Hello!
The easiest (but not the fastest) way to do this could be the following
(untested, never did this...)
Transform A using the stylesheet into a MemoryStream. Reset the stream
to its initial position. Create a new XmlDocument, load B. Create
another XmlDocument, load the transformation's output from the MemoryStream.
XmlDocument bDoc...; XmlDocument xslOutput...;
XmlNode sel = xslOutput.SelectSingleNode("/some/xpath");
sel = bDoc.ImportNode(sel, true);
XmlElement target = bDoc.SelectSingleNode("/target/path") as XmlElement;
target.AppendChild( sel );
But you can also combine two documents using XSLT: use an
XslArgumentList, and add new XPathDocument("b.xml").Select("/") as a
param. In you stylesheet you can then use the contents of a.xml and b.xml:
<stylesheet ...>
<param name="b" />
<template match="/">
<value-of select="count(//*)" /> Nodes in a.
<value-of select="count($b//*)" /> Nodes in b.
</template>
</stylesheet>
Your example could look like:
<stylesheet ...>
<param name="b" />
<template match="/">
<o:plan>
<for-each select="$b/proposal">
<o:proposerdetail>
<copy-of select="node()" />
</o:proposerdetail>
</for-each>
</o:plan>
</template>
</stylesheet>
You then just have to Transform from A.xml to temp.xml, then delete
B.xml and rename [eg. move] temp.xml to B.xml
--
Pascal Schmitt
