Newbie Xpath Question

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Edgardo - 25 Jun 2007 04:00 GMT

Hi everyone, i have a quesiton regarding XML/Xpath

i have a XML document that has let say the following structure

<cases>
<case>
<num>1<num>
<detective>
<name>John</name>
<lastname>Smith</lastname>
<detective>
</case>
<case>
<num>2<num>
<detective>
<name>Roger</name>
<lastname>Wilco</lastname>
<detective>
</case>
</cases>

I want to know, what would be the most "by the book" way to get all the
values from the cases nodes into variables for further work.
I need to "spider" XML files that have a defined structure, they wont change
their structure from one day to another, but i need do get the cleanes way to
get these values into variables som each loop of the
nodelist/nodeiterator/cursor i do some calculations.

I see a lot of examples telling me jus thow to print out the whole content
of a element, but not how to acces a child element or subnode and get its
value.

something that would help me do like

while node.read()
case = node.current
mymun = case.num
mydetname = case.det.name
mydetlastname = case.det.lastname
wend

Oviously i know it wont be that way, im just writng the way i would need to
get the variables.

I hope you can help me, any idea, suggestion, url that you can give me would
be really appreciated.

Thanks

-Ed

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Rick - 25 Jun 2007 12:19 GMT

I haven't done any of this in a while, but this should point you in the
right direction...

If you want to get all the "detective" nodes from the xml document:

nodes = xmlDoc.SelectNodes("cases/")

Completely untested, but that should give you all the "case" nodes so you
can process them further.

Rick

> Hi everyone, i have a quesiton regarding XML/Xpath
>
[quoted text clipped - 50 lines]
>
> -Ed

Reply to this Message

Martin Honnen - 25 Jun 2007 13:02 GMT

> <cases>
> <case>
[quoted text clipped - 19 lines]
> get these values into variables som each loop of the
> nodelist/nodeiterator/cursor i do some calculations.

If you don't want to change the XML and want to use XPath to access
nodes then use XPathDocument/XPathNavigator, for instance as follows in
.NET 2.0 or later:

XPathDocument doc = new XPathDocument(@"file.xml");
XPathNavigator navigator = doc.CreateNavigator();
XPathNodeIterator iterator = navigator.Select(@"cases/case");
while (iterator.MoveNext())
{
int num =
iterator.Current.SelectSingleNode("num").ValueAsInt;
string name =
iterator.Current.SelectSingleNode("detective/name").Value;
string lastname =
iterator.Current.SelectSingleNode("detective/lastname").Value;
Console.WriteLine("Number {0} has name {1} {2}.", num,
name, lastname);
}

You could also consider XML deserialization, that way you do not have to
know XPath.

Signature

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

Reply to this Message

WenYuan Wang [MSFT] - 25 Jun 2007 13:12 GMT

Hello ED,
Thanks for Rick's suggestion.

I also agree with him. The best way is to retrieve all Case nodes and get
the values from its nested node.

System.Xml.XmlDocument xd = new System.Xml.XmlDocument();
xd.Load("XX.xml");
System.Xml.XmlNodeList xnl = xd.SelectNodes(@"/cases/case");
foreach (System.Xml.XmlNode xn in xnl)
{
string mynum= xn["num"].InnerText;
string mydetname = xn["detective"]["name"].InnerText;
string mydelastname = xn["detective"]["lastname"].InnerText;
}

Hope this helps. If you meet any further issue, feel free to update here.
I'm glad to assist you.

Have a great day,
Sincerely,
Wen Yuan
Microsoft Online Community Support
==================================================
This posting is provided "AS IS" with no warranties, and confers no rights.

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Edgardo - 25 Jun 2007 15:30 GMT

Thank you all for the help! ;)

I'll try those tonight and let you know what happens.

Regards

-Ed

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Edgardo - 25 Jun 2007 18:01 GMT

Thanks for the tip WenYuan, im working with VB and framework 1.1 and im not
getting this line to work

string mydelastname = xn["detective"]["lastname"].InnerText;

i tryed xnl("detective")("lastname").innertText

but doesn not work

any ideas?

Best Regards

> Hello ED,
> Thanks for Rick's suggestion.
[quoted text clipped - 21 lines]
> ==================================================
> This posting is provided "AS IS" with no warranties, and confers no rights.

Reply to this Message

Martin Honnen - 25 Jun 2007 18:16 GMT

> Thanks for the tip WenYuan, im working with VB and framework 1.1 and im not
> getting this line to work
[quoted text clipped - 4 lines]
>
> but doesn not work

That should be
xnl("detective")("lastname").InnerText

If you still have problems then tell us exactly which error message you get.

Signature

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

Reply to this Message

Edgardo - 25 Jun 2007 18:42 GMT

Thanks for the help Martin, i get

Object reference not set to an instance of an object.

i and double checked the spelling and its correct

any suggestions?

