Hi Rob,
here's yet another solution:
let $path := "Q{ http://www.w3.org/2005/xpath-functions%7Droot()/Q%7B%7DA%5B1%5D/Q%7B%7DB%5B1...] " return analyze-string($path, '(Q{.*?}.*?)(/|$)')/fn:match/fn:group[1]/string()
@Christian: On the other hand I can imagine that within Basex the
path-expression is built from these elementary steps.
If so, is it in any way possibly to retrieve this elementary steps
sequence instead of this path-expression?
Currently, there is no other way to retrieve paths, so you'll have to manually split the path as indicated above..
Christian
On Wed, Oct 8, 2014 at 2:10 PM, Marco Lettere marco.lettere@dedalus.eu wrote:
Hi Rob, don't know whether your prefix is constant.
If it is then
let $prefix := "Q{http://www.w3.org/2005/xpath-functions%7Droot()" let $inpstr := "Q{http://www.w3.org/2005/xpath-functions }root()/Q{}A[1]/Q{}B[1]/Q{}C[1]/Q{}D[1]" return ($prefix, tokenize(substring-after($inpstr, $prefix),"/"))
should do the job. M.
On 08/10/2014 13:37, Rob Stapper wrote:
Hi,
Basically I want to break down the path-expression of a node in my XML into its elementary steps.
QName: “Q{http://www.w3.org/2005/xpath-functions%7Droot()/Q%7B%7DA%5B1%5D/Q%7B%7DB%5B1... must become QName-sequence: (“Q{http://www.w3.org/2005/xpath-functions%7Droot()%E2%80%9D, ”Q{}A[1] ”, ”Q{}B[1] ”, ”Q{}C[1] ”, ”Q{}D[1]”)
So I need to break the path-expression down on the “/”-characters which aren’t part of the namespace-uries.
I’m using the “fn:tokenize”-function for this. The function uses a “regular expression” for it’s separator.
Unfortunately my regex-experience is zero (well, two days by now). I came up with the next brilliant ;-) solution: fn:tokenize( path( $node), “/(?=Q)”).
Again, unfortunately, this doesn’t work because the xquery-flavor of regex doesn’t support looking around.
Sure, I can use “/Q” as pattern and subsequently stick a “Q” in front of every substring (except the first one) but I’m looking for a solid solution.
I there anyone out there with enough “regex-for-xquery”- experience to help me with this?
Big thanx for that in advance.
@Christian: On the other hand I can imagine that within Basex the path-expression is built from these elementary steps.
If so, is it in any way possibly to retrieve this elementary steps sequence instead of this path-expression?
Thanx,
Rob Stapper
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