XLST Replace

rangeshram · November 23rd, 2009, 07:42 AM

Hi All,

I would like to know how to replace the attribute with '\'(BackwardSlash) with '_'(Underscore). Since we have to assign the path for href purpose.

I am extracting the ref.id attribute with the following xslt:

Code:

 
<xsl:template match="index.level">
<a><xsl:attribute name="name"><xsl:apply-templates select="@ref.id"/></xsl:attribute></a>
 
 
<xsl:template match="xref.index"><a><xsl:attribute name="href"><xsl:text>#</xsl:text>
<xsl:apply-templates select="@ref"/></xsl:attribute><xsl:apply-templates/></a></xsl:template>

Input:

Code:

 
<index.level ref.id="Exemp_254\AFL-CIO"><name>AFL-CIO</name>
....
<xref.index ref="Exemp_254\AFL-CIO">See AFL-CIO</xref.index>

output required:

Code:

 
<a name="Exemp_254_AFL-CIO">AFL-CIO
 
<a href="Exemp_254_AFL-CIO">See AFL-CIO

Please provide me solution for replace.

Thanks

Martin Honnen · November 23rd, 2009, 07:50 AM

Use translate e.g.

Code:

<xsl:template match="index.level">
  <a name="{translate(@ref.id, '\', '_')}">...</a>
</xsl:template>