Filtering on locations in XPath

[Ben Jessel]

All,

I have an rss feed:

<rss>
<channel>

<item>
<title>
some title
</title>
<link>
some link
</link>
<description>
some description
</description>
</item>


<item>
<title>
my
</title>
<link>
some link
</link>
<description>
test
</description>
</item>

<webMaster>benj essel@hotmail.c om</webMaster>
</channel>
</rss>

How could I write an xPath statement that would return me all my data
apart from <item> clauses that don't meed a specific critera?

I'd like the output to be:




I've tried

"//item[( contains(title, 'my') or contains(descri ption,'test') )]"
hoping that it would return:




<rss>
<channel>
<item>
<title>
my
</title>
<link>
some link
</link>
<description>
test
</description>
</item>
<webMaster>benj essel@hotmail.c om</webMaster>
</channel>
</rss>

However it only returns the matching <item>.

I've tried reading trhought the specification, but I really need a
real-world example to help me!

Thanks,

Ben

[Richard Tobin]

Re: Filtering on locations in XPath

In article <2535a6fa.04072 70400.1689ef82@ posting.google. com>,
Ben Jessel <ben.jessel@mor pheme.co.uk> wrote:
[color=blue]
>How could I write an xPath statement that would return me all my data
>apart from <item> clauses that don't meed a specific critera?[/color]

You can't. XPath can only return structures that are present in the
input, and you want a modified version of the input.

You could use a stylesheet to do it.

-- Richard

[Ben Jessel]

Re: Filtering on locations in XPath

> You can't. XPath can only return structures that are present in the[color=blue]
> input, and you want a modified version of the input.[/color]

Without wanting someone else to do my work for me ( :-) ) could
someone come up with a sample stylesheet that gets me some of the way
to where I want to be with this?
Thanks,

Ben

[Graham Shaw]

Re: Filtering on locations in XPath

Hi Ben,

not perfect but the output matches what you want and you can probably use it
as a starter

Graham

<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:styleshe et xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:template match="/">
<xsl:element name="rss">
<xsl:element name="channel">
<xsl:for-each select="/rss/channel/item">
<xsl:if test="current()[contains(title, 'my') or
contains(descri ption,'test')]">
<xsl:element name="item">
<xsl:element name="title">
<xsl:value-of select="title"/>
</xsl:element>
<xsl:element name="link">
<xsl:value-of select="link"/>
</xsl:element>
<xsl:element name="descripti on">
<xsl:value-of select="descrip tion"/>
</xsl:element>
</xsl:element>
</xsl:if>
</xsl:for-each>
<xsl:for-each select="/rss/channel/webMaster">
<xsl:element name="webMaster ">
<xsl:value-of select="."/>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
"Ben Jessel" <ben.jessel@mor pheme.co.uk> wrote in message
news:2535a6fa.0 407280333.33633 edf@posting.goo gle.com...[color=blue][color=green]
> > You can't. XPath can only return structures that are present in the
> > input, and you want a modified version of the input.[/color]
>
> Without wanting someone else to do my work for me ( :-) ) could
> someone come up with a sample stylesheet that gets me some of the way
> to where I want to be with this?
> Thanks,
>
> Ben[/color]

[Ben Jessel]

Re: Filtering on locations in XPath

"Graham Shaw" <Graham@somewhe re.com> wrote in message news:<ckSNc.530 $YG3.476@newsfe 1-gui.ntli.net>.. .[color=blue]
> Hi Ben,
>
> not perfect but the output matches what you want and you can probably use it
> as a starter[/color]

wow! thanks!


Ben