for-each:获取每3个项目(分成3列)

Question

我在xsl中有以下节点:

<foo>
    <bar>1</bar>
    <bar>2</bar>    
    <bar>3</bar>
    <bar>4</bar>        
    <bar>5</bar>
    <bar>6</bar>    
    <bar>7</bar>
    <bar>8</bar>            
    <bar>9</bar>
</foo>

并希望将其转换为以下html:

<ul class="one">
    <li>1</li>
    <li>4</li>
    <li>7</li>
</ul>
<ul class="two">
    <li>2</li>
    <li>5</li>
    <li>8</li>
</ul>
<ul class="three">
    <li>3</li>
    <li>6</li>
    <li>9</li>
</ul>

很难弄清楚如何循环并获得每个第三项,想要做这样的事情:

<ul class="one">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X1]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>

<ul class="two">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X2]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>

<ul class="three">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X3]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>   

Where:
X1 = Every third item starting from position 1
X2 = Every third item starting from position 2
X3 = Every third item starting from position 3

可能需要使用last(),但不能完全正常工作.

Answer 1

在XPath中,条件是:

not(position() mod 3)

要么

position() mod 3 = 0

我不明白为什么你不能使用op:mod.

编辑:关于新问题,只需减去偏移量.所以:

X1:

position() mod 3 = 1

X2:

position() mod 3 = 2

X3:

position() mod 3 = 0

编辑2:现在我理解你的问题.