Question

当我将XML结构更改为具有多个这样的好实体时，如何使工作成为here描述的解决方案：

<?xml version="1.0" encoding="ISO-8859-1"?>
<Work>    
<Good id = "1">
    <list  num="1050" id = "2531" desc="List 1">
        <part  num="1">
            <pos  isKey="0" id="2532" pid="2531" desc="Part 1" />
            <pos  num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2" />
            <pos  num="1.2.6." isKey="1" id="2591" pid="2554" desc="Position 1.2.6" />
          </part>
        </list>
        <list  num="1090" id = "3029" desc="List 2">
          <part  num="2">
            <pos  isKey="0" id="3033" pid="3029" desc="Category 2" />
            <pos  isKey="0" id="3040" pid="3033" desc="Part 9" />
            <pos  num="9.2." isKey="0" id="3333" pid="3040" desc="Position 9.2" />
            <pos  num="9.2.1." isKey="0" id="3334" pid="3333" desc="Position 9.2.1" />
            <pos  num="9.2.1.2" isKey="1" id="3339" pid="3334" desc="Position 9.2.1.2" />
        </part>
    </list>
</Good>
<Good id = "2">
    <list  num="1050" id = "2531" desc="List 3">
        <part  num="1">
            <pos  isKey="0" id="2532" pid="2531" desc="Part 1" />
            <pos  num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2" />
            <pos  num="1.2.6." isKey="0" id="2591" pid="2554" desc="Position 1.2.6" />
            <pos  num="1.2.6.1." isKey="1" id="2592" pid="2591" desc="Position 1.2.6.1" />       
         </part>
     </list>      
 </Good>
</Work>

我尝试为Work / Good实体创建for-each循环，但它没有帮助：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="html" indent="yes" />

  <!-- key to look up any element with an id attribute based on the value of
       that id -->
  <xsl:key name="elementsByPid" match="*[@pid]" use="@pid" />

  <xsl:template match="/">
    <html>
      <body>
      <h2>Lists</h2>
        <xsl:apply-templates select="Work/Goods" />
      </body>
    </html>
  </xsl:template>

<xsl:template match="Work/Goods">
    <xsl:for-each select="Work/Goods">
        <xsl:value-of select="."/>
    </xsl:for-each>
    <xsl:apply-templates select="Work/Goods/list"  />
</xsl:template>

<xsl:template match="Work/Goods/list">
    <xsl:for-each select="Work/Goods/list">
        <xsl:value-of select="."/>
    </xsl:for-each>
    <xsl:apply-templates select="."  mode="table"/>
</xsl:template>

<xsl:template match="*" mode="table">
    <xsl:variable name="shouldOutput">
      <xsl:apply-templates select="." mode="shouldOutput" />
    </xsl:variable>
    <xsl:if test="string-length($shouldOutput)">
      <table>
        <xsl:apply-templates select="." />
      </table>
    </xsl:if>
  </xsl:template>

  <!-- the main recursive logic - first produce output for this row, then
       process any of the children (in the id->pid chain) that need to be
       output -->
  <xsl:template match="*">
    <xsl:apply-templates select="." mode="row" />
    <xsl:for-each select="key('elementsByPid', @id)">
      <xsl:variable name="shouldOutput">
        <xsl:apply-templates select="." mode="shouldOutput" />
      </xsl:variable>
      <xsl:if test="string-length($shouldOutput)">
        <xsl:apply-templates select="." />
      </xsl:if>
    </xsl:for-each>
  </xsl:template>

  <xsl:template match="*" mode="row">
    <tr>
      <td colspan="2"><xsl:value-of select="@desc" /></td>
    </tr>
  </xsl:template>

  <!-- special case for pos elements with a @num - produce two columns -->
  <xsl:template match="pos[@num]" mode="row">
    <tr>
      <td><xsl:value-of select="@num" /></td>
      <td><xsl:value-of select="@desc" /></td>
    </tr>
  </xsl:template>

  <!-- check whether this node should be output by checking whether it, or any
       of its descendants in the id->pid tree, has @out=1.  The template will
       return an empty RTF for nodes that should not be output, and an RTF
       containing a text node with one or more "1" characters for nodes that
       should.  -->
  <xsl:template match="*[@out='1']" mode="shouldOutput">1</xsl:template>
  <xsl:template match="*" mode="shouldOutput">
    <xsl:apply-templates select="key('elementsByPid', @id)"
         mode="shouldOutput"/>
  </xsl:template>

</xsl:stylesheet>
enter code here

模板中没有允许此代码工作我应该改变什么才能使其发挥作用？

Answer 1

您不需要使用任何for-each，只需删除两个模板

即可

<xsl:template match="Work/Goods">

<xsl:template match="Work/Goods/list">

完全

，并将根模板改为简单地说

  <xsl:template match="/">
    <html>
      <body>
      <h2>Lists</h2>
        <xsl:apply-templates select="Work/Good/list" mode="table" />
      </body>
    </html>
  </xsl:template>

