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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: How do I pull out an element name out for use as
ok, i solved my own problem but thought that it might be nice to post how in case someone else needs it. Recap: -I have the following XML structure: -<contacts> -<contact> - <name>xxxx</name> - <title>Program Director</title> - <phone>xxx.xxx.xxxx</phone> - <email> - <link mailto:href="mailto:xxxxxxx@xxxxxxxxxxx>xxxxxxx@xxxxxxxxxxx</link> - </email> -</contact> -<contact>.....</contact> -</contacts> - -The children elements of contact vary slightly between different contact lists. Because of this, -the tables in which they appear should often have different headers (which would just be the -name of that child element... ie name, title, phone, email ... in this case). I tried using the -recommendation in the FAQ under the first table section but i couldn't get it to work for children -elements instead of the attributes. Any help is appreciated. -eric Solution: <xsl:for-each select="contact[1]/*"> <td> <xsl:value-of select="name(.)"/> </td> </xsl:for-each> Eric XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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