Question

In an xquery expression, I have obtained a set of values within a for-expression, and one value is in a separate variable.

Now, I want to subtract the single value from first value of the list, and then subtract consecutive members of the list from each other– and in the resulting set of difference values, I want to obtain the min/max values…

The query upto now looks like this–

let $value1:= 1998
let $rows_citations:= 
         $doc//div[@id="patent_citations"]
                     /div[@id="patent_citations_v"]
                          /table[@class="rel_patent"]
                             /tbody/tr[1]
                                      /following-sibling::tr
for $pos in $rows_citations/position()
let $date2_c := customfn:dateconverter1($rows_citations[$pos]/td[3])

Now the subtraction I want is between first value of date2_c and value 1, and after that between consecutive members of date2_c… And from the resulting list I want the min/max values… How do I go about doing this?

I am esp. confused about creating a new list variable that stores all the differences, esp. when we are already inside a for loop, and are iterating over each value of a list (via variable date2_c)

Answer 1

I. This XQuery 3.0 query (which is also a pure XPath 3.0 expression):

let $pVal := 1,
    $vList := (2,4,7,11,16),
    $vList2 := ($pVal, subsequence($vList, 1, count($vList)-1)),
    $vSubtactedList :=
            map-pairs(function($m as xs:integer, $n as xs:integer) as xs:integer
                        {
                         $m - $n
                        },
                        $vList,
                        $vList2
                      )
    return
      (min($vSubtactedList), max($vSubtactedList))

produces the wanted result the minimum and maximum values from the list of subtractions:

1 5

---

II. XQuery 1.0 solution:

let $pVal := 1,
    $vList := (2,4,7,11,16),
    $vList2 := ($pVal, subsequence($vList, 1, count($vList)-1)),
    $vSubtactedList :=
                for $i in 1 to count($vList)
                 return
                    $vList[$i] - $vList2[$i]

 return
    (min($vSubtactedList), max($vSubtactedList))

This again produces the same correct result:

1 5