Re: Need an XPath expression which returns all xs:patt

On Thu, 4 Apr 2024 at 16:46, Roger L Costello costello@xxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> David Carlisle devised a brilliant approach:
>
> Do a series of replace operations:
>
> remove all whitespace:
>
>     replace(@value,'\s','')
>
> replace \-quoted characters by x:
>
>     replace(@value,'\\.','x')
>
> replace {99,} constructs by *
>
>     replace(@value,'\{[0-9]+,\}','*')
>
> Here are the replaces, inlined:
>
> replace(replace(replace(@value,'\{[0-9]+,\}','*'),'\\.','x'),'\s','')
>
> Here are the results of applying the replaces to some regexes:
>
> A*  --> apply replaces --> A*
> A+  --> apply replaces --> A+
> A\*  --> apply replaces --> Ax
> A\+  --> apply replaces --> Ax
> A{0,} --> apply replaces --> A*
> A{1,} --> apply replaces --> A*
> A{5,} --> apply replaces --> A*
> \\* --> apply replaces --> x*
>
> To implement "Find all xs:pattern elements that permit an unbounded number
> of characters" do this:
>
>         If the string resulting from applying the replaces
>         contains * or +, then the regex permits an
>         unbounded number of characters.
>
> David (or anyone), is this correct?
>

it's what I said but not correct (or I would rather say, I left you some
further cases as an exercise) you need more normalisation eg '[*] matches
one character as does [{0,}]  so you need to replace something like
\[[^\]*\]` before testing

David


> /Roger