Hello,
it may be also 5min, but if you're 20h late and still 10h do work you
don't have time, also it's quite a delicate task to change an xslt with
another xslt, so I don't want to do it in an hurry, even if a nice
idea...
Maybe in the future...
Many thanks
Bye
Il giorno ven, 02/12/2011 alle 17.18 +0100, Bartolomeo Nicolotti ha
scritto:
Hello,
ok, now I've understood, but a little bit long task for my schedule...
maybe some time in the future...
Many thanks
bye
Il giorno ven, 02/12/2011 alle 17.16 +0100, Wolfgang Laun ha scritto:
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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