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

[David Carlisle <davidc@xxxxxxxxx>]

you said
from &#13; to <br/>,

but your code had
              <xsl:with-param name="pattern" select="'&#13;'"/>
              <xsl:with-param name="replacement" select="'"/>

ie replacing #13 by nothing.

Also it's most unlikely that there are any #13's in the file they would
only be there by literally going & # 1 3 ; anh combiation of #10 and 13
get normalised to a single #10 on input, so you want to look for #10 and
replace it by br

              <xsl:with-param name="pattern" select="'&#10;'"/>
              <xsl:with-param name="replacement">
<br/>
</xsl:with-param>


You also need to lose
                 <xsl:when test="$replacement != ''">

as that test the string value of the replacement which is "" in the case
of an empty br element.

There seems to be something strange with the logic of the template, with
each replacement being specified more than once, but I didn't debug that
tonight.

David


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