i have the string, it contains xml nodes, returned from the PHP file.
It's like
<a>1</a><b>0</b><c>4</c>..............
Now i need to find out what value each node have i.e a, b, c......
while loading this string to xmlDocument i'm getting error like "There are multiple root elements".
any solution for this
One of the basic rules for well-formed XML that it has a single root node. With your example, you have multiple roots:
<a>1</a>
<b>0</b>
<c>4</c>
To make it well-formed you will have to make these elements a child of a single root:
<root>
<a>1</a>
<b>0</b>
<c>4</c>
</root>
An XML document that is not well-formed is not really an XML document at all and you will find that no XML parser will be able to read it!
Wrap it in a root element. E.g:
From <a>1</a><b>2</b>...
To <root><a>1</a><b>2</b>...</root>
Then compute as normal.
That is because each element is at the same level. There need to be a "root" that encloses all of them. Wrap them in a arbitrary node, say <root>...</root> then load the new string to the xmlDocument.
This seems like XML, but it's not valid XML. In XML, you have a root element that wraps all of the other elements.
So, if you have that string in str, do this:
str = String.Format("<root>{0}</root>", str);
Related
I have an xml file which has a CDATA section, which has again xml data. I would need to get a specific node from the xml within CDATA and create one more node of the same type and save the xml.
The replace and save functionality works for 1 input. but i want the tag to be appended in the same file. I hope i am clear!
Have o look at this thread XML parsing : Reading CDATA You probably need to read the CDATA value, convert, create node and write it back
I have an XML string like below:
<root>
<Test1>
<Result time="2">ProperEnding</Result>
</Test1>
<Test2></Test2>
I have to operate on these elements. Most of the time the elements are unique within their parent element. I am using XDocument. I can remember that there is a way to access an element like this.
XNode resultTest1 = GetNodes("/root//Test1//result")
But I forgot it. It is possible to access the same using linq:
doc.root.Elements.etc.etc.
But I want it using a single string as shown above. Can anybody say how to make it?
Descendants() will skip any number level of intermediate nodes, e.g. this will skip over root and Test1:
doc.Decendants("Result")
Also note that you can use XPath with Linq2Xml as well, e.g. XPathSelectElements
doc.XPathSelectElements("/root/Test1/Result");
You can skip intermediate levels of the hierarchy with // (or use // at the start of the xpath string to skip the root)
"/root//Result"
One caveat - Xml is case sensitive , so Result and result are not the same element.
The string you're referring to ("/root//Test1//result") is an XPath expression.
You can use it with LINQ to XML classes (like XDocument) using XPathEvaluate, XPathSelectElement, and XPathSelectElements extension methods.
You can find more info about these methods on MSDN: http://msdn.microsoft.com/en-us/library/vstudio/system.xml.xpath.extensions_methods(v=vs.90).aspx
To make them work, you need using System.Xml.XPath at the top of your file and System.Xml.Linq.dll assembly referenced (which is probably already there).
You can try to load your xml using XDocument:
// loads xml file with root element
XDocument xml = XDocument.Load("filename.xml");
Now you can append LINQ statements to your xml variable like this:
var retrieveSomeSpecificDataLikeListOfElementsAsAnonymousObjects = xml.Descendants("parentNodeName").Select(node => new { SomeSpecialValueYouWant = node.Element("elementNameUnderParentNode").Value }).ToList();
You can mix and do whatever you want - above is just an example.
Is this what you looking?
XmlDocument xmlDocument = new XmlDocument();
xmlDocument.LoadXml("YourXML");
XmlNodeList xmlNodes = xmlDocument.SelectNodes("/root/Test1/result");
I am having trouble generating an xml root. I have to match this structure as the elements of the xml use the prefixs throughout.
