Subject: RE: [xsl] How to process a list of files From: "shailesh" <shailesh@xxxxxxxxxxxx> Date: Tue, 8 Nov 2005 13:36:51 +0530 |
Hi, Actually I need to know whether Multiple input to multiple output can be done? How to do this. Here multiple input means my input.xml which contains <files> <file fileloc="a/gen.xml"/> <file fileloc="a/gen1.xml"/> </files> Thanks, Shailesh -----Original Message----- From: Ragulf Pickaxe [mailto:ragulf.pickaxe@xxxxxxxxx] Sent: Tuesday, November 08, 2005 1:24 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] How to process a list of files Shailesh, You have not received any answers to this question (nor to the last one with a different title, which is identical). Rather than just reposting, you should perhaps see if there is something with your mail, that makes it not answered. When I look at the input, the desired output, and the XSL that you give, there is absolutely no correlation between the XSL and the input/output. Taking one thing, a line like this - which is a solution to a question you asked some days previously - gives the impression that you do not read the answers that you are given: <xsl:value-of select="concat(substring-before(substring-after(@fileloc,'/xml/'),'.'),'.htm ')"/> A line like that would not in any way, given "A/gen1.xml" give you gen1.xml. The XSL also contains code that is not relevant to the problem, which 1) makes it harder to read, 2) gives the impression that you have only a small subset of the problem, and a solution to your question would not answer your real problem. As you ask your question, a solution would be: <xsl:template match="/"> <xsl:apply-templates select="Root"/> </xsl:template> <xsl:template match="Root"> <xsl:for-each select="file"/> <xsl:value-of select="substring-after(@fileloc,'A/')"/><xsl:text> </xsl:text> </xsl:for-each> </xsl:template> I suspect that this is not what you really want, but this is the answer to the question that you posed. Regards, Ragulf Pickaxe :-/
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