Hi folks !
I have text in a database with \n characters in it.
I want to produce HTML and keep LF in my presentation.
So I have to convert \n to <BR>
I can't replace with < and > characters because they can be confusing
for the parser , am I right ?
In order to see if my template is ok, I tried it but I have a
StackOverFlowError ! (replacing \n with BR )
Any help is appreciated
Thanks
Sébastien
Here is what I did :
<xsl:template name="remplaceCaracDansChaine">
<xsl:param name="chaineEntrante"/><!--Input String-->
<xsl:param name="caracEntrant"/> <!--what I want to be replaced-->
<xsl:param name="caracSortant"/><!--the new piece of string-->
<xsl:choose>
<xsl:when test="contains($chaineEntrante,caracEntrant)">
<xsl:value-of
select="concat(substring-before($chaineEntrante,$caracEntrant),$caracSortant)"/>
<xsl:call-template name="remplaceCaracDansChaine">
<xsl:with-param name="chaineEntrante"
select="substring-after($chaineEntrante,$caracEntrant)"/>
<xsl:with-param name="caracEntrant" select="$caracEntrant"/>
<xsl:with-param name="caracSortant" select="$caracSortant"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$chaineEntrante"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
Then, I call it like that :
<xsl:variable name="monTexte"><xsl:value-of
select="/page/rowfiche/texte"/></xsl:variable> <!--value coming from my
XML-->
<xsl:variable name="texteBR"><!--the destination variable with modified
text-->
<xsl:call-template name="remplaceCaracDansChaine">
<xsl:with-param name="chaineEntrante" select="string($monTexte)"/>
<xsl:with-param name="caracEntrant" select="n"/>
<xsl:with-param name="caracSortant" select="BR"/><!--I can't use
<BR> because of < and > characters-->
</xsl:call-template>
</xsl:variable>
<xsl:value-of select="$texteBR"/><!--print the transformed string-->
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