Output text containing 


[jeffward]

My xml source file contains text representing code and the software that saves this to xml format has converted the line breaks in the code to &#10.

I am using xsl to output the contents to html and so need to display the code block properly.

for example if the xml contains:-
Code="line 1
line 2
line 1
"

by default this is displayed in html as:-
line 1
line 2
line 1


i would like this displayed as:-
line 1
line 2
line 3

 

[JontyMC]

Have you tried using: disable-output-escaping="yes"

Jon

"Asteroids do not concern me, Admiral. I want that ship, not excuses.

 

[jeffward]

I have managed to find a solution elsewhere and have adapted this into a template which replaces the &amp;#10; entries with <br/> and then uses a pre tab to retain the formating, as follows: -

<pre>
<xsl:call-template name="ReplaceWithNL">
<xsl:with-param name="P_String">
<xsl:value-of select="." />
</xsl:with-param>
</xsl:call-template>
</pre>

<!-- Output P_String with <br /> instead of &amp;#10; -->
<xsl:template name="ReplaceWithNL">
<xslaram name="P_String" />

<!-- Does the string contain '&amp;#10;' ? -->
<xsl:if test="contains($P_String, '&amp;#10;')">

<!-- Output the first part of the string -->
<xsl:value-of select="substring-before($P_String, '&amp;#10;')" />

<!-- Output the line break -->
<br />

<!-- Call the template again to replace any further occurances -->
<xsl:call-template name="ReplaceWithNL">
<xsl:with-param name="P_String">
<xsl:value-of select="substring-after($P_String, '&amp;#10;')" />
</xsl:with-param>
</xsl:call-template>

</xsl:if>

<!-- If the string does not contain the value to replace then simply output it as is -->
<xsl:if test="not(contains($P_String, '&amp;#10;'))">
<xsl:value-of select="$P_String" />
</xsl:if>

</xsl:template>