how to concatenate increased number into href tag - XSLT

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ds 191 posts 223 karma points

Sep 21, 2010 @ 14:14

how to concatenate increased number into href tag?

I have created such loop

 <xsl:template match="/">
  <ul class="mainTabSection">
    <xsl:for-each select="$currentPage/node [string(data [@alias='umbracoNaviHide']) != '1']">
   <li>
        <xsl:variable name="i" select="position()" />
        <xsl:copy>
         <xsl:value-of select="concat('#tab', $i)"/>
        </xsl:copy>
  </li>
    </xsl:for-each>
   </ul>
 </xsl:template>

What I need to achieve is to get the following result

        <ul class="mainTabSection"> 
            <li><a href="#tab1">link name</a></li> 
            <li><a href="#tab2">link name</a></li> 
            <li><a href="#tab3">link name</a></li> 
        </ul>

but I do not know how to wrap "<a href=""></a> with increased tab number.

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Chriztian Steinmeier 2800 posts 8790 karma points MVP 8x admin c-trib

Sep 21, 2010 @ 14:18
0
Hi ds,

position() is your friend:
```
<a href="#tab{position()}"> ... </a>
```
/Chriztian
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ds 191 posts 223 karma points

Sep 21, 2010 @ 14:25

Thanks Chriztian,

I changed code snippet and it is working as it is expected.

 <xsl:template match="/">
  <ul class="mainTabSection">
    <xsl:for-each select="$currentPage/node [string(data [@alias='umbracoNaviHide']) != '1']">
   <li>
        <a href="#tab{position()}"><xsl:value-of select="data [@alias = 'tabLink']"/></a>
   </li>
    </xsl:for-each>
   </ul>
 </xsl:template>

thx,

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