RE: [xsl] How do you get the non-transformed character entity out of MSXML3.0

["Chris Bayes" <chris@xxxxxxxxxxx>]

Jim,
You can't get it except from the .xml property. So you need something
like
var b = xml.selectSingleNode("/Customer/@NumberOfPurchases").xml;
var z = b.substring(b.indexOf("=\"")+2, b.lastIndexOf("\""));
Or rather the vb equivalent
Dim b:b = xml.selectSingleNode("/Customer/@NumberOfPurchases").xml
Dim z:z = Mid(b, InStr(1, b, "=""")+2, InStrRev(b, """")-(InStr(1, b,
"=""")+2))

Ciao Chris

XML/XSL Portal
http://www.bayes.co.uk/xml


> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of 
> jdgarrett@xxxxxxxxxx
> Sent: 15 September 2001 23:37
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] How do you get the non-transformed character 
> entity out of MSXML3.0
> 
> 
> 
> 
> Here is the node
> 
> <Customer  number="125487"   NumberOfPurchases="&lt;4" >Joe 
> &amp; Bill's
> Shop</Customer>
> 
> (I am just loading up an xml document via
> Visual Basic inside of a code module and then accessing
> the xml via manipulation of the DOM via Visual Basic, I am
> not transforming the xml with an xslt file...that
> would happen later and that is why the element and attribute 
> values in the Customer node must stay in their untransformed state)
> 
> What I am trying to retrieve from the node attribute 
> NumberOfPurchases (using MSXML 3.0 ) is :
> 
> &lt;4
> 
> 
> What I have been able to get via one of the properties in MSXML3.0
> 
> accessing the xml property on the node attribute you get 
> NumberOfPurchases="&lt;4"
> 
> accessing the text property on the node attribute you get
> <4
> 
> Anyone know how to return just  &lt;4 using one of the 
> property's or methods of MSXML3.0
> 
> Thanks
> Jim Garrett
> 
> 
> 
> 
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 


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