RE: Multiple search and replace

You want $1 rather than \1 to refer to group 1 in the replace string.

Michael Kay
http://www.saxonica.com/ 

> -----Original Message-----
> From: Pankaj Chaturvedi [mailto:pankaj.chaturvedi@xxxxxxxxx] 
> Sent: 03 April 2008 19:23
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE:  Multiple search and replace
> 
> Hi,
> 
>     <xsl:sequence select="replace(., '\[#x(\d+)\]', 
> '&#xE000;#x\1;')" />
> 
> Gives error: invalid replace string. The problem I believe is 
> at group \1, though I do not see any wrong in regex syntax. 
> May be some problem with "Built-in XSLT engine" of XMLSpy.
> 
> It seems I need to install java in my system to use Saxon 9 
> processor. Any ideas.
> 
> 
> 
> 
> 
> -----Original Message-----
> From: Abel Braaksma [mailto:abel.online@xxxxxxxxx]
> Sent: Wednesday, April 02, 2008 12:03 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re:  Multiple search and replace
> 
> 
> Hi Pankaj,
> 
> see my comments below:
> 
> 
> Pankaj Chaturvedi wrote:
> > Hi all,
> >
> > I am trying to define multiple search and replace in style sheet.
> >
> 
> first thought: consider using XSLT 2.0, which has search and 
> replace built in using the replace() function which can 
> handle regular expression style search and replace in one call.
> 
> > Basically trying to convert [#x02010] (and other Unicode values) to 
> > their corresponding values &#x02010; .
> >
> 
> Second thought consider using XSLT 2.0: getting the numeric 
> value of a character can be done using string-to-codepoints, 
> which is not available in XSLT 1.0. Second thought (b): 
> sorry, I see that you mean the literal string '[#...]'....
> 
> > Below is what I am trying to do:
> >
> > <snip />
> >
> > I have two questions in regard:
> >
> > 1. I am bound to define & as &amp; as XMLSpy giving an error 
> > "character is grammatically unexpected". Is there other way of 
> > overcoming this issue and get & in output.
> >
> 
> Not doing something because your tool limits you is very dangerous...
> However, in this case, XMLSpy is correct. The underlying technology
> (XML) does not allow a literal &amp;, simply because XSLT 
> *is* XML and XML does not allow it. However, if you output as 
> text, the serializer will output '&' when you put &amp; somewhere.
> 
> Third thought: use XSLT 2.0. It has the ability to add character maps.
> In a character map you can say that some character, say '$' 
> (but using something from the Private Area Unicode ranges is 
> recommended) can be mapped to some string. Using character 
> maps you can get a literal '&' in the output.
> 
> > 2. I also need to replace "]" to  ";" for which I was 
> trying to call 
> > the another template with in <xml:template match="text()"> as below 
> > but
> doesn't
> > seems to be working.
> >
> > <snip />
> > Can we do multiple search and replaces in one named 
> template or do I 
> > need
> to
> > define them all separately (I need to call all of them in 
> one template 
> > <xsl:template match="text()">).
> >
> 
> XSLT is a functional language. You will have to call the 
> replace function recursively. I believe there's an example on 
> the exslt.org site which shows how you  can do this for a 
> multiple search and replace in a generic way.
> 
> Fourth thought: use XSLT 2.0. All you'll end up with then is a nested
> replace(replace(....)) call.
> 
> Fifth thought: use XSLT 2.0 for the whole shebang. Your whole 
> solution will look like this:
> 
> <xsl:output use-character-maps="searchreplace" />
> 
> <xsl:character-map name="searchreplace">
>     <xsl:output-character character="&#xE000;" string="&amp;" 
> /> </xsl:character>
> 
> <xsl:template match="text()">
>     <xsl:sequence select="replace(., '\[#(\d+)\]', 
> '&#xE000;#x\1;')" /> </xsl:template>
> 
> 
> Sixth thought: use XSLT 2.0. You seem to be using XMLSpy, 
> which can handle XSLT 2.0. However, its engine is a bit 
> flaky. If you run into problems, consider using either 
> Gestalt XSLT 2.0 or Saxon XSLT 2.0 processors.
> 
> Note: you may think that putting &amp; inside the 
> string-attribute of xsl:output-character creates &amp; in the 
> output, but this is not true.
> Since XSLT is XML, you must put &amp; there. But to get the 
> translation to serialize to literal &amp; instead means 
> double escaping: "&amp;amp;"
> (but that is not what you are after here). Understanding the 
> implications of using character references in XML is vital of 
> headache-free working with XML and XSLT (plus all other XML 
> related technologies in fact), but it can be hard at times to 
> get it right in your head.
> 
> Hope this helps,
> 
> Cheers,
> -- Abel Braaksma
> 
> 
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