this+'.').split('.')
x = x[0]+((p>0)?"."+(x[1]+s).substr(0,p):"");
return x;
}

</script>

Zvonko

Slightly better readable is this one:

<script>
Number.prototype.fixTo = fixToPlaces;

alert((123.456).fixTo(3)+
"\n"+(123.456).fixTo(2)+
"\n"+(123.456789).fixTo(7)+
"\n"+(123).fixTo(2)+
"\n"+(123).fixTo(3)+
"\n"+(123.4).fixTo(0)+
"\n"+(123.4).fixTo(2)+
"\n"+(123.4).fixTo(4));

function fixToPlaces(p){
if(p<1) return parseInt(this);
for(var s="",i=0;i<p;i++) s+="0";
var x = (this+'.').split('.')
x = x[0]+"."+(x[1]+s).substr(0,p);
return x;
}

</script>

Zvonko

I hate to flooding this thread, but I had to remove the last line :)

function fixToPlaces(p){
if(p<1) return parseInt(this);
for(var s="",i=0;i<p;i++) s+="0";
var x = (this+'.').split('.')
return x[0]+"."+(x[1]+s).substr(0,p);
}

GwynforWeb

My function works for -ve's as well in the sense that it cuts off the extra digits. which I believe is what is required here.

Zvonko

I used here your power factor for protopyping :)

<script>
Number.prototype.toX = Number.prototype.toFixed;
Number.prototype.toFixed = fixNoRound;
function fixNoRound(p){
var f = Math.pow(10,p);
var n = Math.floor(this*f)/f;
return n.toX(p);
}

alert((123.456).toFixed(3)+
"\n"+(123.456).toFixed(2)+
"\n"+(123.456789).toFixed(7)+
"\n"+(123).toFixed(0)+
"\n"+(0.01).toFixed(5)+
"\n"+(123).toFixed(3)+
"\n"+(123.4).toFixed(2)+
"\n"+(123.4).toFixed(4));
</script>

GwynforWeb

Zvonko

Sorry, I misspelled the word prototyping.
It shoild read: "I used here your power factor for prototyping."

Anarchon

Ah, the just of type interconversion...this version leaves you with a string. That has the sole virtue of conserving trailing zeroes, if you intend to maintain the significance of such.

---
<html>

<head>
<script>
function truncateToDecimalPlace(amt, dec) {
// Return string, containing amount >amt< truncated to >dec< decimal places.
if (dec < 0) return 'bad call to truncateToDecimalPlace(): dec = ' + dec;
var str = '' + amt; // Force conversion to string type, if numeric.
if (-1 == str.indexOf('.')) str += '.';
str += '0000000000'; // should handle reasonable cases
return str.slice(0, str.indexOf('.') + dec + (dec == 0 ? 0 : 1));
}
alert('fmtToDecimalPlace(123.15678, 4) is >' + truncateToDecimalPlace(123.15678, 4) + '<');
alert('fmtToDecimalPlace(123.15678, 2) is > ' + truncateToDecimalPlace(123.15678, 2) + '<');
alert('fmtToDecimalPlace(123.9, 0) is > ' + truncateToDecimalPlace(123.9, 0) + '<');
</script>
</head>

<body>
</body>

</html>

Niversoft

Why not using parseInt() ?

var value = 123.45;
value = parseInt(value);
alert(value);

Niversoft

ah, never mind.

heartland-tech

The accepted solution is incorrect.
Here is the correction to the accepted solution:

function truncateNumber(inputNumber, digits){
var number = Math.floor(inputNumber);
var fdecimal = (inputNumber - number) + '';
if(fdecimal.length > (digits + 2))
fdecimal = fdecimal.substr(0, (digits + 2));
var returnValue = (number/1 + fdecimal/1);
return returnValue.toFixed(digits);
}

function truncateNumber(inputNumber, digits){        
        var number = Math.floor(inputNumber);
        var fdecimal = (inputNumber - number) + '';
        if(fdecimal.length > (digits + 2)) 
            fdecimal = fdecimal.substr(0, (digits + 2));
        var returnValue = (number/1 + fdecimal/1);        
        return returnValue.toFixed(digits);            
    }

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