CODEWITHSUNDEEP
xmlxquery
John Babson
John Babson

Reputation: 51

XQuery: How to put a variable in a URL?

I'm still relatively new to Xpath.

I'm wondering how exactly I would use a variable to generate a url.

For instance, one application I'm using has a different XML document URL for each date.

So one URL might be http://www.example/2014-09-01.xml.

I need to retrieve a large number of these URLs based on the current date. I'm not sure how one would insert a $variable into a statement like doc("URL")/filters

Upvotes: 0

Views: 691

Answers (2)

Charles Duffy
Charles Duffy

Reputation: 295383

The syntax for variables in XPath is simply $varname. So:

doc($url)/filters

The method for passing in variables in from an out-of-band tool or processing engine varies with that processing engine. With XMLStarlet, for instance, that might be:

xmlstarlet sel --var url "http://www.example.com/$date.xml"

...or, with the BaseX command-line interface:

basex -b'$url='"http://www.example.com/$date.xml"

Upvotes: 2

wst
wst

Reputation: 11771

let $URI := concat('http://www.example/', $date, '.xml')
return doc($URI)

Upvotes: 2

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