Xquery to find the next to last item-CodePudding

with xml something like:

<item > 
   <datapoint>5</datapoint>
</item>
<item > 
   <datapoint>4</datapoint>
</item>
<item > 
   <datapoint>6</datapoint>
</item>
<item > 
   <datapoint>8</datapoint>
</item>
<item > 
   <datapoint>9</datapoint>
</item>

And the number of items is not a fixed number. I am already retrieving the last() data point, however I now also need to grab the next to last data point. So in this example I need to separately return both the 8 and the 9.

CodePudding user response：

This XQuery should work:

for $it in $file/item[position()=last() or position()=last()-1]
return $it/datapoint

Or, you could also use a shorter version:

for $it in $file/item[position()>=last()-1]
return $it/datapoint

Output is:

<datapoint>8</datapoint>
<datapoint>9</datapoint>

CodePudding user response：

First, your XML should be well-formed. That's why I added a root element

XML

<root>
    <item>
        <datapoint>5</datapoint>
    </item>
    <item>
        <datapoint>4</datapoint>
    </item>
    <item>
        <datapoint>6</datapoint>
    </item>
    <item>
        <datapoint>8</datapoint>
    </item>
    <item>
        <datapoint>9</datapoint>
    </item>
</root>

XQuery

for $x in /root/item[last()-1]/datapoint
return $x

Output

<datapoint>8</datapoint>