Hi everyone. I'm in a situation where I need to work with XML data
that I don't have the format for. It's a web service situation where
I hand them a query and an XSLT file. They produce XML fromt the
query, then run the XSLT on it and return the transformed data. But I
need to find out the original structure since the interface is poorly
documented.
Anyone have any suggestions as to how to do this? Or even better a
XSLT file laying around that can generically return the original XML?
Thanks for your time.
-Matt Bradbury
3  4297 
Hi
I have a transformation routine that is supposed transform the xml
file more or less "as it is", and I've tried the template you suggest.
There is , however, one problem I have yet to resolve: is it possible
to copy the DOCTYPE declaration as well? As far as I know this is not
one of the nodes xPath is able to target, so how can I copy it?
I would be very happy if could supply an answer, or point me in the
right direction here...
Best wishes
Vemund
On Wed, 20 Aug 2003 21:16:34 +0200, "Dimitre Novatchev"
<dn********@yah oo.com> wrote: This is the wellknown identity template:
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
"Matt Bradbury" <br******@iense mble.com> wrote in message
news:2d******* *************** ***@posting.goo gle.com... Hi everyone. I'm in a situation where I need to work with XML data
that I don't have the format for. It's a web service situation where
I hand them a query and an XSLT file. They produce XML fromt the
query, then run the XSLT on it and return the transformed data. But I
need to find out the original structure since the interface is poorly
documented.
Anyone have any suggestions as to how to do this? Or even better a
XSLT file laying around that can generically return the original XML?
Thanks for your time.
-Matt Bradbury
> There is , however, one problem I have yet to resolve: is it possible to copy the DOCTYPE declaration as well? As far as I know this is not
one of the nodes xPath is able to target, so how can I copy it?
Sorry -- you can't. There are only "traces" of a DTD in the XML Infoset.
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
"Vemund Olstad" <ve***********@ hit.uib.no> wrote in message
news:3f******** *******@nntp.ui b.no... Hi
I have a transformation routine that is supposed transform the xml
file more or less "as it is", and I've tried the template you suggest.
There is , however, one problem I have yet to resolve: is it possible
to copy the DOCTYPE declaration as well? As far as I know this is not
one of the nodes xPath is able to target, so how can I copy it?
I would be very happy if could supply an answer, or point me in the
right direction here...
Best wishes
Vemund
On Wed, 20 Aug 2003 21:16:34 +0200, "Dimitre Novatchev"
<dn********@yah oo.com> wrote:
This is the wellknown identity template:
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
"Matt Bradbury" <br******@iense mble.com> wrote in message
news:2d******* *************** ***@posting.goo gle.com... Hi everyone. I'm in a situation where I need to work with XML data
that I don't have the format for. It's a web service situation where
I hand them a query and an XSLT file. They produce XML fromt the
query, then run the XSLT on it and return the transformed data. But I
need to find out the original structure since the interface is poorly
documented.
Anyone have any suggestions as to how to do this? Or even better a
XSLT file laying around that can generically return the original XML?
Thanks for your time.
-Matt Bradbury
On Thu, 21 Aug 2003 14:53:11 +0200, "Dimitre Novatchev"
<dn********@yah oo.com> wrote: Sorry -- you can't. There are only "traces" of a DTD in the XML Infoset.
Blegh - I was afraid of that. Thank you for the very quick response -
I guess I'll have to resort to scripting....
Best wishes
Vemund
"Vemund Olstad" <ve***********@ hit.uib.no> wrote in message
news:3f******* ********@nntp.u ib.no... Hi
I have a transformation routine that is supposed transform the xml
file more or less "as it is", and I've tried the template you suggest.
There is , however, one problem I have yet to resolve: is it possible
to copy the DOCTYPE declaration as well? As far as I know this is not
one of the nodes xPath is able to target, so how can I copy it?
I would be very happy if could supply an answer, or point me in the
right direction here...
Best wishes
Vemund
On Wed, 20 Aug 2003 21:16:34 +0200, "Dimitre Novatchev"
<dn********@yah oo.com> wrote:
>This is the wellknown identity template:
>
> <xsl:template match="@* | node()">
> <xsl:copy>
> <xsl:apply-templates select="@* | node()"/>
> </xsl:copy>
> </xsl:template>
>
>
>
>=====
>Cheers,
>
>Dimitre Novatchev.
>http://fxsl.sourceforge.net/ -- the home of FXSL
>
>
>"Matt Bradbury" <br******@iense mble.com> wrote in message
>news:2d******* *************** ***@posting.goo gle.com...
>> Hi everyone. I'm in a situation where I need to work with XML data
>> that I don't have the format for. It's a web service situation where
>> I hand them a query and an XSLT file. They produce XML fromt the
>> query, then run the XSLT on it and return the transformed data. But I
>> need to find out the original structure since the interface is poorly
>> documented.
>>
>> Anyone have any suggestions as to how to do this? Or even better a
>> XSLT file laying around that can generically return the original XML?
>>
>> Thanks for your time.
>>
>> -Matt Bradbury
>
>
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