xsl:sort Question

kwilliams · July 14th, 2005, 04:39 PM

I'm using the transformNode method to convert an XML doc and an XSL stylesheet to XHTML using an ASP.NET/VB.NET script, and it works great. But when I try to add an <xsl:sort... property to the xsl stylesheet, I get the following error message:
Server Error in '/' Application.
Keyword xsl:sort may not be used here.
Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code.

Exception Details: System.Runtime.InteropServices.COMException: Keyword xsl:sort may not be used here.

Source Error:
Line 18:
Line 19: 'Transform file
Line 20: Response.Write(xml.transformNode(xsl))
Line 21: %>
Source File: c:\inetpub\wwwroot\DIRECTORY\FILENAME.aspx Line: 20

I'm a complete newbie to XML and XSL/XSLT, so I'm not sure why I'm receiving this message. So my question is:
Is it not possible to use the <xsl:sort... method when using the transformNode method?

Thanks for any & all help.

P.S. Here's the code that I'm using in case it needs to be referenced:

XML doc:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<formsdocs>
    <form>
        <title>FORM TITLE</title>
        <url>http://DOMAIN/DIRECTORY/FORMNAME.pdf</url>
        <format>pdf</format>
        <target>_blank</target>
        <type>internal</type>
        <size>3053 KB</size>
        <dept>DEPT</dept>
        <div />
    </form>
</formsdocs>

XSL stylesheet:

Code:

<?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 xmlns="http://www.w3.org/1999/xhtml">
            <head>
                <title>PAGETITLE</title>
            </head>
            <body>
                <h1>PAGE HEADER</h1>
                <table class="navyblueborder">
                    <tr bgcolor="#333366" class="whiteheader">
                        <th align="left">Title</th>
                        <th align="left">Format</th>
                        <th align="left">Type</th>
                        <th align="left">Size</th>
                        <th align="left">Dept</th>
                    </tr>
                    <xsl:for-each select="formsdocs/form">
                        <tr>
                            <td><a href="{url}" target="{target}"><xsl:value-of select="title" /></a></td>
                            <td><xsl:value-of select="format" /></td>
                            <td><xsl:value-of select="type" /></td>
                            <td><xsl:value-of select="size" /></td>
                            <td><xsl:value-of select="dept" />
                            <xsl:choose>
                                <xsl:when test="div = ''">
                                </xsl:when>
                                <xsl:otherwise>
                                    (<xsl:value-of select="div" /> Division)
                                </xsl:otherwise>
                            </xsl:choose>
                            </td>
                        </tr>
                    </xsl:for-each>
                </table>
            </body>
        </html>
    </xsl:template>
</xsl:stylesheet>

ASP.NET/VB.NET script:
[code]<%@ Page Language="VB" ContentType="text/html" ResponseEncoding="iso-8859-1" Debug="true" %>
<%@ import Namespace="System.Data" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<%
'Load XML
Dim xml = Server.CreateObject("Microsoft.XMLDOM")
xml.async = false
xml.load("http://DOMAIN/DIRECTORY/XMLFILE.xml")

'Load XSL
Dim xsl = Server.CreateObject("Microsoft.XMLDOM")
xsl.async = false
xsl.load("http://DOMAIN/DIRECTORY/XSLFILE.xsl")

'Transform file
Response.Write(xml.transformNode(xsl))
%>

Code:

KWilliams

mhkay · July 14th, 2005, 04:48 PM

You said:

But when I try to add an <xsl:sort... property to the xsl stylesheet, I get the following error message:
Server Error in '/' Application.
Keyword xsl:sort may not be used here.

This message means you have added xsl:sort in the wrong place.

Since you haven't told us where you added it, it's hard to tell you any more than the error message already says.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

kwilliams · July 14th, 2005, 04:57 PM

...sorry about that. This is where I added the sort method:

<xsl:choose>
<xsl:when test="div != ''">
(<xsl:sort select="div" /> Division)
</xsl:when>
</xsl:choose>

...which I think should work. It's a pretty simple setup, and it's correct from everything that I've read about it. But if you see what I'm doing wrong, please let me know.

