Subject: RE: [xsl] How do you get the non-transformed character entity out of MSXML3.0 From: "Chris Bayes" <chris@xxxxxxxxxxx> Date: Sun, 16 Sep 2001 09:34:39 +0100 |
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="<4" >Joe > & 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 : > > <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="<4" > > accessing the text property on the node attribute you get > <4 > > Anyone know how to return just <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 > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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