Re: XslTransform, XslCompiledTransform Sum Large Number Rounding Probl - XML SOAP

Thanks for the advice dimitre.

I already try your suggestion to use the workaroud below on 1.0.
But it is only work in a browser or using XslCompledTransform class.
If I use XslTransform the result is still: 75,556,694,820,930.00

When I use msxsl.exe as the transformation engine the result is correct
without changing my xsl template file.

So currently I use msxsl.exe as the transformation, but it is not to elegant,
because I have to do transformation process with files, I can't use in
memory process.

thanks in advance.

"Dimitre Novatchev" wrote:

> Numbers in XSLT are of type double and the operations with them follow the
> laws of floating point arithmetics. In XSLT2.0/XPath 2.0 one will choose the
> xs:decimal type and avoid such a problem.
>
> In XSLT 1.0 one can use a workaround as the following:
>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xslutput method="text"/>
>
> <xsl:template match="/">
> <xsl:variable name="vSumm100Times">
> <xsl:call-template name="sumPrecise">
> <xsl:with-param name="pList" select="//a"/>
> </xsl:call-template>
> </xsl:variable>
>
> <xsl:value-of select="$vSumm100Times div 100"/>
> </xsl:template>
>
> <xsl:template name="sumPrecise">
> <xslaram name="pList" select="/.."/>
> <xslaram name="pAccum" select="0"/>
> <xslaram name="pPrec" select="100"/>
>
> <xsl:choose>
> <xsl:when test="$pList">
> <xsl:call-template name="sumPrecise">
> <xsl:with-param name="pList"
> select="$pList[position() > 1]"/>
> <xsl:with-param name="pAccum"
> select="$pAccum + $pList[1] * 100"/>
> </xsl:call-template>
> </xsl:when>
> <xsltherwise>
> <xsl:value-of select="$pAccum"/>
> </xsltherwise>
> </xsl:choose>
> </xsl:template>
> </xsl:stylesheet>
>
>
> When the above transformation is applied on your xml document, the result
> is:
>
>
> 75556694820929.97
>
>
> Cheers,
> Dimitre Novatchev
> "Feiry" <Feiry@discussions.microsoft.com> wrote in message
> news:F43EC524-7461-4F80-A8BF-4A0217F128F4@microsoft.com...
> > I'm trying to sum nodes that have number with minimum 16 digit.
> > when I try using XslTransform to do the transformation I've got a rounding
> > but when I use browser transformation by opening the xml directly through
> > web browser, the sum result is OK.
> > It's looks like XslTransform use double as the sum buffer.
> > I try use XslCompiledTransform but still get the same rounding problem.
> > Is there any body know how to solve this ?
> > I try to search everywhere but seems that no one get the same problem.
> > I use this for large rupiah currency banking transaction reporting, all
> > sample below is just for simplification.
> >
> > thanks for any suggestion in advance.
> >
> > sum result using xsltransform: 75,556,694,820,930.00
> > sum result using browser: 75,556,694,820,929.97
> >
> > transform code .netframework 1.1:
> > XslTransform xslTrans = new XslTransform();
> > xslTrans.Load("sum.xsl");
> > xslTrans.Transform("a.xml","a.txt");
> >
> > transform code .netframework 2.0:
> > XslCompiledTransform xslt = new XslCompiledTransform();
> > xslt.Load("a.xsl");
> > xslt.Transform("a.xml", "a.html");
> >
> > xml example (a.xml):
> > <?xml version='1.0'?>
> > <?xml-stylesheet type="text/xsl" href="sum.xsl"?>
> > <root>
> > <a>6574705.82</a>
> > <a>5350503889475.60</a>
> > <a>13788014503961.10</a>
> > <a>4637582864105.00</a>
> > <a>12291887216661.00</a>
> > <a>11231072464569.00</a>
> > <a>26701607763086.00</a>
> > <a>166575342465.75</a>
> > <a>1343750794520.54</a>
> > <a>45693407380.16</a>
> > </root>
> >
> > xsl sample (sum.xsl):
> > <xsl:stylesheet version="1.0"
> > xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> > <xsl:template match="/">
> > <xsl:value-of select="format-number(sum(//a),'###,###.00')"/>
> > </xsl:template>
> > </xsl:stylesheet>
> >
>
>
>

Dear Feiry,

Could you, please, explain more clearly the contents of your last message?

Sorry, but I didn't understand anything from it. What is really the problem?


My results are:

MSXML3: 75556694820929.97
MSXML4: 75556694820929.97

.Net 1.1 XslTransform: 75556694820929.969
.Net 2.0 XslCompiledTransform: 75556694820929.97


MSXML3 is what IE uses internally to do a transformation as directed by a
xml-stylesheet processing-instruction.

The versions of .NET (for the XslTransform and XslCompiledTransform,
respectively) used are:

..NET 1.1.4322.2032
..NET 2.0.50727.42

Cheers,
Dimitre Novatchev


"Feiry" <Feiry@discussions.microsoft.com> wrote in message
news:BB49A13D-8BB2-4937-B238-AD71DCB06BE1@microsoft.com...
> Thanks for the advice dimitre.
>
> I already try your suggestion to use the workaroud below on 1.0.
> But it is only work in a browser or using XslCompledTransform class.
> If I use XslTransform the result is still: 75,556,694,820,930.00
>
> When I use msxsl.exe as the transformation engine the result is correct
> without changing my xsl template file.
>
> So currently I use msxsl.exe as the transformation, but it is not to
> elegant,
> because I have to do transformation process with files, I can't use in
> memory process.
>
> thanks in advance.
>
> "Dimitre Novatchev" wrote:
>
>> Numbers in XSLT are of type double and the operations with them follow
>> the
>> laws of floating point arithmetics. In XSLT2.0/XPath 2.0 one will choose
>> the
>> xs:decimal type and avoid such a problem.
>>
>> In XSLT 1.0 one can use a workaround as the following:
>>
>> <xsl:stylesheet version="1.0"
>> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
>> <xslutput method="text"/>
>>
>> <xsl:template match="/">
>> <xsl:variable name="vSumm100Times">
>> <xsl:call-template name="sumPrecise">
>> <xsl:with-param name="pList" select="//a"/>
>> </xsl:call-template>
>> </xsl:variable>
>>
>> <xsl:value-of select="$vSumm100Times div 100"/>
>> </xsl:template>
>>
>> <xsl:template name="sumPrecise">
>> <xslaram name="pList" select="/.."/>
>> <xslaram name="pAccum" select="0"/>
>> <xslaram name="pPrec" select="100"/>
>>
>> <xsl:choose>
>> <xsl:when test="$pList">
>> <xsl:call-template name="sumPrecise">
>> <xsl:with-param name="pList"
>> select="$pList[position() > 1]"/>
>> <xsl:with-param name="pAccum"
>> select="$pAccum + $pList[1] * 100"/>
>> </xsl:call-template>
>> </xsl:when>
>> <xsltherwise>
>> <xsl:value-of select="$pAccum"/>
>> </xsltherwise>
>> </xsl:choose>
>> </xsl:template>
>> </xsl:stylesheet>
>>
>>
>> When the above transformation is applied on your xml document, the result
>> is:
>>
>>
>> 75556694820929.97
>>
>>
>> Cheers,
>> Dimitre Novatchev
>> "Feiry" <Feiry@discussions.microsoft.com> wrote in message
>> news:F43EC524-7461-4F80-A8BF-4A0217F128F4@microsoft.com...
>> > I'm trying to sum nodes that have number with minimum 16 digit.
>> > when I try using XslTransform to do the transformation I've got a
>> > rounding
>> > but when I use browser transformation by opening the xml directly
>> > through
>> > web browser, the sum result is OK.
>> > It's looks like XslTransform use double as the sum buffer.
>> > I try use XslCompiledTransform but still get the same rounding problem.
>> > Is there any body know how to solve this ?
>> > I try to search everywhere but seems that no one get the same problem.
>> > I use this for large rupiah currency banking transaction reporting, all
>> > sample below is just for simplification.
>> >
>> > thanks for any suggestion in advance.
>> >
>> > sum result using xsltransform: 75,556,694,820,930.00
>> > sum result using browser: 75,556,694,820,929.97
>> >
>> > transform code .netframework 1.1:
>> > XslTransform xslTrans = new XslTransform();
>> > xslTrans.Load("sum.xsl");
>> > xslTrans.Transform("a.xml","a.txt");
>> >
>> > transform code .netframework 2.0:
>> > XslCompiledTransform xslt = new XslCompiledTransform();
>> > xslt.Load("a.xsl");
>> > xslt.Transform("a.xml", "a.html");
>> >
>> > xml example (a.xml):
>> > <?xml version='1.0'?>
>> > <?xml-stylesheet type="text/xsl" href="sum.xsl"?>
>> > <root>
>> > <a>6574705.82</a>
>> > <a>5350503889475.60</a>
>> > <a>13788014503961.10</a>
>> > <a>4637582864105.00</a>
>> > <a>12291887216661.00</a>
>> > <a>11231072464569.00</a>
>> > <a>26701607763086.00</a>
>> > <a>166575342465.75</a>
>> > <a>1343750794520.54</a>
>> > <a>45693407380.16</a>
>> > </root>
>> >
>> > xsl sample (sum.xsl):
>> > <xsl:stylesheet version="1.0"
>> > xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
>> > <xsl:template match="/">
>> > <xsl:value-of select="format-number(sum(//a),'###,###.00')"/>
>> > </xsl:template>
>> > </xsl:stylesheet>
>> >
>>
>>
>>