Also, I tried to "<>" or "#60;#62;" instead of "!=", but I get this error message:
Expression expected. div <-->><-- ''

Thanks.

KWilliams

mhkay · July 14th, 2005, 05:45 PM

xsl:sort has to be a child of xsl:for-each or xsl:apply-templates. These are the two instructions that iterate over a node-set, and xsl:sort says in what order the nodes in the node-set are processed. Not only can't you use it like this:

<xsl:choose>
<xsl:when test="div != ''">
(<xsl:sort select="div" /> Division)
</xsl:when>
</xsl:choose>

but I have no idea what you imagine it might mean.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

kwilliams · July 15th, 2005, 08:11 AM

Ok, I would ask how you might go about doing that, but you don't seem like you're in the mood to answer my question anyway. So don't worry about it.

KWilliams

mhkay · July 16th, 2005, 10:01 PM

If you can tell us what you want to achieve then we can try and help you achieve it. So far, I don't know what you are trying to do. Your code is incorrect, but I can't correct it for you because I don't know what you thought it might do. I think the fact that I'm replying at all is evidence enough that I'm trying to help.

I've just noticed this remark of yours:

>Also, I tried to "<>" or "#60;#62;" instead of "!="

This is really a bit worrying. It suggests you're trying to work by trial and error. This is not a good way forward. There are specs, books, and tutorials to teach the language that suit every possible taste. They will all tell you that the XPath "not-equals" operator is "!=".

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

kwilliams · July 18th, 2005, 08:39 AM

Quote:

quote:If you can tell us what you want to achieve then we can try and help you achieve it. So far, I don't know what you are trying to do. Your code is incorrect, but I can't correct it for you because I don't know what you thought it might do. I think the fact that I'm replying at all is evidence enough that I'm trying to help.

I appreciate you responding to my post, but I thought that I was pretty clear in posing my question "Is it not possible to use the <xsl:sort... method when using the transformNode method?". I thought that this question would only have a yes or no answer, but I understand now that the answer to my question is "Yes it is possible, if it's done correctly".

Quote:

quote:I've just noticed this remark of yours:

>Also, I tried to "<>" or "#60;#62;" instead of "!="

This is really a bit worrying. It suggests you're trying to work by trial and error. This is not a good way forward. There are specs, books, and tutorials to teach the language that suit every possible taste. They will all tell you that the XPath "not-equals" operator is "!=".

As I stated in my original post, "I'm a complete newbie to XML and XSL/XSLT". I am learning as I'm going along, since I've not received any training on these languages. I have found several tutorials on XML and XSLT, but I've found that almost all of these so-called tutorials are basically copied & pasted from an original tutorial. Which is not a great help if you have a question that's outside of that tutorial's realm. Concerning the "trial & error" method of development, I would think that most developers without resources such as training will try their best to create correct code the first time. But if I ever meet a developer (except for an expert developer) that reviews some code online, and is able to reproduce that code without any issues arrising, then I will be in shock. I agree that it's best to do correct code, but if errors weren't meant to come up for developers, I don't believe that they would have created error messages for those same developers to troubleshoot/debug their code. But that's my own opinion. Please Note: My pages are located on a test-server internally, so they're of course not live for users to view.

Before your similar post that contained an answer, I had received this answer to my original question through another forum:
"<xsl:sort> can only be the child of <xsl:for-each> or <xsl:apply-templates>, for example:

Code:

<xsl:for-each select="formsdocs/form"> <xsl:sort select="div" /> ... etc. ... </xsl:for-each> <xsl:apply-templates select="formsdocs/form"> <xsl:sort select="div" /> </xsl:apply-templates>

Make sure that the sort instruction is the first instruction after the opening <xsl:for-each ... >.

If <div> is empty (in the XML) then it won't effect the sort anyway so you don't really need to instruct the XSLT to sort by it only if it's not empty."[quote]

So I've changed my code to reflect this correction, and it works fine. I'm now working on making that same xsl:sort method dynamic with the use of the xsl:param method, and it's going well so far. But thanks again for your reply, and I hope that you have a great week.

KWilliams

KWilliams