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

2003-10-01 02:02:38
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(_at_)lists(_dot_)mulberrytech(_dot_)com 
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf 
Of 
Sergiu Ignat
Sent: 30 September 2003 16:30
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
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

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