# DevOps, CI/CD and Automation

## Count consecutive repetitive element in a sequence using XQuery

KaushikBoseFeb 24 2015 — edited Feb 25 2015

If the Sequence = [a a b c c c a d d e e e f g h h]

then the Output = [1 2 1 1 2 3 1 1 2 1 2 3 1 1 1 2]

Have tried to use recursion but no luck...Please Help..Thanks in Anticipation

**Note: Using XQuery implementation 1.0**

One of my failed implementation looks like:

`declare function local:test($sequence,$count){`

for $counter in (1 to count($sequence))

let $maxIndex := count($sequence)

return

if (matches(subsequence($sequence,1,$maxIndex)[$counter],subsequence($sequence,1,$maxIndex)[$counter + +1])) then let $count := $count + 1 return $count[last()]

else let $count := 1 return $count[last()]

};

It seems that as $counter is a sequence hence it returns a sequence in matches condition and hence this implementation fail

This post has been answered by

Jump to Answer**odie_63**on Feb 24 2015Comments

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