Question

我有一个HTML文件，我想只使用XSLT转换为XML ..

我想要 1.要保留的所有节点。 2.所有元素都已排序。 3.代码应该是动态的。

我有一个巨大的文件所以我想要一个简单的代码来处理所有的html节点。我用xslt解释了我的编码如果你明白了，请帮助我..

我更新的HTML文件是..

<div id="2196" class="one_tail">
<span id="2197" class="one_biblio">
    <span id="2198" class="one_section-title"><b>Title</b></span>
    <ol id="2199" class="one_biblio-sec">
        <li id="2200" class="one_bib-reference">
            <span id="2202" class="one_reference">
                <span id="2203" class="one_contribution">
                    <span id="2204" class="one_authors">
                        <span id="2205" class="one_author">

                            <!-- here the id value misplaced -->
                               <span id="2207" class="one_surname">Surname</span>
                            <span id="2206" class="one_given-name">GivenName</span>

                        </span>
                    </span>
                    <span id="2208" class="one_title">
                        <span id="2209" class="one_maintitle">technology</span>
                    </span>
                </span>
                <span id="2210" class="one_host">
                    <span id="2211" class="one_book">
                        <span id="2213" class="one_publisher">Publisher </span>
                    </span>
                </span>.
            </span>
        </li>
    </ol>
</span>
</div>

我希望XML文件为：这里，属性类值用作元素名称。

<tail id="2196">
<biblio id="2197" >
    <section-title id="2198" ><b>Title</b></section-title>
    <biblio-sec id="2199" >
        <bib-reference id="2200" >
            <reference id="2202" >
                <contribution id="2203" >
                    <authors id="2204" >
                        <author id="2205" >
                                           <!-- correrct the id -->
                              <given-name id="2206" >GivenName </given-name>
                            <surname id="2207" >Surname</surname>
                             </author>
                    </authors>
                        <title id="2208" >
                                <maintitle id="2209" >technology</maintitle>
                            </title>
                </contribution>
                </reference>
            <host id="2210" >
                <book id="2211" >
                    <publisher id="2213" >Publisher </publisher>
                    </book>
                </host>
            </bib-reference>
               </biblio-sec>
        </biblio>
</tail>

我写的XSLT并没有给出我想要的东西.. XSLT代码是：

 <xsl:template match="*|/"
    <xsl:for-each select=".">
            <xsl:for-each select="current()/*">
              <xsl:sort select="@id" order="ascending"/>
                     </xsl:for-each>
             <xsl:if test="starts-with(@class,'one_') ">
                  <xsl:variable name="nodename" select="substring-after(@class,'one_')"/>
                       <xsl:element name="{$nodename}">
                          <xsl:attribute name="id" select="@id"/>
                       <xsl:apply-templates/>
                    </xsl:element>
                  </xsl:if>
                  <xsl:if test="not(@class)">
                         <xsl:apply-templates/>
                  </xsl:if>
                  </xsl:for-each>                       
          </xsl:template>

任何人都可以帮助我..

Answer 1

以下是样式表示例：

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">

  <xsl:output indent="yes" method="xml"/>
  <xsl:strip-space elements="*"/>

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

  <xsl:template match="*[@class]">
    <xsl:element name="{substring-after(@class, 'one_')}">
      <xsl:copy-of select="@id"/>
      <xsl:apply-templates select="*">
        <xsl:sort select="@id" data-type="number"/>
      </xsl:apply-templates>
    </xsl:element>
  </xsl:template>

</xsl:stylesheet>

Answer 2

此转化：

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

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

 <xsl:template match="*[@class]">
  <xsl:element name="{substring-after(@class, 'one_')}">
   <xsl:copy-of select="@*[not(name()='class')]"/>
   <xsl:apply-templates>
     <xsl:sort select="@id" data-type="number"/>
   </xsl:apply-templates>
  </xsl:element>
 </xsl:template>
</xsl:stylesheet>

应用于提供的XML文档：

<div id="2196" class="one_tail">
    <span id="2197" class="one_biblio">
        <span id="2198" class="one_section-title">
            <b>Title</b>
        </span>
        <ol id="2199" class="one_biblio-sec">
            <li id="2200" class="one_bib-reference">
                <span id="2202" class="one_reference">
                    <span id="2203" class="one_contribution">
                        <span id="2204" class="one_authors">
                            <span id="2205" class="one_author">
                                <!-- here the id value misplaced -->
                                <span id="2207" class="one_surname">Surname</span>
                                <span id="2206" class="one_given-name">GivenName</span>
                        </span>
                    </span>
                    <span id="2208" class="one_title">
                            <span id="2209" class="one_maintitle">technology</span>
                    </span>
                </span>
                    <span id="2210" class="one_host">
                        <span id="2211" class="one_book">
                            <span id="2213" class="one_publisher">Publisher </span>
                        </span>
                    </span>.
                </span>
            </li>
        </ol>
    </span>
</div>

生成想要的正确结果：

<tail id="2196">
   <biblio id="2197">
      <section-title id="2198">
         <b>Title</b>
      </section-title>
      <biblio-sec id="2199">
         <bib-reference id="2200">
            <reference id="2202">.
                <contribution id="2203">
                  <authors id="2204">
                     <author id="2205"><!-- here the id value misplaced -->
                        <given-name id="2206">GivenName</given-name>
                        <surname id="2207">Surname</surname>
                     </author>
                  </authors>
                  <title id="2208">
                     <maintitle id="2209">technology</maintitle>
                  </title>
               </contribution>
               <host id="2210">
                  <book id="2211">
                     <publisher id="2213">Publisher </publisher>
                  </book>
               </host>
            </reference>
         </bib-reference>
      </biblio-sec>
   </biblio>
</tail>

按id动态排序节点并使用XSLT处理每个节点

2 个答案: