克服 xsl 中的未闭合标签 [英] Overcoming unclosed tags in xsl

问题描述

我正在开发一个下拉菜单，该菜单从 Sharepoint 列表中提取数据并使用 xsl 进行样式设置.html结构是一个简单的无序列表

I'm working on a drop down menu that pulls data from a Sharepoint list and is styled with xsl. The html structure is a simple unordered list

<ul><li>parent page</li><ul><li>sub page 1</li><li>sub page2</li></ul></ul>

xsl 测试列表中的项目是父页面还是子页面以及子页面编号是什么.父页面有一个编号，因此可以将所需的子页面附加到正确的父页面.

The xsl tests if an item in the list is a parent page or a sub page and what the sub page number is. Parent pages have a number so the desired sub pages can be attached to the correct parent page.

每个项目的基本 xsl 是这样的:

The basic xsl for each item is this:

<xsl:for-each select="//Data/Row">
    <xsl:if test="./@Page_x0020_type = 0">
        <li >
            <a>
                <xsl:attribute name="href">
                    <xsl:value-of
                     select="./@Page_x002f_link_x0020_url"/>
                </xsl:attribute>
                <xsl:value-of select="./@Title0"/>
            </a>
        </li>
    </xsl:if>

我希望 xsl 等同于:如果是子页面，则在项目前添加 ul 标签，如果是父页面下的最后一个子页面，则添加结束 ul 标签.

I want the xsl to equate to this: If it's a sub page, precede the item with a ul tag and if it's the last sub page under a parent, add a closing ul tag.

如何解决 xsl 不允许我添加未关闭的 ul 标签的事实，因为它(正确地)不知道该标签是否会关闭?

How do I get around the fact that xsl won't let me add an unclosed ul tag because it (rightly) doesn't know if the tag is going to get closed?

XML

    <Field Type="Text" 
           DisplayName="Page/link url" 
           Required="FALSE" 
           MaxLength="255" 
           Name="Page_x002f_link_x0020_url"/>
    <Field ReadOnly="TRUE" 
           Type="Computed" 
           Name="LinkTitle" 
           DisplayName="Page number"/>
    <Field Type="Text" 
           DisplayName="Title" 
           Required="FALSE" 
           MaxLength="255" 
           Name="Title0"/>  
    <Field Type="Choice" 
           DisplayName="Page type" 
           Required="FALSE" 
           Format="RadioButtons" 
           FillInChoice="FALSE" 
           Name="Page_x0020_type">    
           <CHOICES>  
               <CHOICE>0</CHOICE>  
               <CHOICE>1</CHOICE>  
           </CHOICES>  
           <DefaultFormula>=0</DefaultFormula>  
           <DefaultFormulaValue/>  
    </Field>
    <Field Type="Text" 
           DisplayName="Sub page number" 
           Required="FALSE" 
           MaxLength="2" 
           Name="Sub_x0020_page_x0020_number"/>
    <Field Type="Text" 
           DisplayName="Parent page number" 
           Required="FALSE" 
           MaxLength="1" 
           Name="Parent_x0020_page_x0020_number">
           <Default>0</Default>
    </Field>
</Schema>
<Data ItemCount="1">
    <Row Page_x002f_link_x0020_url="" 
         LinkTitle="" 
         Title0="" 
         Page_x0020_type="" 
         Sub_x0020_page_x0020_number="" 
         Parent_x0020_page_x0020_number="" 
         ul="" 
         _x003c_li_x003e__x003c_a_x003e_="" 
         _x003c__x002f_a_x003e__x003c__x0="" 
         _x003c_ul_x003e__x003c_li_x003e_="" 
         _x003c__x002f_a_x003e__x003c__x00="" 
         _x003c__x002f_a_x003e__x003c__x01=""/>

---

更新:

页面类型:

• 父
• parent1
• 子页面 1 - 需要以
  ...
  • 子页面 1
  开头
  • 中间子页面-(编号设置排序顺序-html与父页面相同)
  • 子页面#
  • 组中的最后一个子页面 - 需要以 </ul>
  结尾

  ---

  UPDATE:

  Page types:
  • Parent <li>parent1</li>
  • Sub page 1 - needs to start with a <ul>...<li>sub page 1</li>
  • Middle sub pages-(numbered to set sort order-html same as parent page) <li>sub page #</li>
  • Last sub page in group - needs to end with a </ul>

  推荐答案

  我希望 xsl 等同于这个:如果它是一个子页面，在项目之前添加一个 ul 标签，如果它是父页面下的最后一个子页面，添加一个结束 ul 标签.

  I want the xsl to equate to this: If it's a sub page, precede the item with a ul tag and if it's the last sub page under a parent, add a closing ul tag.

  你想错了.XSLT 不编写标签，它编写节点树.您不能将半个节点写入结果树.

  You're thinking the wrong way. XSLT doesn't write tags, it writes a tree of nodes. You can't write half a node to the result tree.

  在 XSLT 2.0 中，您可能可以使用带有 group-starting-with 属性的 xsl:for-each-group 来做您想做的事 - 我不能具体说明，因为您还没有足够清楚地说明问题，但它是大概是这样的:

  In XSLT 2.0 you can probably do what you want using xsl:for-each-group with the group-starting-with attribute - I can't be specific, because you haven't specified the problem clearly enough, but it's probably something like this:

  <xsl:template match="parent">
    <xsl:for-each-group select="*" group-starting-with="li[....]">
      <ul>
        <xsl:copy-of select="current-group()"/>
      </ul>
    </xsl:for-each-group>
  </xsl:template>
  

  如果您坚持使用 XSLT 1.0 那么它会更困难:我将使用一种称为兄弟递归"的技术，在该技术中，您将模板应用于第一个孩子，然后将模板应用于下一个兄弟，并且很快.父级的模板规则是这样的:

  If you're stuck with XSLT 1.0 then it will be more difficult: I would use a technique called "sibling recursion" in which you apply-templates to the first child, which in turn applies templates to the next sibling, and so on. The template rule for the parent does this:

  <xsl:template match="parent">
    <xsl:apply-templates select="*[1]"/>
  </xsl:template>
  

  作为组中第一个兄弟姐妹的模板规则是这样的:

  The template rule for a sibling that is to be the first in a group does this:

  <xsl:template match="parent/*[ (: where this is the first in a group :) ]">
    <group>
      <xsl:copy-of select="."/>
      <xsl:apply-templates select="following-sibling::*[1]"/>
    </group>
    <xsl:apply-templates select="following-sibling::*[(* start of next group *)][1]"/>
  </xsl:template>
  

  其他兄弟姐妹的模板规则确实如此:

  and the template rule for other siblings does:

  <xsl:template match="parent/*">
    <xsl:copy-of select="."/>
    <xsl:apply-templates select="following-sibling::*[1]"/>
  </xsl:template>
  

  即使对于有经验的 XSLT 开发人员来说，细节也可能很棘手.

  The details can be tricky even for experienced XSLT developers.

  这篇关于克服 xsl 中的未闭合标签的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！