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  #11  
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Martin Honnen
 
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Default Re: Extracting hyperlink data from XML file - 07-30-2007 , 10:31 AM





Jonathan Attree wrote:

Quote:
These are the declarations at the top of the file:

?xml version="1.0" encoding="UTF-8"?
soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><soapenv:Body
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
getOrdersForDateRangeResponse
xmlns="http://obom/service/FormatB"><FilteredOrders href="#id0"
xmlns=""/></getOrdersForDateRangeResponse

multiRef id="id0" soapenc:root="0"
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xsi:type="ns1829687187:FormatBFilter"
xmlns:ns1829687187="http://obom/service/types" xmlns=""
ns1829687187:filterId xsi:type="xsd:int">94</ns1829687187:filterId
ns1829687187:Orders
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("file.xml");
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace("ns1829687187", "http://obom/service/types");
foreach (XmlElement part in xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)) {
Console.WriteLine("href is \"{0}\".", part.GetAttribute("href"));
}

If you only want to access the first Part element then use
SelectSingleNode instead of SelectNodes.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/


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  #12  
Old   
=?Utf-8?B?Sm9uYXRoYW4gQXR0cmVl?=
 
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Default Re: Extracting hyperlink data from XML file - 08-17-2007 , 07:30 AM







"Martin Honnen" wrote:

Quote:
Jonathan Attree wrote:

These are the declarations at the top of the file:

?xml version="1.0" encoding="UTF-8"?
soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><soapenv:Body
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
getOrdersForDateRangeResponse
xmlns="http://obom/service/FormatB"><FilteredOrders href="#id0"
xmlns=""/></getOrdersForDateRangeResponse

multiRef id="id0" soapenc:root="0"
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xsi:type="ns1829687187:FormatBFilter"
xmlns:ns1829687187="http://obom/service/types" xmlns=""
ns1829687187:filterId xsi:type="xsd:int">94</ns1829687187:filterId
ns1829687187:Orders

XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("file.xml");
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace("ns1829687187", "http://obom/service/types");
foreach (XmlElement part in xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)) {
Console.WriteLine("href is \"{0}\".", part.GetAttribute("href"));
}

If you only want to access the first Part element then use
SelectSingleNode instead of SelectNodes.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/
Hi again
Thanks for this. I need to do this in VB and have got this far but am having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.

Thanks
Jonathan Attree


Reply With Quote
  #13  
Old   
=?Utf-8?B?Sm9uYXRoYW4gQXR0cmVl?=
 
Posts: n/a

Default Re: Extracting hyperlink data from XML file - 08-17-2007 , 07:30 AM




"Martin Honnen" wrote:

Quote:
Jonathan Attree wrote:

These are the declarations at the top of the file:

?xml version="1.0" encoding="UTF-8"?
soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><soapenv:Body
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
getOrdersForDateRangeResponse
xmlns="http://obom/service/FormatB"><FilteredOrders href="#id0"
xmlns=""/></getOrdersForDateRangeResponse

multiRef id="id0" soapenc:root="0"
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xsi:type="ns1829687187:FormatBFilter"
xmlns:ns1829687187="http://obom/service/types" xmlns=""
ns1829687187:filterId xsi:type="xsd:int">94</ns1829687187:filterId
ns1829687187:Orders

XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("file.xml");
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace("ns1829687187", "http://obom/service/types");
foreach (XmlElement part in xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)) {
Console.WriteLine("href is \"{0}\".", part.GetAttribute("href"));
}

If you only want to access the first Part element then use
SelectSingleNode instead of SelectNodes.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/
Hi again
Thanks for this. I need to do this in VB and have got this far but am having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.

Thanks
Jonathan Attree


Reply With Quote
  #14  
Old   
John Saunders [MVP]
 
Posts: n/a

Default Re: Extracting hyperlink data from XML file - 08-17-2007 , 09:40 PM


"Jonathan Attree" <JonathanAttree (AT) discussions (DOT) microsoft.com> wrote in
message news:9A5801CD-A42E-4369-A2EC-F01435CB9534 (AT) microsoft (DOT) com...
Quote:

"Martin Honnen" wrote:

Jonathan Attree wrote:

These are the declarations at the top of the file:

