Subject: RE: [xsl] Writing a stylesheet to create another XML Document From: "Kunal H. Parikh" <kunal@xxxxxxxxxx> Date: Mon, 8 Apr 2002 09:13:05 +1000 |
G'Day Jeni ! Thanks a million for all the help that you have provided ! I have managed to get the merging of my XML files working. Now, another question along the same lines. What I am trying to do is access Author.xml/AuthorList/Author/Name by using the following: TutorialWithAuthorName.xsl ************************** <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:import href="Identity.xsl"/> <xsl:output method="xml" indent="yes" /> <xsl:variable name="authors" select="document('Author.xml')" /> <xsl:key name="authors" match="Author" use="@id" /> <xsl:template match="Author"> <xsl:variable name="id" select="@id" /> <xsl:for-each select="$authors"> <xsl:copy-of select="key('authors', $id)" /> </xsl:for-each> </xsl:template> <!-- Beginning of EXTRA STUFF --> <xsl:template match="/"> <xsl:for-each select="Tutorial/AuthorList/Author" > <xsl:value-of select="Name" /> </xsl:for-each> </xsl:template> <!-- End of EXTRA STUFF --> </xsl:stylesheet> ************************** However, in VS.NET which uses MSXML 3.0, there is no output at all if the "EXTRA STUFF" is there in the stylesheet, else all works out well. As always, TAM...IA (Thanx a million .... in advance) Kunal -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Jeni Tennison Sent: Sunday, 7 April 2002 03:26 To: Kunal H. Parikh Cc: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Writing a stylesheet to create another XML Document Hi Kunal, > According to your email, I have modified the files (Included) > > However, I don't get any output .... in IE 6. > > Do I need to use another XML Parser instead of MSXML ? Sorry, I accidentlly missed off a $ -- in the template, you need to access the value of the $authors global variable: <xsl:template match="Author"> <xsl:variable name="id" select="@id" /> <xsl:copy-of select="$authors[@id = $id]" /> </xsl:template> Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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