Equivalent to like in XQuery?

Question

I am trying to get all titles that contain a certain word from my database in XQuery in SQL I would use WHERE title LIKE '%England%' - I am wondering how to do this in XQuery?

My XQuery query looks like this:

(:Selects document:)
for $x in doc ("tmdb_5000_movies(1).xml")/movies/movie 
(: :)
where $x/title (:NOT SURE WHAT TO PUT HERE:)
(:0:)
return $x/ *

Does this line of code do the same as like? for $x in doc ("tmdb_5000_movies(1).xml")/movies/movie [matches(title, '.*England')] (:0:) return $x/ * — BradleighJL, Nov 29 '19 at 16:07

score 1 · Answer 1 · answered Nov 29 '19 at 16:10

1

This does it.

for $x in doc ("tmdb_5000_movies(1).xml")/movies/movie [matches(title, '.*England')] 

return $x/ *

Much prefer SQL, However, I can see why this is the future

answered Nov 29 '19 at 16:10

BradleighJL

43
1
8

score 1 · Answer 2 · answered Dec 01 '19 at 10:13

I have no SQL DB to test with at hand but

doc("movies.xml")//movie[matches(title, '^.*England.*$')]/*

does work in other xquery runtimes. You should test, if you need the FLOWR-expression.

The regular expression @BradleighJL uses in his answer could be faster (.*England) and should also match all the cases. I just wanted to make a distinction between just "%England" ('^.*England$') and %England%.

Equivalent to like in XQuery?

2 Answers2