Learn how to efficiently filter unique year instances in XQuery and XPath using the `for loop` as a predicate. --- This video is based ...
In XQuery and XPath, using a for
loop as a predicate can be a bit tricky, but you can achieve filtering unique years using a combination of loops and the distinct-values
function. Here’s how you can do it step-by-step:
Assuming you have the following XML structure containing years:
<records>
<record>
<year>2020</year>
<event>Event A</event>
</record>
<record>
<year>2021</year>
<event>Event B</event>
</record>
<record>
<year>2020</year>
<event>Event C</event>
</record>
<record>
<year>2022</year>
<event>Event D</event>
</record>
</records>
You can use distinct-values
to get unique years from the XML data. Here’s how to do it in XQuery:
let $records :=
<records>
<record>
<year>2020</year>
<event>Event A</event>
</record>
<record>
<year>2021</year>
<event>Event B</event>
</record>
<record>
<year>2020</year>
<event>Event C</event>
</record>
<record>
<year>2022</year>
<event>Event D</event>
</record>
</records>
let $uniqueYears := distinct-values($records/record/year)
return
<unique-year-list>
{
for $year in $uniqueYears
return <year>{$year}</year>
}
</unique-year-list>
$records
variable.distinct-values
function extracts unique years from the records.for
loop iterates over each unique year, creating a new <year>
element for each one.The output will be:
<unique-year-list>
<year>2020</year>
<year>2021</year>
<year>2022</year>
</unique-year-list>
If you want to use XPath to filter unique years, you can use a similar approach with XPath 2.0 or higher:
distinct-values(/records/record/year)
This XPath expression will return a sequence of unique years directly.
By combining the distinct-values
function with a for
loop, you can effectively filter and display unique years from your XML data in XQuery. This method leverages the strengths of both XQuery and XPath for efficient data manipulation.