I have the below fragement of XML, notice that the Reference node holds a URI which links to the Id attribute of the Body node.
<Reference URI="#Body">
<SOAP-ENV:Body Id="Body" xmlns:SOAP-ENV="http://www.dingo.org">
<ns0:Add xmlns:ns0="http://www.moo.com">
<ns0:a>2</ns0:a>
<ns0:b>3</ns0:b>
</ns0:Add>
</SOAP-ENV:Body>
If I had the value of the URI attribute how would I then get the whole Body XMLNode? I presume this would be best done via an XPath epression but haven't any clue on XPath. Note that the XML will not always be so simple. I'm doing this in c# btw :)
Any ideas?
Thanks
Jon
EDIT: I wouldn't know the XML structure or namespaces before hand, all I would know is that the reference element has the ID of the xmlNode i want to retrieve, hope this is sligtly clearer.
You can add a condition that applies to a relative (or absolute node) to any step of an XPath expression.
In this case:
//*[#id=substring-after(/Reference/#URI, '#')]
The //* matches all elements in the document. The part in [] is a condition. Inside the condition the part of the URI element of the root References node is taken, but ignoring the '#' (and anything before it).
Sample code, assuming you have loaded your XML into XPathDocument doc:
var nav = doc.CreateNavigator();
var found = nav.SelectSingleNode("//*[#id=substring-after(/Reference/#URI, '#')]");
If you have the value of the URI attribute in a variable you could use
myXmlDocument.DocumentElement.SelectSingleNode("//SOAP-ENV:Body[ID='pURI']")
where pURI is the value of the URI attribute and myXmlDocument is the Xml Document object
Something like this:
XmlDocument requestDocument = new XmlDocument();
requestDocument.LoadXml(yourXmlString);
String someXml = requestDocument.SelectSingleNode(#"/*[local-name()='Reference ']/*[local-name()='Body']").InnerXml;
Related
I'm trying to capture the attribute "description" in this XML:
<ProductoModel xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://schemas.datacontract.org/2004/07/WebApi.Models">
<descripcion>descripcion 1</descripcion>
<fecha_registro>2016-03-01</fecha_registro>
<id_producto>1</id_producto>
<id_proveedor>1</id_proveedor>
<nombre_producto>producto 1</nombre_producto>
<precio>200</precio>
</ProductoModel>
My Code :
XmlDocument xDoc = new XmlDocument();
xDoc.LoadXml(content);
XmlNamespaceManager manager = new XmlNamespaceManager(xDoc.NameTable);
manager.AddNamespace("MYNS", "http://schemas.datacontract.org/2004/07/WebApi.Models");
XmlNode node = xDoc.DocumentElement.SelectSingleNode("MYNS:ProductoModel", manager);
MessageBox.Show(node.Attributes.GetNamedItem("descripcion").Value);
The problem is I can not capture the attribute "descripcion" and get the following error:
Object reference not set to an instance of an object.
As I can capture the required attribute?
<descripcion> is not attribute. It is element.
You can get any element (or attribute) with a single xpath query.
XmlNode node = xDoc.DocumentElement.SelectSingleNode("/MYNS:ProductoModel/MYNS:descripcion", manager);
MessageBox.Show(node.InnerText);
Note the character / at the beginning of the xpath expression.
If you want another easy way operate XML, check this out. This is a little tool for xml operate, it's much easier to use and understand than XmlNode.
Assume this simple XML fragment in which there may or may not be the xml declaration and has exactly one NodeElement as a root node, followed by exactly one other NodeElement, which may contain an assortment of various number of different kinds of elements.
<?xml version="1.0">
<NodeElement xmlns="xyz">
<NodeElement xmlns="">
<SomeElement></SomeElement>
</NodeElement>
</NodeElement>
How could I go about selecting the inner NodeElement and its contents without the namespace? For instance, "//*[local-name()='NodeElement/NodeElement[1]']" (and other variations I've tried) doesn't seem to yield results.
As for in general the thing that I'm really trying to accomplish is to Deserialize a fragment of a larger XML document contained in a XmlDocument. Something like the following
var doc = new XmlDocument();
doc.LoadXml(File.ReadAllText(#"trickynodefile.xml")); //ReadAllText to avoid Unicode trouble.
var n = doc.SelectSingleNode("//*[local-name()='NodeElement/NodeElement[1]']");
using(var reader = XmlReader.Create(new StringReader(n.OuterXml)))
{
var obj = new XmlSerializer(typeof(NodeElementNodeElement)).Deserialize(reader);
I believe I'm missing just the right XPath expression, which seem to be rather elusive. Any help much appreciated!
Try this:
/*/*
It selects children of the root node.
Or
/*/*[local-name() = 'NodeElement']
It selects children with local-name() = 'NodeElement' of the root node.
