RE: [xsl] Replace characters in a node and all of its descendents

["Sergiu Ignat" <sergiu@xxxxxxxxxx>]

Thank you Michael Kay and Tom Passin.
I do have to serialize xml with angle brackets and attributes.
I must use Microsoft .NET parser and processor.
Postprocessing with multiple transformation, I think, is the only viable solution.
I hope there will be a serialization capability defined for XSLT 2.0 .

Sergiu

> 
> Try saxon:serialize(), followed by one of the many ways of 
> doing string
> replacement to double the quotes. Or you could write a 
> similar extension
> function for a different processor. People have in the past 
> posted code
> that does serialization in XSLT itself, and then uses
> disable-output-escaping to prevent the markup being escaped, but I
> wouldn't recommend it myself.
> 
> > -----Original Message-----
> > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of 
> > Sergiu Ignat
> > Sent: 30 September 2003 16:30
> > To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: [xsl] Replace characters in a node and all of its 
> descendents
> > 
> > 
> > I need to generate an SQL "INSERT INTO" statement from an xml 
> > document. One of the columns must hold the string 
> > representation of an XML node with all of its descendents. To 
> > insert this node as a string I must replace each single quote 
> > with other two single quotes for conformance with SQL syntax.
> > 
> > I can not send a deep copy of the node taken with 
> > <xsl:copy-of> to the $text parameter of the "replace-string" 
> > template posted by Evan Lenz at 
> > http://www.dpawson.co.uk/xsl/sect2/replace.htm> l
> > How can I get 
> > a deep copy of a node as a string to process 
> > it? Multiple transformations are not recomended. Thanks. 
> Sergiu Ignat
> > 
> >  XSL-List info and archive:  
> http://www.mulberrytech.com/xsl/xsl-list
> > 
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 

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