Subject: [xsl] listing elements & sub-elements in a seqence - skiping elemts if sub-elemts are present From: Karl <call14@xxxxxxxxx> Date: Mon, 24 Jul 2006 15:20:45 +0100 (BST) |
The problem is I have various <section> elemts, with some having <subsection>'s. So, I need to create a sequence order/pagebreak at <subsection>'s SKIPING 'section', if <subsection>'s are available under section. Otherwise pagebreak at <section>. e.g. <section/> will become 1.Section title in list + creat pg 1 <section/> .. <section/> will become list num 3.Section title & pg 3 <section> ** [No pagebreaks/pagenums SINCE IT HAS <subsection> Pg 4 moved to subsection <subsection/> So, this become 4. Subsection title (pg 4) <subsection/> <subsection/> will become 6. Subsection title (pg 6) <section> <section/> will become 7.Section title (pg 7) <section> <subsection/> will become 8. Subsection title (pg 8) <section> and so on... so, looks like if i write something like <for-each - //section[with no subsection] | //section/subsection..> and then using position(), will i get the correct order>. 1) Whether the above code is correct or not, pls can forward me the syntax code equivalent to <for-each - //section[with no subsection] | //section/subsection..> 2) If the above is not correct, pls can you suggest some effective solution? Thanks in adv. karl ___________________________________________________________ Try the all-new Yahoo! Mail. "The New Version is radically easier to use" The Wall Street Journal http://uk.docs.yahoo.com/nowyoucan.html
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