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
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There could be simpler solution than this one, but it's a shot:
<script type="text/javascript">
function truncateNumber(inputNumber
var number = Math.floor(inputNumber);
var decimal = (inputNumber - number) + "";
if(decimal.length>(digits+
var returnValue = (number + decimal)/1;
return returnValue.toFixed(digits
}
var number = 123.456 ;
alert(truncateNumber(numbe
</script>
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).s
return x;
}
</script>
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
return x;
}
</script>
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(
"\n"+(123).toFixed(0)+
"\n"+(0.01).toFixed(5)+
"\n"+(123).toFixed(3)+
"\n"+(123.4).toFixed(2)+
"\n"+(123.4).toFixed(4));
</script>
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
// 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(1
alert('fmtToDecimalPlace(1
alert('fmtToDecimalPlace(1
</script>
</head>
<body>
</body>
</html>
The accepted solution is incorrect.
Here is the correction to the accepted solution:
function truncateNumber(inputNumber
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
}
Business Accounts
Answer for Membership
by: ZvonkoPosted on 2005-07-01 at 11:01:45ID: 14349981
Use toFixed() method of Number objects.
Like this:
<script>
alert((123.456).toFixed(3)
"\n"+(123.456).toFixed(2)+
"\n"+(123.456789).toFixed(
"\n"+(123).toFixed(2)+
"\n"+(123).toFixed(3)+
"\n"+(123.4).toFixed(2)+
"\n"+(123.4).toFixed(4));
</script>