XSLT條件排序兩次的N值

Question

XSLT文件

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
xmlns:fn="http://www.w3.org/2005/xpath-functions"> 

    <xsl:template match="/"> 
     <xsl:call-template name="content"> 
     </xsl:call-template> 
    </xsl:template> 

    <xsl:template name="content"> 
     <movies> 
     <xsl:for-each select="/movies/movie"> 

      <xsl:for-each select="movie[starts-with(name,'Ka')]"> 
       <xsl:apply-templates select="name" /> 
       <xsl:apply-templates select="vdate" />   
      </xsl:for-each> 

      <xsl:for-each select="/movies/movie"> 
       <xsl:sort select="date" data-type="number" order="descending"/> 
       <xsl:if test="position() &lt;= 50"> 
        <xsl:value-of select="vdate"/> 
       </xsl:if> 
      </xsl:for-each> 

      <xsl:sort select="date" data-type="number" order="'.($this->_boolAscending?'ascending':'descending').'"/> 
      <movie>    
       <vdate><xsl:value-of select="vdate"/></vdate> 
       <no><xsl:value-of select="no"/></no> 
       <name><xsl:value-of select="name"/></name> 
       <link1><xsl:value-of select="link1"/></link1> 
       <link2><xsl:value-of select="link2"/></link2> 
       <link3><xsl:value-of select="link3"/></link3> 
       <year><xsl:value-of select="year"/></year> 
      </movie> 


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

</xsl:stylesheet> 

XML文件

<?xml version="1.0"?> 
    <movies> 
    <movie><no>2013001</no> 
    <date>20131213</date> 
    <vdate>13-12-2013</vdate> 
    <name>Kalyana Kanavugal</name> 
    <mlink1>http://www.moviebunk.com/?t7f4vm15zza5xv8</mlink1> 
    <mlink2>http://www.moviebunk.com/?mvz1tjks2z571fj</mlink2> 
    <mlink3>http://www.moviebunk.com/?k64p6yjfl5b24z4</mlink3> 
    <link1><![CDATA[<a href="index2.php?q=K176AZ0DN8LW930D0ZYJ8T74P">320KBPS</a>]]></link1> 
    <link2><![CDATA[<a href="index3.php?q=5245C918Z99SOHBRU6298G13M">256KBPS</a>]]></link2> 
    <link3><![CDATA[<a href="index4.php?q=3FZ0MJ8S65749XBSI0670ARH2">128KBPS</a>]]></link3> 
    <year>2011</year> 
    </movie> 

    <movie><no>2013002</no> 
    <date>20131110</date> 
    <vdate>10-11-2013</vdate> 
    <name>Kandaudan</name> 
    <mlink1>http://www.moviebunk.com/?d2411v82s08x1b2</mlink1> 
    <mlink2>http://www.moviebunk.com/?gd0ume8bdo6lazg</mlink2> 
    <mlink3>http://www.moviebunk.com/?gyry7qb81kvnind</mlink3> 
    <link1><![CDATA[<a href="index2.php?q=6A29SM5TI180BJD50E97VTMQM">320KBPS</a>]]></link1> 
    <link2><![CDATA[<a href="index3.php?q=1MGEFRH6H675H6599G0APR0KZ">256KBPS</a>]]></link2> 
    <link3><![CDATA[<a href="index4.php?q=83GDW3645XHF296HY3XUEO96A">128KBPS</a>]]></link3> 
    <year>2011</year> 
    </movie> 

    <movie><no>2013003</no> 
    <date>20121019</date> 
    <vdate>19-10-2012</vdate> 
    <name>Koyambedu Perundhu Nilayam</name> 
    <mlink1>http://www.moviebunk.com/?bdb6a76twx4kfnb</mlink1> 
    <mlink2>http://www.moviebunk.com/?n56o5xuqc7euqq6</mlink2> 
    <mlink3>http://www.moviebunk.com/?wksvaa965pp6a1e</mlink3> 
    <link1><![CDATA[<a href="index2.php?q=5R9B6916IN5H5R6G4O37G3XB3">320KBPS</a>]]></link1> 
    <link2><![CDATA[<a href="index3.php?q=TM7ZFY8619W371P972P92R51F">256KBPS</a>]]></link2> 
    <link3><![CDATA[<a href="index4.php?q=HQ6YHH751Y7M0H8ZK60V773U3">128KBPS</a>]]></link3> 
    <year>2011</year> 
    </movie> 
    </movies> 

我是新來的XSLT編程。請給我一種排序XML文件的方法。 我有排序問題，如下所示：名稱以'K'字母開頭，然後按降序排序。僅顯示前50個值。我試着不用'Ka'開始排序並限制結果 - 這工作正常。

這樣

 

Movie Information 
================= 
Kalyana Kanavugal 
2011 
Download-320KBPS 
Download-256KBPS 
Download-128KBPS 


Kandaudan 
2011 
Download-320KBPS 
Download-256KBPS 
Download-128KBPS 


Koyambedu Perundhu Nilayam 
2011 
Download-320KBPS 
Download-256KBPS 
Download-128KBPS 

Answer 1

名HTML輸出與 'K' 字母開頭，然後排序[按日期？]以降序 的順序。僅顯示前50個值。

如果我理解正確的話，那麼這應該是足夠了：

<?xml version="1.0" encoding="UTF-8"?> 
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 

<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes" 
cdata-section-elements="link1 link2 link3"/> 

<xsl:template match="/movies"> 
<xsl:copy> 
    <xsl:apply-templates select="movie[starts-with(name, 'K')]"> 
     <xsl:sort select="date" data-type="number" order="descending"/> 
    </xsl:apply-templates> 
</xsl:copy> 
</xsl:template> 

<xsl:template match="movie"> 
<xsl:if test="position() &lt;= 50"> 
<xsl:copy> 
    <xsl:copy-of select="vdate"/> 
    <xsl:copy-of select="no"/> 
    <xsl:copy-of select="name"/> 
    <xsl:copy-of select="link1"/> 
    <xsl:copy-of select="link2"/> 
    <xsl:copy-of select="link3"/> 
    <xsl:copy-of select="year"/> 
</xsl:copy> 
</xsl:if> 
</xsl:template> 

</xsl:stylesheet> 

---

增加：
對於HTML輸出，你可以試試：

<?xml version="1.0" encoding="UTF-8"?> 
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 

<xsl:output method="html" version="1.0" encoding="utf-8" indent="yes"/> 

<xsl:template match="/movies"> 
<html> 
<body> 
<h1>Movie Information</h1> 
    <xsl:apply-templates select="movie[starts-with(name, 'K')]"> 
     <xsl:sort select="date" data-type="number" order="descending"/> 
    </xsl:apply-templates> 
</body> 
</html> 
</xsl:template> 

<xsl:template match="movie"> 
<xsl:if test="position() &lt;= 50"> 
    <h4> 
     <xsl:value-of select="name"/><br/> 
     <xsl:value-of select="year"/> 
    </h4> 
    <p> 
     Download - <xsl:value-of select="link1" disable-output-escaping="yes"/><br/> 
     Download - <xsl:value-of select="link2" disable-output-escaping="yes"/><br/> 
     Download - <xsl:value-of select="link3" disable-output-escaping="yes"/> 
    </p> 
</xsl:if> 
</xsl:template> 

</xsl:stylesheet>