I see what you mean, but I guess the case I gave was a little too
simple. What I really want to do is take any nodeset and copy it n times
and replace a string in each copy. Like this:
Input:
<anything>
<innerAnyThing>
<anyValue>replaceMe</ anyValue >
</ innerAnyThing >
<innerAnyThing>
<another>
<anyValue>replaceMe</ anyValue >
</another>
</ innerAnyThing >
<innerAnyThing>
<anyValue>replaceMe</ anyValue >
</ innerAnyThing >
</anything>
Assuming input is 3 iterations the output of last one would be:
<anything>
<innerAnyThing>
<anyValue>replaced1</ anyValue >
</ innerAnyThing >
<innerAnyThing>
<another>
<anyValue> replaced1</ anyValue >
</another>
</ innerAnyThing >
<innerAnyThing>
<anyValue> replaced1</ anyValue >
</ innerAnyThing >
</anything>
<anything>
<innerAnyThing>
<anyValue>replaced2</ anyValue >
</ innerAnyThing >
<innerAnyThing>
<another>
<anyValue> replaced2</ anyValue >
</another>
</ innerAnyThing >
<innerAnyThing>
<anyValue> replaced2</ anyValue >
</ innerAnyThing >
</anything>
<anything>
<innerAnyThing>
<anyValue>replaced3</ anyValue >
</ innerAnyThing >
<innerAnyThing>
<another>
<anyValue> replaced3</ anyValue >
</another>
</ innerAnyThing >
<innerAnyThing>
<anyValue> replaced3</ anyValue >
</ innerAnyThing >
</anything>
That's why I thought that replacing the string using the XPath string
replace would be the best idea, but I didn't know how to process a node
set as a string and then convert it back to a node set. (Unless there is
a better way to do it.)
Roshan Punnoose
Phone: 301-497-6039
Roshan Punnoose
Phone: 301-497-6039
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: Tuesday, January 30, 2007 12:57 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Duplicate Nodes in XSL and transform them
-A template that takes any node and copies it -Each copy has
to replace a string "replaceMe" with the string "replaced" +
i. ( 'i' being the iteration of the copy)
Generally, it's reasonably easy to generate numbers that are computed by
reference to the position of a node in the source tree. In some special
cases you can also use position() to generate a number that reflects the
position in the result tree. But in the general case, this kind of
problem
is best tackled as a two-pass transformation, in which you generate the
numbers in the second pass. For example, rather than inserting a
sequential
number, insert <n/>, and then in the second pass do a "modified copy"
transformation with
<xsl:template match="n">
<xsl:number level="any"/>
</xsl:template>
(You can do a two-pass transform with two stylesheets, or within a
single
stylesheet, whichever is most convenient. Using two stylesheets is a bit
harder to set up but gives you more reusable code.)
Michael Kay
http://www.saxonica.com/
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