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Re: [xsl] __LINE__ equivalent in XSLT

2011-12-02 10:16:39
Michael suggested the implementation of an equivalent of a C preprocessor
by replacing processing instructions *in your XSLT" by *another XSLT* which
transforms the input XML (your actual XSLT) into one where "<?line?> is
replaced by the current line number within that XML (your actual XSLT).

-W

On 2 December 2011 16:02, Bartolomeo Nicolotti 
<bnicolotti(_at_)siapcn(_dot_)it> wrote:
Sorry,

this

(line 34)<xsl:comment>line <xsl:value-of  select="saxon:line-number()"/>

gives this

<!--line -1

Many thanks

Bye

Il giorno ven, 02/12/2011 alle 16.01 +0100, Bartolomeo Nicolotti ha
scritto:
hello,

I've tried this:

zxsl:value-of select="saxon:line-number(.)"/>
</xsl:template>

buy gives me the line of the input xml

I need to know the line number in the xslt itself

Many thanks

Bye

Il giorno ven, 02/12/2011 alle 13.01 +0000, Michael Kay ha scritto:
You could implement the preprocessor easily enough if you are using
Saxon. For example if you use <?line?> then you can write the
preprocessor as

<xsl:template match="*">
   .. identity template ..
</xsl:template>

<xsl:template match="processing-instruction(line)">
<xsl:value-of select="saxon:line-number(.)"/>
</xsl:template>

Michael Kay
Saxonica

On 02/12/2011 12:14, Bartolomeo Nicolotti wrote:
To whom it may concern,

in C there's a pre-processor directive

__LINE__

that gives you the line of source where the directive is.

Is there an equivalent in XSLT?

Many thanks

Best regards

Bartolomeo




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