Subject: Re: [xsl] newbie needs help From: "Brian Benson" <bbenson@xxxxxxxxxxxx> Date: Thu, 11 Sep 2003 13:41:14 -0500 |
Thanks David. The code you wrote for me, did exactly what I asked for. Unfortunately, I didn't ask a specific enough question. Not all documents in the source xml file contain links and I want the output xml to only contain documents that have links. How can the template below be changed to only output documents that contain links. <xsl:template match="document"> <document> <xsl:copy-of select="form|.//link"/> </document> </xsl:template> David ======================================= This is probably easy for someone who knows what they are doing, but I haven't been able to figure this out. I'm trying to transform XML to XML. The end result I want is something like: <document> <form>forminfo</form> <link>linkinfo</link> </document> The problem that I'm having is this: in the source xml file, the <form> information is always at the same level in the source document, but the <link> information can be any number of levels deep. For example it could look like this: <document> <form>forminfo</form> <link>linkinfo</link> </document> or it could look like this: <document> <form>forminfo</form> <table> <link>linkinfo</link> </table> </document> or like this: <document> <form>forminfo</form> <table> <table> <link>linkinfo</link> </table> </table> </document> I need to find a way to ignore all of the levels of table information and just get the form and the link information in my output file. Any help is appreciated. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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