不需要显式迭代，因为单个select表达式将提取所有list元素中的所有Good元素（请注意，您的XML示例具有名为{{的元素）在Good元素中添加{}}而不是Goods。

Answer 2

我不确定，这是否是您的意思，但我希望以下内容对您有用。我的建议是两次处理文件。一次使用step1.xsl，其结果为step2.xsl。在我的答案的最后给出了关于转换的内容的解释。

为输入文件指定上面提供的输入文件的形式，但将属性out="1"添加到<part />元素之一

<?xml version="1.0" encoding="ISO-8859-1"?>
<Work>    
<Good id = "1">
<list  num="1050" id = "2531" desc="List 1">
    <part  num="1">
        <pos  isKey="0" id="2532" pid="2531" desc="Part 1" />
        <pos  num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2" />
        <pos  num="1.2.6." isKey="1" id="2591" pid="2554" desc="Position 1.2.6" />
      </part>
    </list>
    <list  num="1090" id = "3029" desc="List 2">
      <part  num="2">
        <pos  isKey="0" id="3033" pid="3029" desc="Category 2" />
        <pos  isKey="0" id="3040" pid="3033" desc="Part 9" />
        <pos  num="9.2." isKey="0" id="3333" pid="3040" desc="Position 9.2" />
        <pos  num="9.2.1." isKey="0" id="3334" pid="3333" desc="Position 9.2.1" out="1" />
        <pos  num="9.2.1.2" isKey="1" id="3339" pid="3334" desc="Position 9.2.1.2" />
    </part>
</list>
</Good>
<Good id = "2">
<list  num="1050" id = "2531" desc="List 3">
    <part  num="1">
        <pos  isKey="0" id="2532" pid="2531" desc="Part 1" />
        <pos  num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2" />
        <pos  num="1.2.6." isKey="0" id="2591" pid="2554" desc="Position 1.2.6" />
        <pos  num="1.2.6.1." isKey="1" id="2592" pid="2591" desc="Position 1.2.6.1" />       
     </part>
 </list>      
 </Good>
</Work>

以下转化step1.xsl

<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes" />
    <xsl:strip-space elements="*" />

    <xsl:key name="id" match="pos" use="@id" />
    <xsl:key name="pid" match="pos" use="@pid" />

    <xsl:template match="@*|node()">
        <xsl:copy><xsl:apply-templates select="@*|node()" /></xsl:copy>
    </xsl:template>

    <xsl:template match="pos[count(key('id', @pid)) = 0]">
        <xsl:variable name="id" select="@id" />
        <xsl:copy>
            <xsl:copy-of select="@*" />
            <xsl:apply-templates select="ancestor::Good//pos[@pid=$id]" mode="nested" />
        </xsl:copy>
    </xsl:template>

    <xsl:template match="pos" />
    <xsl:template match="pos" mode="nested">
        <xsl:variable name="id" select="@id" />
        <xsl:copy>
            <xsl:copy-of select="@*" />
            <xsl:apply-templates select="ancestor::Good//pos[@pid=$id]" mode="nested" />
        </xsl:copy>
    </xsl:template>
</xsl:transform>

产生以下输出

<?xml version="1.0"?>
<Work>
  <Good id="1">
    <list num="1050" id="2531" desc="List 1">
      <part num="1">
        <pos isKey="0" id="2532" pid="2531" desc="Part 1">
          <pos num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2">
            <pos num="1.2.6." isKey="1" id="2591" pid="2554" desc="Position 1.2.6"/>
          </pos>
        </pos>
      </part>
    </list>
    <list num="1090" id="3029" desc="List 2">
      <part num="2">
        <pos isKey="0" id="3033" pid="3029" desc="Category 2">
          <pos isKey="0" id="3040" pid="3033" desc="Part 9">
            <pos num="9.2." isKey="0" id="3333" pid="3040" desc="Position 9.2">
              <pos num="9.2.1." isKey="0" id="3334" pid="3333" desc="Position 9.2.1" out="1">
                <pos num="9.2.1.2" isKey="1" id="3339" pid="3334" desc="Position 9.2.1.2"/>
              </pos>
            </pos>
          </pos>
        </pos>
      </part>
    </list>
  </Good>
  <Good id="2">
    <list num="1050" id="2531" desc="List 3">
      <part num="1">
        <pos isKey="0" id="2532" pid="2531" desc="Part 1">
          <pos num="1.2." isKey="0" id="2554" pid="2532" desc="Position 1.2">
            <pos num="1.2.6." isKey="0" id="2591" pid="2554" desc="Position 1.2.6">
              <pos num="1.2.6.1." isKey="1" id="2592" pid="2591" desc="Position 1.2.6.1"/>
            </pos>
          </pos>
        </pos>
      </part>
    </list>
  </Good>
</Work>

