с использованием xslt и XML для создания нужного HTML

Question

Я использую XSLT и XML, чтобы он генерировал мой желаемый HTML.

У меня есть два XMLS, ниже приведены подробности.

1) Destinations.XML

<?xml version="1.0"?> 
<list type="Destinations"> 
<resources location="include/xml/locations.xml"> 
<publication>481</publication> 
</resources> 
<destination id="594051" title="Sydney" url="/asiapacific/australia/sydney.aspx" > 
<country id="192395" /> 
</destination> 
<destination id="594088" title="Brisbane" url="/asiapacific/australia/brisbane.aspx" > 
<country id="192395" /> 
</destination> 
<destination id="594579" title="Dubai" url="/middleeast/uae/dubai.aspx" > 
<country id="192849" /> 
</destination> 
<destination id="594580" title="Abu Dhabi" url="/middleeast/uae/abudhabi.aspx" > 
<country id="192849" /> 
</destination> 
</list> 

2) Locations.xml

<?xml version="1.0"?> 
<list type="Locations"> 
<region id="192393" code="ASIA" name="Asia &amp; the Pacific" shortname="Asia &amp; the Pacific"> 
<country id="192395" code="AU" name="Australia" shortname="Australia"> 
<city id="192397" code="BNE" name="Brisbane" shortname="Brisbane"> 
<airport id="192399" code="BNE" name="Brisbane International Airport" shortname="Brisbane"></airport> 
</city> 
<city id="192409" code="SYD" name="Sydney" shortname="Sydney"> 
<airport id="192411" code="SYD" name="Kingsford Smith Airport" shortname="Sydney"></airport> 
</city> 
</country> 
</region> 
<region id="192847" code="MEAF" name="The Middle East &amp; Africa" shortname="The Middle East &amp; Africa"> 
<country id="192849" code="AE" name="United Arab Emirates" shortname="United Arab Emirates"> 
<city id="192851" code="AUH" name="Abu Dhabi" shortname="Abu Dhabi"> 
<airport id="192853" code="AUH" name="Abu Dhabi" shortname="Abu Dhabi"></airport> 
</city> 
<city id="192855" code="DXB" name="Dubai" shortname="Dubai"> 
<airport id="192857" code="DXB" name="Dubai International Airport" shortname="Dubai"></airport> 
</city> 
</country> 
</region> 
</list> 

Если вы видите destinations.xml мы получили название, например, "СИДНЕЙ", а также URL = "/ asiapacific/australia/sydney.aspx ", а также получил идентификатор страны = 192395, и когда вы видите файл Locations.xml, есть также идентификатор страны = 192395 и name =" Австралия "выше, чем его название региона =" Азия & Pacific », теперь я хочу использовать эти xmls и написать XSLT, чтобы весь список пунктов назначения из destination.xml появится название страны и название региона с URL-адресами, для страны этот URL-адрес станет (/asiapacific/australia/index.aspx), а для региона он станет (/asiapacific/index.aspx), ниже будет HTML генерируется

<table width="100%" cellspacing="0" cellpadding="0" border="0" class="displayTable"> 
    <tbody> 
     <tr> 
      <th scope="col" class="first sortSelected"> 
       <div class="thPadding"> 
        <a class="iconDownSortArrow" href="#">Destination</a></div> 
      </th> 
      <th scope="col" class="sortHover"> 
       <div class="thPadding"> 
        <a class="iconSortArrowOff" href="#">Country</a></div> 
      </th> 
      <th scope="col" class="sortHover"> 
       <div class="thPadding"> 
        <a class="iconSortArrowOff" href="#">Region</a></div> 
      </th> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/sydney.aspx">Sydney</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/index.aspx">Australia</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/index.aspx">Asia & Pacific</a></td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/brisbane.aspx">Brisbane</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/index.aspx">Australia</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/index.aspx">Asia & Pacific</a></td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/middleeast/uae/dubai.aspx">Dubai</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/uae/index.aspx">UAE</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/index.aspx">Middle East</a></td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/middleeast/uae/abudhabi.aspx">Abu Dhabi</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/uae/index.aspx">UAE</a></td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/index.aspx">Middle East</a></td> 
     </tr> 
    </tbody> 
</table> 

Пожалуйста, рекомендуем использовать XSLT, я хочу использовать нумерацию страниц, также, когда это более 20 направлений, ниже HTML-код для вёрстка.

<div class="continueBar"> 
    <div class="continueBarLeft"> 
     <strong>Displaying destinations 1-20 of 100</strong></div> 
    <div class="continueBarRight"> 
     <ul class="paginationLinks"> 
      <!--<li class="noBorder"><a class="iconButtonBackBar" href="#">&nbsp;</a></li>--> 
      <li class="noBorder"><span class="iconButtonBackBarOff">&nbsp;</span></li> 
      <li><strong class="thisPage">1</strong></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">2</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">3</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">4</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">5</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">6</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">7</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">8</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">9</a></li> 
      <li class="separatorLine">|</li> 
      <li><a href="#">10</a></li> 
      <!--<li class="last"><span class="iconButtonForwardBarOff">&nbsp;</span></li>--> 
      <li class="last"><a class="iconButtonForwardBar" href="#">&nbsp;</a></li> 
     </ul> 
    </div> 
    <div class="clearBoth"> 
    </div> 
</div> 

