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Re: [xsl] sequential numbering in xslt
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Subject: Re: [xsl] sequential numbering in xslt From: a kusa <akusa8@xxxxxxxxx> Date: Mon, 4 Jan 2010 11:23:23 -0600 |
Hi Ken:
Thanks for your response. I use saxon to read a directory of files,
and using xsl:result-document, I output these xml files woth different
names. I started with a random number and tried incrementing it but it
obviously failed because of the xsl:variable limitations.
Also, position() does not increment is that correct? Will it not be always 1?
On Sun, Jan 3, 2010 at 2:48 PM, G. Ken Holman
<gkholman@xxxxxxxxxxxxxxxxxxxx> wrote:
> At 2010-01-03 14:41 -0600, a kusa wrote:
>>
>> I have a number of XML files as input, and I transform them into
>> another XML format using XSLT.
>> ...
>> So, for example, if I had 5 input xml files a.xml, b.xml, c.xml,
>> d.xml, e.xml, and I transformed them into Aout.xml, Bout.xml,
>> Cout.xml, Dout.xml and Eout.xml ,
>
> How are you accessing those 5 input files?
>
>> the output of which looks something
>> like below:
>> Output XML:
>>
>> <car seq="1">
>> <req>....<req>
>>
>> <body id="c1">
>>
>> </body>
>>
>> </car>
>>
>> for every input file, in the transformation, I want to increment
>> attribute 'seq' in the output file by 1 when I transform the input
>> files using xslt.
>
> If you are accessing them in a for-each construct, then the position()
> function gives you what you need:
>
>> I have tried <xsl:number>
>
> That is only for use in a single tree.
>
>> and tried writing a function. But the
>> problem is that since variables in xslt are constants, there is no way
>> to increment a number, store it in a temp variable and increment it
>> for the next time in xslt 2.0 unlike procedural languages like C or
>> C++.
>
> Correct, but you have blinders on thinking that is the way to approach the
> problem.
>
>> Is there any other way of achieving this in XSLT 2.0?
>
> It depends on how you are accessing your 5 input files.
>
> The following would work if this is the way you are doing it:
>
> <xsl:for-each select="('a.xml','b.xml','c.xml','d.xml','e.xml')">
> <xsl:result-document href="{replace(.,'\.','out.')}">
> <car seq="{position()}">
> <xsl:apply-templates select="doc(.)"/>
> ...
>
> Or something like:
>
> <xsl:for-each select="('a.xml','b.xml','c.xml','d.xml','e.xml')">
> <xsl:variable name="seq" select="position()"/>
> <xsl:result-document href="{replace(.,'\.','out.')}">
> <xsl:for=each select="doc(.)">
> <car seq="{$seq}">
> ...
>
> (both untested, so please forgive any typos)
>
> I hope this helps.
>
> . . . . . . . . . . Ken
>
> --
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