[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: XPath expression to check that there are no inter
I like Mike's answer for XSLT 2.
Your XSLT 1 solution can be improved by reducing the number of counting actions. Your solution has three uses of count() ... this has only two: count(B[last()]/preceding-sibling::*[not(self::B)]) = count(B[1]/preceding-sibling::*[not(self::B)]) Avoiding the use of counting you could do the following: string( generate-id( B[1]/preceding-sibling::*[not(self::B)][1] ) ) = string( generate-id( B[last()]/preceding-sibling::*[not(self::B)][1] ) ) The use of string() is necessary in case the sequence of children starts with B. Come to think of it, neither your answer, Mike's answer nor my answer will work if there are no B children at all. You will get a false positive. Does your test need to accommodate that situation? . . . . . . . . Ken At 2016-07-19 15:44 +0000, Costello, Roger L. costello@xxxxxxxxx wrote: Hi Folks, -- Check our site for free XML, XSLT, XSL-FO and UBL developer resources | Streaming hands-on XSLT/XPath 2 training @US$45: http://goo.gl/Dd9qBK | Crane Softwrights Ltd. _ _ _ _ _ _ http://www.CraneSoftwrights.com/s/ | G Ken Holman _ _ _ _ _ _ _ _ _ _ mailto:gkholman@xxxxxxxxxxxxxxxxxxxx | Google+ blog _ _ _ _ _ http://plus.google.com/+GKenHolman-Crane/posts | Legal business disclaimers: _ _ http://www.CraneSoftwrights.com/legal |
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|