[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: create xml from incoming xml and load it into a v
Thank you. This is exactly what I want. --- On Tue, 7/29/08, Michael Kay <mike@xxxxxxxxxxxx> wrote: > From: Michael Kay <mike@xxxxxxxxxxxx> > Subject: RE: create xml from incoming xml and load it into a variable > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Date: Tuesday, July 29, 2008, 6:52 AM > Unfortunately the output of xsl:message is > implementation-defined. Your XSLT > processor appears to be flattening it to a string, which I > don't think is a > particularly good idea, but it's allowed by the spec. > (Some processors > discard xsl:message output completely, which is also > permitted.) > > You can also view the variable by copying it temporarily to > the result tree: > > <debug variable="var"> > <xsl:copy-of select="$var"/> > </debug> > > Michael Kay > http://www.saxonica.com/ > > > -----Original Message----- > > From: sudheshna iyer [mailto:sudheshnaiyer@xxxxxxxxx] > > Sent: 29 July 2008 11:30 > > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > > Subject: Re: create xml from incoming xml and > load it > > into a variable > > > > Thank you for the reply. > > > > If I use > > <xsl:message> > > var: <xsl:copy-of select="$var"/> > > </xsl:message> > > It only prints the following for the below mentioned > xml: > > > > > 112005-08-26aaaaaaaaaa11111222005-08-26bbbbb112005-08-26ccccc1 > > 12005-08-26dddddddddd11111 > > > > 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@xxxxxxxxxxxx> wrote: > > > > > From: Florent Georges <lists@xxxxxxxxxxxx> > > > Subject: Re: create xml from incoming xml > and load it into a > > > variable > > > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > > > 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 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > ______________________________________________________________________ > > > _______ > > > > > > Envoyez avec Yahoo! Mail. Une boite mail plus > intelligente > > > http://mail.yahoo.fr
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