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Re: [xsl] construct dynamic replacement value in replace()?

2017-10-31 05:40:44
Dear Martin (cc xsl-list),


Thank you for the quick and helpful response! This does the job elegantly.

Best,

David

________________________________
From: Martin Honnen <martin(_dot_)honnen(_at_)gmx(_dot_)de>
Sent: Monday, October 30, 2017 1:47 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] construct dynamic replacement value in replace()?

On 30.10.2017 18:33, Birnbaum, David J djbpitt(_at_)pitt(_dot_)edu wrote:
Dear xsl-list,


I know how to accomplish this with XSLT string surgery, but is there an
XPath or XQuery way to calculate the replacement value of the replace()
function? The following (broken) XQuery expresses the
general aspiration, although not the reality:


declare function local:stuff($input) {
      let $result := number($input) + 10
      return xs:string($result)
      };
let $initial := '1 Tim. 4:123'
return replace($initial, '\d+', local:stuff('$0'))

The desired output would be '11 Tim. 14:133', that is, each sequence of
digits would be regarded as a discrete decimal numerical value, captured
as the match with '$0', and passed to the local:stuff() function, where
it would be converted to a number. augmented by 10, converted back to a
string (since the replacement part of the replace() function must be a
string), and returned. The actual output, alas, is 'NaN Tim. NaN:NaN'.
Before I give up and do it in XSLT, I would be grateful for any pointers
toward an XPath or XQuery solution.

XPath 3.1 can do it with analyze-string:

let $initial := '1 Tim. 4:123'
return string-join(analyze-string($initial, '[0-9]+')/*!(if
(local-name() eq 'match') then . + 10 else string()))
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