Answer 1

Я не знаю, какой "пагинацией" Вы хотите, но этот стиль:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:key name="kCountryById" match="country" use="@id"/> 
    <xsl:variable name="vLocations" select="document(/*/resources/@location)"/> 
    <xsl:template match="list"> 
     <table width="100%" cellspacing="0" cellpadding="0" border="0" 
                  class="displayTable"> 
      <tbody> 
       <tr> 
        <th scope="col" class="first sortSelected"> 
         <div class="thPadding"> 
          <a class="iconDownSortArrow" href="#">Destination</a> 
         </div> 
        </th> 
        <th scope="col" class="sortHover"> 
         <div class="thPadding"> 
          <a class="iconSortArrowOff" href="#">Country</a> 
         </div> 
        </th> 
        <th scope="col" class="sortHover"> 
         <div class="thPadding"> 
          <a class="iconSortArrowOff" href="#">Region</a> 
         </div> 
        </th> 
       </tr> 
       <xsl:apply-templates select="destination"/> 
      </tbody> 
     </table> 
    </xsl:template> 
    <xsl:template match="text()"/> 
    <xsl:template match="destination"> 
     <xsl:variable name="vReverseURL"> 
      <xsl:call-template name="reverse"> 
       <xsl:with-param name="string" select="@url"/> 
      </xsl:call-template> 
     </xsl:variable> 
     <xsl:variable name="vCountryURL"> 
      <xsl:call-template name="reverse"> 
       <xsl:with-param name="string" 
           select="substring-after($vReverseURL,'/')"/> 
      </xsl:call-template> 
      <xsl:text>/index.asp</xsl:text> 
     </xsl:variable> 
     <xsl:variable name="vRegionURL"> 
      <xsl:call-template name="reverse"> 
       <xsl:with-param name="string" 
       select="substring-after(substring-after($vReverseURL,'/'),'/')"/> 
      </xsl:call-template> 
      <xsl:text>/index.asp</xsl:text> 
     </xsl:variable> 
     <xsl:variable name="me" select="."/> 
     <xsl:for-each select="$vLocations"> 
      <tr> 
       <td class="detail first"> 
        <a class="arrowSmallFront" href="{$me/@url}"> 
         <xsl:value-of select="$me/@title"/> 
        </a> 
       </td> 
       <td class="detail noLeftBorder"> 
        <a class="arrowSmallFront" href="{$vCountryURL}"> 
         <xsl:value-of 
          select="key('kCountryById',$me/country/@id)/@name"/> 
        </a> 
       </td> 
       <td class="detail noLeftBorder"> 
        <a class="arrowSmallFront" href="{$vRegionURL}"> 
         <xsl:value-of 
         select="key('kCountryById',$me/country/@id)/../@name"/> 
        </a> 
       </td> 
      </tr> 
     </xsl:for-each> 
    </xsl:template> 
    <xsl:template name="reverse"> 
     <xsl:param name="string" select="''"/> 
     <xsl:if test="$string != ''"> 
      <xsl:call-template name="reverse"> 
       <xsl:with-param name="string" select="substring($string,2)"/> 
      </xsl:call-template> 
      <xsl:value-of select="substring($string,1,1)"/> 
     </xsl:if> 
    </xsl:template> 
</xsl:stylesheet> 

Выход:

<table width="100%" cellspacing="0" cellpadding="0" border="0" class="displayTable"> 
    <tbody> 
     <tr> 
      <th scope="col" class="first sortSelected"> 
       <div class="thPadding"> 
        <a class="iconDownSortArrow" href="#">Destination</a> 
       </div> 
      </th> 
      <th scope="col" class="sortHover"> 
       <div class="thPadding"> 
        <a class="iconSortArrowOff" href="#">Country</a> 
       </div> 
      </th> 
      <th scope="col" class="sortHover"> 
       <div class="thPadding"> 
        <a class="iconSortArrowOff" href="#">Region</a> 
       </div> 
      </th> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/sydney.aspx">Sydney</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/index.asp">Australia</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/index.asp">Asia &amp; the Pacific</a> 
      </td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/brisbane.aspx">Brisbane</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/australia/index.asp">Australia</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/asiapacific/index.asp">Asia &amp; the Pacific</a> 
      </td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/middleeast/uae/dubai.aspx">Dubai</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/uae/index.asp">United Arab Emirates</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/index.asp">The Middle East &amp; Africa</a> 
      </td> 
     </tr> 
     <tr> 
      <td class="detail first"> 
       <a class="arrowSmallFront" href="/middleeast/uae/abudhabi.aspx">Abu Dhabi</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/uae/index.asp">United Arab Emirates</a> 
      </td> 
      <td class="detail noLeftBorder"> 
       <a class="arrowSmallFront" href="/middleeast/index.asp">The Middle East &amp; Africa</a> 
      </td> 
     </tr> 
    </tbody> 
</table> 

Примечание: Reverse Строка шаблона, чтобы относительный URL для изменения. Ключи с несколькими источниками документов.