?xml version="1.0" encoding="UTF-8"?
soapenv:Envelope
xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><soapenv:Body
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
getOrdersForDateRangeResponse
xmlns="http://obom/service/FormatB"><FilteredOrders href="#id0"
xmlns=""/></getOrdersForDateRangeResponse

multiRef id="id0" soapenc:root="0"
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xsi:type="ns1829687187:FormatBFilter"
xmlns:ns1829687187="http://obom/service/types" xmlns=""
ns1829687187:filterId xsi:type="xsd:int">94</ns1829687187:filterId
ns1829687187:Orders

XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("file.xml");
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace("ns1829687187",
"http://obom/service/types");
foreach (XmlElement part in xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)) {
Console.WriteLine("href is \"{0}\".", part.GetAttribute("href"));
}

If you only want to access the first Part element then use
SelectSingleNode instead of SelectNodes.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/


Hi again
Thanks for this. I need to do this in VB and have got this far but am
having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.
The syntax for the For loop would be:

For Each part As XmlElement In xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)
Console.WriteLine("href is ""{0}"".", part.GetAttribute("href"))
Next

--
John Saunders [MVP]



Reply With Quote
  #15  
Old   
John Saunders [MVP]
 
Posts: n/a

Default Re: Extracting hyperlink data from XML file - 08-17-2007 , 09:40 PM


"Jonathan Attree" <JonathanAttree (AT) discussions (DOT) microsoft.com> wrote in
message news:9A5801CD-A42E-4369-A2EC-F01435CB9534 (AT) microsoft (DOT) com...
Quote:

"Martin Honnen" wrote:

Jonathan Attree wrote:

These are the declarations at the top of the file:

?xml version="1.0" encoding="UTF-8"?
soapenv:Envelope
xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><soapenv:Body
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
getOrdersForDateRangeResponse
xmlns="http://obom/service/FormatB"><FilteredOrders href="#id0"
xmlns=""/></getOrdersForDateRangeResponse

multiRef id="id0" soapenc:root="0"
soapenv:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xsi:type="ns1829687187:FormatBFilter"
xmlns:ns1829687187="http://obom/service/types" xmlns=""
ns1829687187:filterId xsi:type="xsd:int">94</ns1829687187:filterId
ns1829687187:Orders

XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("file.xml");
XmlNamespaceManager namespaceManager = new
XmlNamespaceManager(xmlDoc.NameTable);
namespaceManager.AddNamespace("ns1829687187",
"http://obom/service/types");
foreach (XmlElement part in xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)) {
Console.WriteLine("href is \"{0}\".", part.GetAttribute("href"));
}

If you only want to access the first Part element then use
SelectSingleNode instead of SelectNodes.


--

Martin Honnen --- MVP XML
http://JavaScript.FAQTs.com/


Hi again
Thanks for this. I need to do this in VB and have got this far but am
having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.
The syntax for the For loop would be:

For Each part As XmlElement In xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)
Console.WriteLine("href is ""{0}"".", part.GetAttribute("href"))
Next

--
John Saunders [MVP]



Reply With Quote
  #16  
Old   
=?Utf-8?B?Sm9uYXRoYW4gQXR0cmVl?=
 
Posts: n/a

Default Re: Extracting hyperlink data from XML file - 08-20-2007 , 10:32 AM


Quote:
Hi again
Thanks for this. I need to do this in VB and have got this far but am
having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.

The syntax for the For loop would be:

For Each part As XmlElement In xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)
Console.WriteLine("href is ""{0}"".", part.GetAttribute("href"))
Next

--
John Saunders [MVP]
Thanks very much.

JOnathan Attree


Reply With Quote
  #17  
Old   
=?Utf-8?B?Sm9uYXRoYW4gQXR0cmVl?=
 
Posts: n/a

Default Re: Extracting hyperlink data from XML file - 08-20-2007 , 10:32 AM


Quote:
Hi again
Thanks for this. I need to do this in VB and have got this far but am
having
trouble with For loop:

Dim XMLResponseDoc As New XmlDocument
Dim nsManager As New XmlNamespaceManager(XMLResponseDoc.NameTable)
Dim element As New XmlElement

XMLResponseDoc.Load(strXMLResponseFile)
nsManager.AddNamespace("ns1829687187", "http://obom/service/types")
For Each

Next

A VB version would be much appreciated.

The syntax for the For loop would be:

For Each part As XmlElement In xmlDoc.SelectNodes("//ns1829687187:Part",
namespaceManager)
Console.WriteLine("href is ""{0}"".", part.GetAttribute("href"))
Next

--
John Saunders [MVP]
Thanks very much.

JOnathan Attree


Reply With Quote
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