CODEWITHSUNDEEP
xquery
Ohhh
Ohhh

Reputation: 435

XQuery Comparing 2 lists of number

Given 2 xml file that contain a list of number in each. 

<list1>
    <item>1</item>
    <item>2</item>
    ...
</list1>

list1 = [1,2,3,4,5,7,8,9,10]
list2 = [1,4,6,2,8]

I only want the number in list1 that are greater than all of the number in list2, how can I do that?

I tried this, but it only return the number if it is greater than one of the number in list2 and not the entire list.

for $x in doc("list1.xml")/item
    for $y in doc("list2.xml")/item
        where ($x > $y) 
return $x

Thanks in advance, newbie in xquery.

Upvotes: 1

Views: 1206

Answers (2)

adamretter
adamretter

Reputation: 3517

Perhaps I have misunderstood the question, but it seems to me that there is a much simpler and more efficient solution using a single predicate which does not require any explicit loops:

xquery version "1.0";

let $list1 := (1, 2, 3, 4, 5, 7, 8, 9, 10)
let $list2 := (1, 4, 6, 2, 8)

let $max2 := max($list2)

return

    $list1[. gt $max2]

I would also like to make it clear to @ou-ye a little about XQuery terminology. At present there is no such thing as a List structure in XQuery, the closest thing is a Sequence which is constructed using ( and ).

Upvotes: 0

dirkk
dirkk

Reputation: 6218

Using XQuery 3.0 you can use fn:for-each, which makes a nicer syntax in my opinion. So for each item in $list1 it checks whether there is an item in $list2 which is larger.

let $list1 := (1 to 10)
let $list2 := (1, 4, 6, 2, 8)
return for-each($list1, function($l) {
  if (for-each($list2, function($x) {$l > $x}) = false())
  then ()
  else $l
})

If you just have XQuery 1.0, this should work as well and does basically the same:

let $list1 := (1 to 10)
let $list2 := (1, 4, 6, 2, 8)
for $l in $list1
where not((
  for $f in $list2
  return $l > $f
) = false())
return $l

Upvotes: 2

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