作为以下转化的输入step2.xsl

<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes" />
    <xsl:strip-space elements="*" />

    <xsl:template match="/">
        <xsl:apply-templates match="pos[@out=1]" />
    </xsl:template>

    <xsl:template match="pos[@out=1]">
        <table>
        <tr><td><xsl:value-of select="ancestor::list/@desc" /></td></tr>
        <xsl:for-each select="ancestor-or-self::pos[count(ancestor::pos) &lt; 2]">
            <xsl:sort select="position()" order="descending" />
            <tr><td><xsl:value-of select="@desc" /></td></tr>
        </xsl:for-each>
        <xsl:for-each select="ancestor-or-self::pos[count(ancestor::pos) >= 2]">
            <tr>
                <td><xsl:value-of select="@num" /></td>
                <td><xsl:value-of select="@desc" /></td>
            </tr>
        </xsl:for-each>
        </table>
    </xsl:template>
</xsl:transform>

产生的输出看起来像我想要的那样

<?xml version="1.0"?>
<table>
  <tr>
    <td>List 2</td>
  </tr>
  <tr>
    <td>Part 9</td>
  </tr>
  <tr>
    <td>Category 2</td>
  </tr>
  <tr>
    <td>9.2.</td>
    <td>Position 9.2</td>
  </tr>
  <tr>
    <td>9.2.1.</td>
    <td>Position 9.2.1</td>
  </tr>
</table>

请澄清是否没有。

说明：第一个转换step1.xsl根据pos属性从@pid元素的平面列表构建嵌套层次结构。这使得在第二轮中更容易通过父母链。

第二个转换step2.xsl然后匹配应该生成输出的每个pos元素，由其@out属性指示，并写出您在ASCII中作为ASCII艺术起草的结构表原来的问题。

尚未完成的是在相同的情况下合并相同的表格。两个pos[@out=1]元素包含在同一list。

中

Answer 3

问题在于@pid不足以正确识别<pos>元素，因为多个商品中可能会出现相同的@pid。

这意味着您需要将每个@pid与<Good>中所属的<xsl:key>相关联（然后，在每次使用密钥时）。这可以通过构建@pid和封闭Good/@id的唯一字符串来完成。

concat(@pid, '|', ancestor::Good/@id)

以下内容基本上是来自previous question的XSLT。我突出了这些变化。

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="html" indent="yes" />

  <xsl:key name="elementsByPid" match="*[@pid]" use="concat(@pid, '|', ancestor::Good/@id)" />
  <!--  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^  -->

  <xsl:template match="/">
    <html>
      <body>
        <h2>Lists</h2>
        <xsl:apply-templates select="/Work/Good/list" mode="table" />
      </body>
    </html>
  </xsl:template>

  <xsl:template match="*" mode="table">
    <xsl:variable name="shouldOutput">
      <xsl:apply-templates select="." mode="shouldOutput" />
    </xsl:variable>
    <xsl:if test="string-length($shouldOutput)">
      <table>
        <xsl:apply-templates select="." />
      </table>
    </xsl:if>
  </xsl:template>

  <xsl:template match="*">
    <xsl:apply-templates select="." mode="row" />
    <xsl:for-each select="key('elementsByPid', concat(@id, '|', ancestor::Good/@id))">
    <!--  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^  -->
      <xsl:variable name="shouldOutput">
        <xsl:apply-templates select="." mode="shouldOutput" />
      </xsl:variable>
      <xsl:if test="string-length($shouldOutput)">
        <xsl:apply-templates select="." />
      </xsl:if>
    </xsl:for-each>
  </xsl:template>

  <xsl:template match="*" mode="row">
    <tr>
      <td colspan="2"><xsl:value-of select="@description" /></td>
    </tr>
  </xsl:template>

  <xsl:template match="pos[@num]" mode="row">
    <tr>
      <td class="num"><xsl:value-of select="@num" /></td>
      <td><xsl:value-of select="@description" /></td>
    </tr>
  </xsl:template>

  <xsl:template match="*[@out='1']" mode="shouldOutput">1</xsl:template>

  <xsl:template match="*" mode="shouldOutput">
    <xsl:apply-templates select="key('elementsByPid', concat(@id, '|', ancestor::Good/@id))" mode="shouldOutput"/>
    <!--  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^  -->
  </xsl:template>

</xsl:stylesheet>

另见http://www.xmlplayground.com/9iG5oz。

PS：由于您似乎在了解<xsl:apply-templates>的工作原理方面存在问题，因此您可能会发现an older answer of mine explaining it有用。我还写了an answer that explains <xsl:key>。

多种商品的类别树的XSL结构

3 个答案: