Simple XSLT Setup w/Strange Result

kwilliams · August 5th, 2005, 10:08 AM

I'm not sure why this is happening, but here it goes.
I have a source XML doc, like this:
[code]<?xml version="1.0" encoding="UTF-8"?>
<mostrequested>
    <request>
        <mr_title>...</mr_title>
        <mr_url>...</mr_url>
        <mr_target>...</mr_target>
        <mr_display>yes</mr_display>
    </request>
</mostrequested>[code]

...and an XSLT stylesheet like this:

Code:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="UTF-8" indent="yes"/>

<xsl:param name="par_mr" select="document('mostrequested.xml')" />
<xsl:template match="/">[list]
    <xsl:for-each select="$par_mr">
        <xsl:if test="mostrequested/request/mr_display = 'yes'">
            <li><xsl:element name="a"><xsl:attribute name="href"><xsl:value-of select="mostrequested/request/mr_url" /></xsl:attribute><xsl:attribute name="target">_blank</xsl:attribute><xsl:value-of select="mostrequested/request/mr_title" /></xsl:element></li>
        </xsl:if>
    </xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>

...which should result a list of elements where the node's <mr_display> value = "yes".

But instead, it's resulting in only the first element displaying. I've also tried this setup:

Code:

<xsl:for-each select="$par_mr[mr_display = 'yes'">

...to replace the <xsl:if> statement, but it worked exactly the same.

And finally, if I change it to be:

Code:

<xsl:for-each select="$par_mr">
    Most Requested: <xsl:value-of select="*" />
</xsl:for-each>

..all of the node data DOES display properly. So I know that it's pulling in the data correctly. It's just not displaying that node for all of the elements for some reason.

I've reviewed several online sources concerning what should be a simple setup, but I'm stumped as to why this is not working. If anyone can tell me what I'm doing wrong, that would be great. Thanks.

KWilliams

joefawcett · August 5th, 2005, 10:41 AM

The document function returns the root node so xsl:for-each will only find one node, the root.
If you want to process each request then use:

Code:

<xsl:for-each select="$par_mr/mostrequested/request">

Personally I'd have thought if you want all the requests where the mr_display equals "yes" then I'd do:

Code:

<xsl:apply-templaes select="$par_mr/mostrequested/request[mr_display = 'yes']"/>

and then catch them with:

Code:

<xsl:template match="request">



</xsl:template>

--

Joe (Microsoft MVP - XML)

kwilliams · August 5th, 2005, 11:18 AM

Hello Joe,

You wrote:

Quote:

quote:The document function returns the root node so xsl:for-each will only find one node, the root.
If you want to process each request then use:
<xsl:for-each select="$par_mr/mostrequested/request">
Personally I'd have thought if you want all the requests where the mr_display equals "yes" then I'd do:
<xsl:apply-templaes select="$par_mr/mostrequested/request[mr_display = 'yes']"/>
and then catch them with:
<xsl:template match="request">

</xsl:template>

Thanks for the info, as I was able to create this working solution:

Code:

[list]
<xsl:for-each select="$par_mr/mostrequested/request">
    <xsl:if test="mr_display = 'yes'">
    <li><xsl:element name="a"><xsl:attribute name="href"><xsl:value-of select="mr_url" /></xsl:attribute><xsl:attribute name="target">_blank</xsl:attribute><xsl:value-of select="mr_title" /></xsl:element></li>
    </xsl:if>
</xsl:for-each>
</ul>

Concerning the other option that you mentioned using <xsl:apply-templates, what would be the benefit of using that method vs. what I'm using? To use that method with other code within my stylesheet, I'd have to do this:
<xsl:template match="/">

</xsl:template>
<xsl:template match="request">
...
</xsl:template>
<xsl:template match="/">

</xsl:template>

Wouldn't this setup be more complicated than what I'm using. Thanks for your help and advice on this subject.

KWilliams

mhkay · August 5th, 2005, 11:20 AM

Firstly, why do you do this:

<xsl:element name="a"><xsl:attribute name="href"><xsl:value-of select="mostrequested/request/mr_url" /></xsl:attribute><xsl:attribute name="target">_blank</xsl:attribute><xsl:value-of select="mostrequested/request/mr_title" /></xsl:element>

when you could do this:

<a href="{mostrequested/request/mr_url}" target="_blank">
<xsl:value-of select="mostrequested/request/mr_title" />
</a>

Secondly, document() selects one node, so your for-each loop is going to be executed exactly once. I'm not sure what you wanted to happen, so I can't correct it for you.

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

kwilliams · August 5th, 2005, 11:36 AM

Quote:

quote:Firstly, why do you do this:
<xsl:element name="a"><xsl:attribute name="href"><xsl:value-of select="mostrequested/request/mr_url" /></xsl:attribute><xsl:attribute name="target">_blank</xsl:attribute><xsl:value-of select="mostrequested/request/mr_title" /></xsl:element>

when you could do this:

<a href="{mostrequested/request/mr_url}" target="_blank">
<xsl:value-of select="mostrequested/request/mr_title" />
</a>

Good point. I'll change that.

Quote:

quote:Secondly, document() selects one node, so your for-each loop is going to be executed exactly once. I'm not sure what you wanted to happen, so I can't correct it for you.

Actually, with joefawcett's help, I was able to create this set of code that pull all results using the document() method:

Code:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="UTF-8" indent="yes"/>
<xsl:param name="par_mr" select="document('mostrequested.xml')" />
<xsl:template match="/">[list]
<xsl:for-each select="$par_mr/mostrequested/request">
    <xsl:if test="mr_display = 'yes'">
        <li><xsl:element name="a"><xsl:attribute name="href"><xsl:value-of select="mr_url" /></xsl:attribute><xsl:attribute name="target">_blank</xsl:attribute><xsl:value-of select="mr_title" /></xsl:element></li>
    </xsl:if>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>

And the resulting XHTML is:[list]
<li><a href="URL" target="_blank">LINKTITLE1</a></li>
<li><a href="URL2" target="_blank">LINKTITLE2</a></li>
<li><a href="URL3" target="_blank">LINKTITLE3</a></li>
<li><a href="URL4" target="_blank">LINKTITLE4</a></li>
<li><a href="URL5" target="_blank">LINKTITLE5</a></li>
</ul>

From this XML doc:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<mostrequested>
    <request>
        <mr_title>LINKTITLE1</mr_title>
        <mr_url>URL1</mr_url>
        <mr_display>yes</mr_display>
    </request>
    <request>
        <mr_title>LINKTITLE2</mr_title>
        <mr_url>URL2</mr_url>
        <mr_display>yes</mr_display>
    </request>
    <request>
        <mr_title>LINKTITLE3</mr_title>
        <mr_url>URL3</mr_url>
        <mr_display>yes</mr_display>
    </request>
    <request>
        <mr_title>LINKTITLE4</mr_title>
        <mr_url>URL4</mr_url>
        <mr_display>yes</mr_display>
    </request>
    <request>
        <mr_title>LINKTITLE5</mr_title>
        <mr_url>URL5</mr_url>
        <mr_display>yes</mr_display>
    </request>
</mostrequested>

KWilliams

kwilliams · August 5th, 2005, 11:47 AM

Hi Michael,

I just tried your simpler link method of:

Code:

<a href="{mostrequested/request/mr_url}" target="_blank">
  <xsl:value-of select="mostrequested/request/mr_title" />
</a>

..but no results were displayed. And when I use my method on that same page, they are displayed. I'm not sure why, but thanks again for your input.

KWilliams