XSL FO属性值匹配

Question

由于XML：

<expdur-entry><itemno id="exp-item1">1</itemno><maintenance lvl="f"/> 
<nsn><fsc>1130</fsc><niin>00-2X3-1</niin></nsn><name>Beeswax</name><desc></desc><partno>Pure Beeswax</partno><cageno>X1148</cageno><ui>LB</ui></expdur-entry> 

<expdur-entry><itemno id="exp-item2">2</itemno><maintenance lvl="f"/><nsn><fsc>X3X0</fsc> 
<niin>00-221-082</niin></nsn><name>Cloth, Abrasive</name><desc></desc><partno>L9-C-4X8</partno><cageno>81348</cageno> 
<ui>EA</ui></expdur-entry> 

<expdur-entry><itemno id="exp-item6">3</itemno> 
<maintenance lvl="f"/><nsn><fsc>130</fsc><niin>00-13-1802</niin></nsn> 
<name>Cloth, Duck</name><desc></desc><partno>L9-C-433XCL 2 FGX04</partno><cageno></cageno><ui>YD</ui></expdur-entry> 

<expdur-entry><itemno id="exp-item49">4</itemno><maintenance lvl="f"/> 
<nsn><fsc>2X81</fsc><niin>01-331-2212</niin></nsn><name>Tape, Textile, 
Nylon</name><desc></desc><partno>123-T-X38</partno> 
<cageno></cageno><ui>YD</ui></expdur-entry> 

<expdur-entry> 
<itemno id="exp-item46">5</itemno><maintenance lvl="f"/><nsn><fsc></fsc><niin></niin></nsn><name>Cloth, Nylon Diamond Weave</name><desc></desc><partno>11-1-13</partno> 
<cageno></cageno><ui></ui></expdur-entry> 

<expdur-entry><itemno id="exp-item45">6</itemno><maintenance lvl="f"/><nsn><fsc></fsc><niin></niin></nsn> 
<name>Cloth, Parachute, Nylon</name><desc></desc> 
<partno>L9-C-4438, TY V</partno><cageno></cageno><ui></ui> 
</expdur-entry> 

<expdur-entry><itemno id="exp-item117">7</itemno> 
<maintenance lvl="f"/><nsn><fsc>TBD</fsc><niin></niin></nsn><name>Cord, Nylon, Type X1</name><desc></desc><partno>L9-C-X1X</partno> 
<cageno></cageno><ui>YD</ui></expdur-entry> 

和

<bulk_itemswp> 
<pi.item><qty>.08 YD</qty><common_part_ref idref="exp-item49"/> 
</pi.item></bulk_itemswp> 

我在pi.item，我想拉<fsc>和<niin>从<expdur-entry>与他们@id的<itemno>匹配@idref从<common_part_ref>。在这种情况下，它是<itemno id="exp-item49">

我想我可以做这样的事情在<pi.item>模板：

<xsl:apply-templates select="//expdur-entry/itemno[@id=common_part_ref/@idref]" mode="bulk"> 

，但没有奏效。下面的XSL-FO的作品，但我怀疑这是最好的方式。如果可能的话，我宁愿更正上面的select子句的语法。

这是我的XSLFO。

<xsl:template match="pi.item"> 
     <fo:table-row> 
      <xsl:choose> 
      <xsl:when test="common_part_ref"> 
       <xsl:apply-templates select="//expdur-entry/itemno" mode="bulk"> 
       <xsl:with-param name="id"> 
        <xsl:value-of select="common_part_ref/@idref"/> 
       </xsl:with-param> 
       </xsl:apply-templates> 
      </xsl:when> 
      <xsl:otherwise /> 
      </xsl:choose> 
     </fo:table-row> 
    </xsl:template> 

<xsl:template match="expdur-entry/itemno" mode="bulk"> 
<xsl:param name="id"/> 
    <xsl:if test="@id = $id"> 
    <fo:table-cell> 
     <fo:block> 
      <xsl:value-of disable-output-escaping="no" select="../nsn/fsc"/> 
      <xsl:text disable-output-escaping="no">-</xsl:text> 
      <xsl:value-of disable-output-escaping="no" select="normalize-space(../nsn/niin)"/> 
     </fo:block> 
    </fo:table-cell> 
    </xsl:if> 
</xsl:template> 

Answer 1

的current() XSLT的功能恰恰解决了这个问题：

<xsl:template match="pi.item"> 
    <fo:table-row> 
     <xsl:apply-templates select="//expdur-entry/itemno[ 
      @id = current()/common_part_ref/@idref 
     ]" mode="bulk" /> 
    </fo:table-row> 
</xsl:template> 

<xsl:template match="expdur-entry/itemno" mode="bulk"> 
    <fo:table-cell> 
     <fo:block> 
      <xsl:value-of select="concat(../nsn/fsc, '-', normalize-space(../nsn/niin))" /> 
     </fo:block> 
    </fo:table-cell> 
</xsl:template> 

Answer 2

另一种选择是使用xsl:key ...

<xsl:key name="expdurByID" match="expdur-entry" use="itemno/@id"/> 

<xsl:template match="pi.item"> 
    <fo:table-row> 
    <xsl:apply-templates 
     select="key('expdurByID',common_part_ref/@idref)" mode="bulk"/> 
    </fo:table-row> 
</xsl:template> 

<xsl:template match="expdur-entry" mode="bulk"> 
    <fo:table-cell> 
    <fo:block> 
     <xsl:value-of select="concat(normalize-space(nsn/fsc), 
     '-',normalize-space(nsn/niin))"/> 
    </fo:block> 
    </fo:table-cell> 
</xsl:template>