-Ed

> > Thanks for the tip WenYuan, im working with VB and framework 1.1 and im not
> > getting this line to work
[quoted text clipped - 9 lines]
>
> If you still have problems then tell us exactly which error message you get.

Reply to this Message

Edgardo - 26 Jun 2007 03:46 GMT

I think i saw something that might be causing the error..

Guys, will it affect if the structure is like the following?

<cases>
<case>
<num>1<num>
<detective>
<name>John1</name>
<lastname>Smith1</lastname>
<detective>
<detective>
<name>John2</name>
<lastname>Smith2</lastname>
<detective>
</case>
<case>
<num>2<num>
<detective>
<name>Roger1</name>
<lastname>Wilco1</lastname>
<detective>
<detective>
<name>Roger2</name>
<lastname>Wilco2</lastname>
<detective>
</case>
</cases>

Reply to this Message

Edgardo - 26 Jun 2007 04:01 GMT

Ok, dont ask me how, now its working but the problem is how can i get the
values of the secons instance of "detective"??

xnl("detective")("lastname").InnerText = "Smith1"
????? = "Smith2"

Best Regards

-Ed

Reply to this Message

WenYuan Wang [MSFT] - 26 Jun 2007 06:21 GMT

Hello Edgardo
Thanks for Martin's help.

I think you have found the key point.
The structure of XML could affect the result.

Please notice there are two DETECTIVE node nested in same Case node
<case>
<num>1</num>
* <detective>
<name>John1</name>
<lastname>Smith1</lastname>
</detective>
*<detective>
<name>John2</name>
<lastname>Smith2</lastname>
</detective>
</case>

xnl("detective") could only get the FIRST node named DETECTIVE. Thus, if
the number of DETECTIVE node is greater than ONE, we would have to retrieve
each node by xnl.selectNodes("detective") mothed rather than
xnl("detective").

Please check the following code snippet.

Dim xd As System.Xml.XmlDocument = New System.Xml.XmlDocument
xd.Load("XMLFile1.xml")
Dim xnlCa As System.Xml.XmlNodeList = xd.SelectNodes("/cases/case")
For Each xnCa As System.Xml.XmlNode In xnlCa
Dim mynum As String = xnCa("num").InnerText

Dim xnlDet As System.Xml.XmlNodeList = xnCa.SelectNodes("detective")
For Each xnDet As System.Xml.XmlNode In xnlDet
Dim mydetName As String = xnDet("name").InnerText
Dim mydetLastName As String = xnDet("lastname").InnerText
Next

Next

BTW, you could aslo retrieve each node by array. For eaxample:

Dim xd As System.Xml.XmlDocument = New System.Xml.XmlDocument
xd.Load("XMLFile1.xml")
Dim xnlCa As System.Xml.XmlNodeList = xd.SelectNodes("/cases/case")
For Each xnCa As System.Xml.XmlNode In xnlCa
Dim mynum As String = xnCa("num").InnerText

Dim xnlDet As System.Xml.XmlNodeList = xnCa.SelectNodes("detective")
Dim mydetName1 As String = xnlDet(0)("name").InnerText ' = "John1"
Dim mydetLastName1 As String = xnlDet(0)("lastname").InnerText ' =
"Smith1"
Dim mydetName2 As String = xnlDet(1)("name").InnerText ' = "John2"
Dim mydetLastName2 As String = xnlDet(1)("lastname").InnerText ' =
"Smith2"

Next

Hope this helps.Please let us know if you meet any further issue. We are
glad to assist you.
Have a great day,
Wen Yuan
Microsoft Online Community Support
==================================================
This posting is provided "AS IS" with no warranties, and confers no rights.

Reply to this Message

Edgardo - 26 Jun 2007 18:31 GMT

Thanks a lot WenYuan, i really appreciate your help, i've read a lot on the
web, but on ly simple examples come up, this is really helpful, i'll give it
a try tonigh.

-Ed

> Hello Edgardo
> Thanks for Martin's help.
[quoted text clipped - 61 lines]
> ==================================================
> This posting is provided "AS IS" with no warranties, and confers no rights.

Reply to this Message

WenYuan Wang [MSFT] - 27 Jun 2007 03:54 GMT

The pleasure is mine.:)
Welcome, Ed.
I will wait for you. Please let me know if you meet any further issue.

Have a great day,
Sincerely,
Wen Yuan
Microsoft Online Community Support
==================================================
This posting is provided "AS IS" with no warranties, and confers no rights.

Reply to this Message

WenYuan Wang [MSFT] - 29 Jun 2007 13:08 GMT

Hello Ed

This is Wen Yuan. Have you tried the method so far?
If you meet any further issue on this, feel free to update here again.
We are glad to assist.:)

Have a great weekend.
Sincerely,
Wen Yuan
Microsoft Online Community Support
==================================================
This posting is provided "AS IS" with no warranties, and confers no rights.

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