I have a node <msg>
which contains a message such as
string1
string 2
sting3
however when it renders, it renders all on one line.
How can i replace all \n
with <br /
> elements?
I've tried
<xsl:value-of select='replace(msg, "&#xA;", "<br/>")' />
but I get this error
Error loading stylesheet: Invalid XSLT/XPath function.
How do I do this?
You are working with an XSLT 1.0 processor, whereas replace()
is a function that has been introduced with XSLT/XPath 2.0.
Call this template on the string you want to process instead:
<xsl:template name="break">
<xsl:param name="text" select="string(.)"/>
<xsl:choose>
<xsl:when test="contains($text, '
')">
<xsl:value-of select="substring-before($text, '
')"/>
<br/>
<xsl:call-template name="break">
<xsl:with-param
name="text"
select="substring-after($text, '
')"
/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$text"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
Like this (it will work on the current node):
<xsl:template match="msg">
<xsl:call-template name="break" />
</xsl:template>
or like this, explicitly passing a parameter:
<xsl:template match="someElement">
<xsl:call-template name="break">
<xsl:with-param name="text" select="msg" />
</xsl:call-template>
</xsl:template>