Re: construct dynamic replacement value in replace()?

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.