Try the following construction:
<xsl:template match="/">
<xsl:call-template name="base">
<xsl:with-param name="pos" select="39"/>
</xsl:call-template>
</xsl:template>
<xsl:template name="base">
<xsl:param name="pos"/>
<xsl:variable name="lastpos" select="number($pos) + 100"/>
<xsl:apply-templates select="item[((position() < $lastpos) and
(position() > $pos)) or (position() = $pos) ]" mode="iteration"/>
</xsl:template>
<xsl:template match="item" mode="iteration">
. . .
</xsl:template>
Best regards,
Roman
Guy De Schepper wrote:
> ok, this query will indeed do the job for this example, but what about
> large documents / collections ?let's say the document contains 10000
> items and I want to display them with 100 at a time.for each page
> (which will contain 100 items) I have to launch 100 queries !
>
> -----Original Message-----
> From: Jeff Greif [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, June 18, 2002 22:24
> To: [email protected]; [EMAIL PROTECTED]
> Subject: Re: nr of results && starting position
> You would need to use the preceding-sibling construct of
> XPath. To get three items starting with the one with a
> partiicular timestamp, you would return /mydoc/myitem
> elements such that the timestamp of the context node, or its
> preceding sibling, or the preceding sibling of its preceding
> sibling was as specified. Jeff
>
> From:[EMAIL PROTECTED]
> To: Xindice-Users
> Sent: Tuesday, June 18, 2002 5:35 AM
> Subject: nr of results && starting position
> ...Now, I want to query this document for the next
> 2 items, starting with item which
> timestamp='20020514124420645'the query I am
> thinking of is something like this:xindice xpath
> -c /db/mytestcol -q
> "/mydoc/[EMAIL PROTECTED]'20020514124420645']"but
> this will return only that one elementGuy
>
