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OriginalCopy · Data înscrierii: 25/Mai/2008 Mesaje: 7 Locaţie: Slatina

Salut si bine am venit :)

Fiind primul meu post aici, tin sa mentionez ca vin de la concurenta, unde stiu ca nu voi primi ajutor. Deci daca ma ajutati poate ma si instalez aici :lol:

Acum asupra problemei: transform un fisier xml cu xslt in xhtml (PHP5, deci DOMDocument si XSLTProcessor). Fisierul xml e relativ simplu (prezint ca PoC):

Cod: <?xml version="1.0" encoding="UTF-8"?>
<article keywords="article keywords" description="article desc" subtitle="article subtitle">
<title>xml article title</title>
<authorgroup>
   <author>
      <surname>author surname</surname>
      <forename>author forename</forename>
      <email>test</email>
   </author>
   <author>
      <surname>author surname</surname>
      <forename>author forename</forename>
      <email>test</email>
   </author>
</authorgroup>

<section title="section title" keywords="section keyword" description="section desc">
   <author surname="section author surname" forename="section author forename" email="a@b.c"/>
   <content>
      section content
   </content>
   <subsection title="hello" keywords="foo" description="bar">
      <author surname="subsection author surname" forename="subsection author forename" email="a@b.c"/>
      <content>
         subsection content
      </content>
      <quizz>
      subsection quizz
      </quizz>
   </subsection>
   <quizz>
      section quizz
   </quizz>
</section>
<section title="section 2 title" keywords="section keyword" description="section desc">
   <author surname="section author surname" forename="section author forename" email="a@b.c"/>
   <content>
      section content
   </content>
   <subsection title="hello" keywords="foo" description="bar">
      <author surname="subsection author surname" forename="subsection author forename" email="a@b.c"/>
      <content>
         subsection content
      </content>
      <quizz>
      subsection quizz
      </quizz>
   </subsection>
   <quizz>
      section quizz
   </quizz>
</section>
</article>
insa in cel xsl am o problema:
Cod: <?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fn="http://www.w3.org/2005/02/xpath-functions">
<xsl:output method="xml"
doctype-system="http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"
doctype-public="-//W3C//DTD XHTML 1.0 Transitional//EN"
encoding="UTF-8"
indent="yes"/>
<xsl:template match="/">
   <html>
      <head>
         <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/>
         <meta name="keywords" content="{/*/@keywords}"/>
         <meta name="description" content=""/>
         <title><xsl:value-of select="./*/@title"/></title>
      </head>
      <body>
      <h1>
         <xsl:choose>
            <xsl:when test="./*/@title">
               <xsl:value-of select="./*/@title"/>
            </xsl:when>
            <xsl:otherwise test="./*/title">
               <xsl:value-of select="./*/title"/>
            </xsl:otherwise>
         </xsl:choose>
      </h1>
      <h2><xsl:value-of select="./*/@subtitle"/></h2>

      <xsl:choose>
         <xsl:when test="./*/authorgroup">
            <h3>Authors</h3>
            <xsl:apply-templates select="./*/authorgroup"/>
         </xsl:when>
         <xsl:when test="./*/author">
            By <xsl:apply-templates select="./*/author"/><br/>
         </xsl:when>
      </xsl:choose>
      <h2>In this article:</h2>
      <xsl:for-each select="article/section">
         <xsl:value-of select="@title"/>
         <br />
      </xsl:for-each>
      </body>
   </html>
</xsl:template>

<xsl:template name="authorgroup" match="authorgroup">
      <xsl:for-each select="author">
      <xsl:apply-templates select="."/><br/>
   </xsl:for-each>
</xsl:template>
<xsl:template name="author" match="author">
   <xsl:value-of select="surname"/> <xsl:value-of select="forename"/> <<xsl:value-of select="email"/>>
</xsl:template>
</xsl:stylesheet>
si anume selectorul XPath din template-ul "authorgroup", aici "XXX", insa bineinteles ca am incercat diferite XPath-uri, bineinteles mai intai cele care mi se pareau logice:
"author", "./author" si apoi si orbeste. Bineinteles ca "//author" functioneaza, dar nu e ce vreau, din motive evidente: vreau doar autorii articolului, nu si al fiecarei sectiuni a documentului.

Deci ce XPath sa folosesc? Cred ca e simplu pentru cei ce-au mai lucrat intensiv cu XSLT/XPath. O eventuala explicatie ar fi deasemenea apreciata, daca nu deduc eu explicatia.

REZOLVARE: foloseste <xsl:apply-templates>

Si tot fiind la XPath, stiti vr-o extensie ori pentru firefox in afara de XPather sau pentru eclipse (gratis, open-source, nimic proprietary, vreau sa-mi pastrez sistemul linux curat :) ) cu care sa pot inspecta in termeni de XPath documente XML? In XPather nu e posibil sa introduc "parent" (FF 2.0.0.14), si asta nu-mi permite sa intru in contextul /*/authorgroup/ si sa inspectez copii <author> relativ la /*/authorgroup.

Multumesc anticipat.

PHPRomania Bot · Mie Dec 26, 2007 7:01 pm