Subject: Re: [xsl] __LINE__ equivalent in XSLT From: Wolfgang Laun <wolfgang.laun@xxxxxxxxx> Date: Fri, 2 Dec 2011 17:16:26 +0100 |
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@xxxxxxxxx> 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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