Thank you for the reply.
If I use
<xsl:message>
var: <xsl:copy-of select="$var"/>
</xsl:message>
It only prints the following for the below mentioned xml:
112005-08-26aaaaaaaaaa11111222005-08-26bbbbb112005-08-26ccccc112005-08-26dddddddddd11111
For the xml:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<subroot id="11111">
<ccc>11</ccc>
<ddd>2005-08-26</ddd>
<eee>aaaaa</eee>
<eee>aaaaa11111</eee>
</subroot>
<subroot id="11111">
<ccc>22</ccc>
<ddd>2005-08-26</ddd>
<eee>bbbbb</eee>
</subroot>
<subroot id="11111">
<ccc>11</ccc>
<ddd>2005-08-26</ddd>
<eee>ccccc</eee>
</subroot>
<subroot id="11111">
<ccc>11</ccc>
<ddd>2005-08-26</ddd>
<eee>ddddd</eee>
<eee>ddddd11111</eee>
</subroot>
</root>
I want to see the complete xml, just like the input xml along with element
names.
For eg:
<root>
<ccc>11</ccc>
<ddd>2005-08-26</ddd>
<eee>aaaaa</eee>
<eee>aaaaa11111</eee>
...
with the hiearchy preserved. I am debugging larger xsl file and I don't know
whether result-tree in the memory is constructed right or not.
Thank you.
--- On Tue, 7/29/08, Florent Georges <lists(_at_)fgeorges(_dot_)org> wrote:
From: Florent Georges <lists(_at_)fgeorges(_dot_)org>
Subject: Re: [xsl] create xml from incoming xml and load it into a variable
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Date: Tuesday, July 29, 2008, 5:57 AM
sudheshna iyer wrote:
Hi
In the below xsl, how do I see the entire xml that is
in
variable "var"?
What do you mean exactly? If you want to "see"
it in the result
tree, use xsl:sequence or xsl:copy-of:
<bla-bla>
<xsl:copy-of select="$var"/>
</bla-bla>
If you want to "see" it on the console (or
whatever your environment
directs messages to), for debugging purpose, you can use
xsl:messages:
<xsl:message select="$var"/>
<!-- or... -->
<xsl:message>
VAR: <xsl:copy-of select="$var"/>
</xsl:message>
Regards,
--drkm
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