andrew welch wrote:
On 4/5/06, Paull <paullus4mlist(_at_)gmail(_dot_)com> wrote:
Hello All,
following xml:
<data>
<item name="1" id="i1">v1</item>
<item name="2" id="i1">v2</item>
<item name="3" id="i1">v1</item>
<item name="4" id="i1">v1</item>
<item name="5" id="i1">v2</item>
<item name="6" id="i1">v1</item>
<item name="7" id="i1">v1</item>
<item name="8" id="i1">v2</item>
<item name="9" id="i1">v1</item>
<item name="10" id="i2">v2</item>
<item name="11" id="i2">v2</item>
<item name="12" id="i2">v2</item>
<group name="g1" id="i1"/>
<group name="g2" id="i2"/>
</data>
should be transformed to the xml, where items are grouped by id, and
divided to pages whith 5 items per page. Result should be like following:
g1
1. i1v1
2. i1v2
3. i1v1
4. i1v1
5. i1v2
EOP
1. i1v1
2. i1v1
3. i1v2
4. i1v1
g2
5. i2v2
EOP
1. i2v2
2. i2v2
I can group it, but how to divide for pages - no idea ...
It's difficult to give an exact answer without seeing how you are
grouping, but this can be achieved using mod, eg position() mod 5 = 0
will give you every fifth item.
I'm grouping with following:
<xsl:template match="/data">
<xsl:for-each select="group">
<xsl:variable name="id" select="@id"/>
<b><xsl:value-of select="$id"/></b><br/>
<xsl:apply-templates select="/data/item[(_at_)id=$id]" />
</xsl:for-each>
</xsl:template>
<xsl:template match="item">
<xsl:value-of select="@id"/> <xsl:value-of select="."/><br/>
<xsl:if test="position() mod 5 = 0">
<!-- EOP -->
</xsl:if>
</xsl:template>
But using <xsl:if test="position() mod 5 = 0"> is not correct because it
not handles situation with 5 groups each with one item should be pasted
into one page.
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