Subject: Re: [xsl] Pattern notation for preceding-sibling From: "Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 11 Aug 2014 11:31:45 -0000 |
I would write /foo/E/preceding-sibling::*[self::A | self::N] If A and N have something in common that leads to this kind of expression being meaningful, then in a schema-aware world it's quite likely that you can make them members of the same substitution group G, in which case you can write /foo/E/preceding-sibling::schema-element(G) And another possibility is /foo/E/preceding-sibling::*[not(self::B)] Michael Kay Saxonica mike@xxxxxxxxxxxx +44 (0) 118 946 5893 On 11 Aug 2014, at 10:49, Heiko Niemann kontakt@xxxxxxxxxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > Hi, > > a rather basic question, but I was not able to find samples/answers for it > yet. > > Elements A,B,B,N,E,B,C are contained in element foo. > > /foo/E/preceding-sibling::* will return A,B,B,N; > > /foo/E/preceding-sibling::N will return N; - easy - > > > Now I want A,N to be returned, so my first guess is, that this might work: > > /foo/E/preceding-sibling::(A,N) > > But it does not, so I either have to write > > /foo/E/(preceding-sibling::A, preceding-sibling::N) or > > /foo/E/*[name() = ('A','N')] > > > So, is there a (short) notation that I missed so far, that does what I > expected /foo/E/preceding-sibling::(A,N) to do? > > > Regards, > Heiko
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