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xmlxquerywhere-clausewildcardflwor
Markus
Markus

Reputation: 21

Wildcard in XQuery corresponding to [where LIKE '%.mp3'] in SQL

I´m wondering how to use wildcards in XQuery.

I have two lines of codes which gives me the return that I want from a xml file:

for $phrase in //phrases[@pPhrase="Phrase1.mp3"]
return concat(string($phrase/@pPhrase), ", ",$phrase/ancestor::entrance/string(@name))

The problem I have is that sometimes I don´t know the exact name of the mp3 file I´m looking for. So, if this would be a SQL query I´d type: where pPhrase LIKE '%.mp3'.

for $phrase in //phrases[@pPhrase LIKE ='%.mp3']

The above line doesn´t work of course, so what is the correct syntax?

Upvotes: 2

Views: 352

Answers (1)

AdamL
AdamL

Reputation: 13191

You could use

//phrases[contains(@pPhrase,'.mp3']

Or to be more precise (since xPath 2.0)

//phrases[ends-with(@pPhrase,'.mp3']

Upvotes: 3

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