Subject: [xsl] sort and order xml data according to what user selects from a form From: "Sergey Demidenko" <sergeyd@xxxxxxx> Date: Fri, 18 Jul 2003 21:36:54 -0400 |
I want to be able to sort and order my XML data according to what the user selects from a form. I am quite a novice to XSL and haven't worked with forms that much at all. So, when the user selects "size" from the menu, the XML data would be sorted by "size" and when the user selects the ascending radio button, I want the XML data order by "ascending" How would I acomplish this? here is the code that I have: --------------------------------------------------------------------- <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <html><body> <form id="form" method="post" action=""> Sort by: <select id="sortby"> <option selected="selected">title</option> <option>author</option> <option>date</option> <option>citation</option> <option>pages</option> <option>size</option> </select> Order of: <input id="descending" name="sortorder" type="radio" value="descending" checked="checked" /> <label for="descending">Descending</label> <input id="ascending" name="sortorder" type="radio" value="ascending" /> <label for="ascending">Ascending</label> </form> <xsl:for-each select="catalog/publication"> <xsl:sort select="date" order="descending" /> ... here where the xml data is displayed ... </xsl:for-each> </body></html> </xsl:template> </xsl:stylesheet> --------------------------------------------------------------------- XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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