Anyway in your case both expressions select <NodeElement xmlns="">.
walk the tree
foreach(XmlNode node in doc.DocumentElement.childnodes[0].childnodes)
{
// do something with node
}
hideously fragile of course might want to check for nulls here and there.
I've found a lot of articles about how to get node content by using simple XPath expression and C#, for example:
XPath:
/bookstore/author/first-name
C#:
string xpathExpression = "/bookstore/author/first-name";
nodes = navigator.Select(xpathExpression);
I wonder how to get content that is inside of an element, and the same element is inside another element and another and another.
Just take a look on below code:
<Cell>
<CellContent>
<Para>
<ParaLine>
<String>ABCabcABC abcABC abc ABCABCABC.</string>
</ParaLine>
</Para>
</CellContent>
</Cell>
I only want to extract content ABCabcABC abcABC abc ABCABCABC. from String element.
Do you know how to resolve problem by use XPath expression and .Net C#?
After googling c# .net xpath for few seconds you'll find this article, which provides example which you can easily modify to use XPathDocument, XPathNavigator and XPathNavigator::SelectSingleNode():
XPathNavigator nav;
XPathDocument docNav;
string xPath;
docNav = new XPathDocument("c:\\books.xml");
nav = docNav.CreateNavigator();
xPath = "/Cell/CellContent/Para/ParaLine/String/text()";
string value = nav.SelectSingleNode(xPath).Value
I recommend more reading on xPath syntax. Much more.
navigator.SelectSingleNode("/Cell/CellContent/Para/ParaLine/String/text()").Value
You can use Linq to XML as well to get value of specified element
var list = XDocument.Parse("xml string").Descendants("ParaLine")
.Select(x => x.Element("string").Value).ToList();
From above query you will get value of all the string element which are inside ParaLine tag.
I am using c# console app to get xml document. Now once xmldocument is loaded i want to search for specific href tag:
href="/abc/def
inside the xml document.
once that node is found i want to strip tag completly and just show Hello.
Hello
I think i can simply get the tag using regex. But can anyone please tell me how can i remove the href tag completly using regex?
xml & html same difference: tagged content. xml is stricter in it's formatting.
for this use case I would use transformations and xpath queries rebuild the document. As #Yahia stated, regex on tagged documents is typically a bad idea. the regex for parsing is far to complex to be affective as a generic solution.
The most popular technology for similar tasks is called XPath. (It is also a key component of XQuery and XSLT.) Would the following perhaps solve your task, too?
root.SelectSingleNode("//a[#href='/abc/def']").InnerText = "Hello";
You could try
string x = #"<?xml version='1.0'?>
<EXAMPLE>
<a href='/abc/def'>Hello</a>
</EXAMPLE>";
System.Xml.XmlDocument doc = new XmlDocument();
doc.LoadXml(x);
XmlNode n = doc.SelectSingleNode("//a[#href='/abc/def']");
XmlNode p = n.ParentNode;
p.RemoveChild(n);
System.Xml.XmlNode newNode = doc.CreateNode("element", "a", "");
newNode.InnerXml = "Hello";
p.AppendChild(newNode);
Not really sure if this is what you are trying to do but it should be enough to get you headed in right direction.
Is it possible to get the open tag from a XmlNode with all attributes, namespace, etc?
eg.
<root xmlns="urn:..." rattr="a">
<child attr="1">test</child>
</root>
I would like to retrieve the entire opening tag, exactly as retrieved from the original XML document if possible, from the XmlNode and later the closing tag. Both as strings.
Basically XmlNode.OuterXml without the child nodes.
EDIT
To elaborate, XmlNode.OuterXml on a node that was created with the XML above would return the entire XML fragment, including child nodes as a single string.
XmlNode.InnerXml on that same fragment would return the child nodes but not the parent node, again as a single string.
But I need the opening tag for the XML fragment without the children nodes. And without building it using the XmlAttribute array, LocalName, Namespace, etc.
This is C# 3.5
Thanks
Is there some reason you can't simply say:
string s = n.OuterXml.Substring(0, n.OuterXml.IndexOf(">") + 1);
I think the simplest way would be to call XmlNode.CloneNode(false) which (according to the docs) will clone all the attributes but not child nodes. You can then use OuterXml - although that will give you the closing tag as well.
For example:
using System;
using System.Xml;
public class Test
{
static void Main()
{
XmlDocument doc = new XmlDocument();
doc.LoadXml(#"<root xmlns='urn:xyz' rattr='a'>
<child attr='1'>test</child></root>");
XmlElement root = doc.DocumentElement;
XmlNode clone = root.CloneNode(false);
Console.WriteLine(clone.OuterXml);
}
}
Output:
<root xmlns="urn:xyz" rattr="a"></root>
Note that this may not be exactly as per the original XML document, in terms of the ordering of attributes etc. However, it will at least be equivalent.
How about:
xmlNode.OuterXML.Replace(xmlNode.InnerXML, String.Empty);
Poor man's solution :)