Create XPath expression dynamically

[FaensenD]

Consider the following XML document

<root>
<PersonStreet>24 Miller Street</PersonStreet>
<PersonZIP>12345</PersonZIP>
<PersonCity>Munich</PersonCity>
<CompanyStreet>24 Miller Street</CompanyStreet>
<CompanyZIP>12345</CompanyZIP>
<CompanyCity>Munich</CompanyCity>
</root>

In a template I want to access the nodes that are prefixed by a given
parameter.

The following doesn't work but should illustrate what I mean.

<xsl:template name="address">
<xslaram name="prefix/>

<xsl:element name="{$prefix}">
<xsl:value-of select=$prefix + "Street" />
<xsl:value-of select=$prefix + "ZIP" />
<xsl:value-of select=$prefix + "City" />
</xsl:element>

</xsl:template>

Is it possible to create such an XPath expression dynamically?

Thank you

Daniel Faensen

 

[Joris Gillis]

Hi,

The following doesn't work but should illustrate what I mean.

<xsl:template name="address">
<xslaram name="prefix/>

<xsl:element name="{$prefix}">
<xsl:value-of select=$prefix + "Street" />
<xsl:value-of select=$prefix + "ZIP" />
<xsl:value-of select=$prefix + "City" />
</xsl:element>

</xsl:template>

Is it possible to create such an XPath expression dynamically?

No it is not.
You can however, work around it:

<xsl:value-of select="*[local-name()=concat($prefix,'City')]" />


regards,