从最低成员获取所有祖先的公共 ID

Question

正在尝试组装正确的 XSLT 代码，以将从递归链的最低成员（从其 "uniq-id" 中）获取的公共 ID 分配给上面的所有元素。无法获取祖先的节点。

来源xml

<root>
  <Object id="top"                        uniq-id="001" status="0" rank="1"/>
  <Object id="middle"  id-parent="top"    uniq-id="020" status="0" rank="3"/>
  <Object id="middle"  id-parent="top"    uniq-id="021" status="1" rank="3"/>

  <!--   only the element with status="0" is considered -->

  <Object id="bottom-1"  id-parent="middle" uniq-id="111" status="1" rank="6" />
  <Object id="bottom-2"  id-parent="middle" uniq-id="222" status="1" rank="6" />
  <Object id="bottom-3"  id-parent="middle" uniq-id="333" status="0" rank="6" />  <!-- will be taken -->
  <Object id="bottom-4"  id-parent="middle" uniq-id="444" status="1" rank="6" />
  <Object id="bottom-5"  id-parent="middle" uniq-id="555" status="1" rank="7" />
  <Object id="bottom-6"  id-parent="middle" uniq-id="666" status="0" rank="7" />  <!-- will be taken -->
  <Object id="bottom-7"  id-parent="middle" uniq-id="777" status="1" rank="6" />
  <Object id="bottom-8"  id-parent="middle" uniq-id="888" status="1" rank="6" />
  <Object id="bottom-9"  id-parent="middle" uniq-id="999" status="0" rank="6" />  <!-- will be taken -->
</root>

starting-xslt

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   xmlns:exslt="http://exslt.org/common">
  <xsl:key name="Object-By-id"        match="Object[(@status='0')]"  use="@id"/>
  <xsl:key name="Object-By-id-parent" match="Object[(@status='0')]"  use="string(@id-parent)"/>

  <xsl:variable name="fold-rtf">
    <xsl:apply-templates select="/" mode="fold"/>
  </xsl:variable>
  <xsl:variable name="folded-tree" select="exslt:node-set($fold-rtf)"/>

  <xsl:template match="@*|node()[(@status='0')]">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()[(@status='0')]"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="Object[@status=0]/@*[last()]">
    <xsl:variable name="current" select=".."/>
    <xsl:copy/>
    <xsl:for-each select="$folded-tree">
      <xsl:for-each select="key('Object-By-id',$current/@id)">

        <!-- ====== !!  =======-->

        <xsl:if test="position()!=last()">
          <xsl:for-each select="ancestor-or-self::*">
            <xsl:attribute name="chain-id">
              <xsl:value-of select="@uniq-id"/>
            </xsl:attribute>
          </xsl:for-each>
        </xsl:if>

        <!-- ======= !! =======-->

      </xsl:for-each>
    </xsl:for-each>
  </xsl:template>

  <xsl:template match="/|*" mode="fold">
    <xsl:copy>
      <xsl:copy-of select="@*"/>
      <xsl:apply-templates select="key('Object-By-id-parent',string(@id))" mode="fold">
      </xsl:apply-templates>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

结果（不正确）https://xsltfiddle.liberty-development.net/94Acsm2/1

假设输出

<root>
  <Object id="top"                          uniq-id="001" status="0" rank="1" chain-id="20"/>
  <Object id="middle"    id-parent="top"    uniq-id="020" status="0" rank="3" chain-id="20"/>

  <Object id="bottom-3"  id-parent="middle" uniq-id="333" status="0" rank="6" chain-id="333"/>
  <Object id="middle"    id-parent="top"    uniq-id="020" status="0" rank="3" chain-id="333"/>
  <Object id="top"                          uniq-id="001" status="0" rank="1" chain-id="333"/>

  <Object id="bottom-6"  id-parent="middle" uniq-id="666" status="0" rank="7" chain-id="666"/>
  <Object id="middle"    id-parent="top"    uniq-id="020" status="0" rank="3" chain-id="666"/>
  <Object id="top"                          uniq-id="001" status="0" rank="1" chain-id="666"/>

  <Object id="bottom-9"  id-parent="middle" uniq-id="999" status="0" rank="6" chain-id="999"/>
  <Object id="middle"    id-parent="top"    uniq-id="020" status="0" rank="3" chain-id="999"/>
  <Object id="top"                          uniq-id="001" status="0" rank="1" chain-id="999"/>
</root>

欢迎您提出有关如何改进 XSLT 代码的所有解决方案

Answer 1

从非工作代码逆向工程你的需求总是很困难，但我认为你想做的是用 @status="0" 对 select 元素进行处理，对于其中的每一个，显示逻辑祖先，即关系 parent(X) 的递归展开，其中 parent(X) selects @id 等于 X/@parent-id.[=17= 的元素]

我用 XSLT 1.0 编码已经有很长一段时间了（现在已经被取代很长时间了......）如果我犯了任何错误，我深表歉意。

首先定义一个key，selects objects by id:

<xsl:key name="by-id" match="Object" use="@id"/>

现在定义一个递归模板来打印节点的逻辑祖先：

<xsl:template match="Object" mode="ancestors">
  <xsl:copy-of select="."/>
  <xsl:apply-templates select="key('by-id', @parent-id)" mode="ancestors"/>
</xsl:template>

现在为 selected 项调用它：

<xsl:template match="/">
 <root>
  <xsl:apply-templates select="root/Object[@status='0']" mode="ancestors"/>
 </root>
</xsl:template>

未测试。

从最低成员获取所有祖先的公共 ID

Get a common id to all ancestors from the lowest member

xml

xslt

xslt-1.0