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

2017-10-30 12:33:28
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.


Best,


David

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