Re: [xsl] Replace Special Characters: return, blank and tab

Subject: Re: [xsl] Replace Special Characters: return, blank and tab
From: "Christina" <lechaelnah@xxxxxxxxxxx>
Date: Thu, 5 Aug 2004 19:00:47 -0700
I changed the:<xsl:with-param name="pattern" select="'&#13;'" />
              <xsl:with-param name="replacement" select="''" />
into : <xsl:with-param name="pattern" select="'&#13;'" />
          <xsl:with-param name="replacement">
           <p>hereIAm</p>
         </xsl:with-param>
still only "hereIAm" is inserted.
I checked the faq, the replacement that only one layer: it only calls the
template once,
If I called the template only once in my code to only replace the #13, then
it works fine, the whole node is inserted instead of the text in it.

here is my input file, which is an output from an xml based servlet:


package milkway.mybean.rect;&#13;
import java.beans.PropertyChangeSupport;&#13;
import java.beans.PropertyChangeListener;&#13;
import java.awt.Font;&#13;
import java.io.Serializable;&#13;
import java.awt.Canvas;&#13;
&#13;
public class MyButton extends Canvas &#13;
		implements Serializable{&#13;
	private PropertyChangeSupport changes=new
PropertyChangeSupport(this);&#13;
	&#13;.....
thank you for your kind help anyway, have a good night!
----- Original Message -----
From: "David Carlisle" <davidc@xxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Thursday, August 05, 2004 3:25 PM
Subject: Re: [xsl] Replace Special Characters: return, blank and tab


>
> > Otherwise, I don't know how to insert "<" and ">" without escaping them,
>
> you don't want to insert those characters you want to insert a br node,
> just put a br node as the content of with-param not using the select
> attribute, as I showed.
>
> > It is a good idea to match the "&#10;" too, but now I cannot even make
the
> > "&#13;" working
>
> It is _most_ unlikely that there are any #13 in the XML file, the XML
> spec specifies that all line endings (dos/mac/unix format) are all
> reported as #10 howeverthey are in the source.
>
> If you are losing element markup then usual reasons are applying
> xsl:value-of  which gives the string value (or string ops such as
> substring-before which also force a string value) or using the text
> output method. Your replace template would get rid of any markup in its
> input (as that passes through substring but I thought you had it nested
> such that adding <br/> was the last thing you do.
>
> Sorry its too late for debugging your code tonight, the faq for this
> list has a replace template that does work, as does Jeni's site.
>
> David
>
>
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