使用XSLT 1.0

Question

我用下面通过一个节点组迭代有条件分组节点项目（40个字符或更少）这样的事情：

<div class="largeItems"> 
    <div class="large">Large Item</div> 
    <div class="large">Large Item</div> 
    <div class="large">Large Item</div> 
    <div class="large">Large Item</div> 
    <div class="large">Large Item</div> 
</div> 
<div class="smallItems"> 
    <div class="small">Small Item</div> 
    <div class="small">Small Item</div> 
    <div class="small">Small Item</div> 
    <div class="small">Small Item</div> 
    <div class="small">Small Item</div> 
</div> 

谢谢。

Answer 1

尝试：

<div class="largeItems"> 
    <xsl:for-each select="NewDataSet/VehicleDetail/Options/Option[string-length() > 40]"> 
     <div class="large"> 
      <xsl:value-of select="."/> 
     </div> 
    </xsl:for-each> 
</div> 
<div class="smallItems"> 
    <xsl:for-each select="NewDataSet/VehicleDetail/Options/Option[string-length() &lt;= 40]"> 
     <div class="small"> 
      <xsl:value-of select="."/> 
     </div> 
    </xsl:for-each> 
</div> 

---

或者，如果你喜欢更少的代码重复：

<div class="largeItems"> 
    <xsl:apply-templates select="NewDataSet/VehicleDetail/Options/Option[string-length() > 40]"> 
     <xsl:with-param name="class" select="'large'"/> 
    </xsl:apply-templates> 
</div> 
<div class="smallItems">  
    <xsl:apply-templates select="NewDataSet/VehicleDetail/Options/Option[string-length() > 40]"> 
     <xsl:with-param name="class" select="'small'"/> 
    </xsl:apply-templates> 
</div> 

然后：

<xsl:template match="Option"> 
    <xsl:param name="class"/> 
    <div class="{$class}"> 
     <xsl:value-of select="."/> 
    </div> 
</xsl:template> 

---

未经测试，因为既没有提供输入也没有提供上下文。