<ShipmentReceiptNotification
xmlns="urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02"
xmlns:dacc="urn:rosettanet:specification:domain:Procurement:AccountClassification:xsd:codelist:01.03"
xmlns:dbpq="urn:rosettanet:specification:domain:Procurement:BookPriceQualifier:xsd:codelist:01.04"
xmlns:dccc="urn:rosettanet:specification:domain:Procurement:CreditCardClassification:xsd:codelist:01.03"
xmlns:dcrt="urn:rosettanet:specification:domain:Procurement:CustomerType:xsd:codelist:01.03"
..\..\XML\Interchange\ShipmentReceiptNotification_02_02.xsd">
if I do something like
XmlNode ShipmentReceiptNotification0Node = xmlDoc.CreateElement("ShipmentReceiptNotification", "xmlns=\"urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02\"xmlns:dacc=\"urn:rosettanet:specification:domain:Procurement:AccountClassification:xsd:codelist:01.03\"");
I get
-ShipmentReceiptNotification
xmlns="xmlns="urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02"xmlns:dacc=&
quot;urn:rosettanet:specification:domain:Procurement:AccountClassification:xsd:codelist:01.03"">
The second argument of CreateElement accepts the URI of the namespace that the element being created, that is ShipmentReceiptNotification, belongs to. Not the whole bunch of xmlns attributes. This code:
XmlElement e = xmlDoc.CreateElement(
"ShipmentReceiptNotification",
"urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02");
Produces this XML:
<ShipmentReceiptNotification
xmlns="urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02" />
To produce what you want, you need to add attributes to the element. Like this:
XmlElement e = xmlDoc.CreateElement("ShipmentReceiptNotification");
e.SetAttribute("xmlns", "urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02");
e.SetAttribute("xmlns:dacc", "urn:rosettanet:specification:domain:Procurement:AccountClassification:xsd:codelist:01.03");
Produces this XML:
<ShipmentReceiptNotification
xmlns="urn:rosettanet:specification:interchange:ShipmentReceiptNotification:xsd:schema:02.02"
xmlns:dacc="urn:rosettanet:specification:domain:Procurement:AccountClassification:xsd:codelist:01.03" />
Note that this is the “manual” way. You should play with XmlNamespaceManager to do it “right”. However, that may be a bit more complex task which need not be necessary for your scenario.
Somewhere in a C# program, I need to get an attribute value from an xml structure. I can reach this xml structure directly as an XElement and have a simple xpath string to get the attribute. However, using XPathEvaluate, I get an empty array most of the time. (Yes, sometimes, the attribute is returned, but mostly it isn't... for the exact same XElement and xpath string...)
However, if I first convert the xml to string and reparse it as an XDocument, I do always get the attribute back. Can somebody explain this behavior ? (Am using .NET 3.5)
Code that mostly returns an empty IEnumerable:
string xpath = "/exampleRoot/exampleSection[#name='test']/#value";
XElement myXelement = RetrieveXElement();
((IEnumerable)myXElement.XPathEvaluate(xpath)).Cast<XAttribute>().FirstOrDefault().Value;
Code that does always work (I get my attribute value):
string xpath = "/exampleRoot/exampleSection[#name='test']/#value";
string myXml = RetrieveXElement().ToString();
XDocument xdoc = XDocument.Parse(myXml);
((IEnumerable)xdoc.XPathEvaluate(xpath)).Cast<XAttribute>().FirstOrDefault().Value;
With the test xml:
<exampleRoot>
<exampleSection name="test" value="2" />
<exampleSection name="test2" value="2" />
</exampleRoot>
By suggestion related to a surrounding root, I did some 'dry tests' in a test program, using the same xml structure (txtbxXml and txtbxXpath representing the xml and xpath expression described above):
// 1. XDocument Trial:
((IEnumerable)XDocument.Parse(txtbxXml.Text).XPathEvaluate(txtbxXPath.Text)).Cast<XAttribute>().FirstOrDefault().Value.ToString();
// 2. XElement trial:
((IEnumerable)XElement.Parse(txtbxXml.Text).XPathEvaluate(txtbxXPath.Text)).Cast<XAttribute>().FirstOrDefault().Value.ToString();
// 3. XElement originating from other root:
((IEnumerable)(new XElement("otherRoot", XElement.Parse(txtbxXml.Text)).Element("exampleRoot")).XPathEvaluate(txtbxXPath.Text)).Cast<XAttribute>().FirstOrDefault().Value.ToString();
Result : case 1 and 3 produce the correct result, while case 2 throws a nullref exception.
If case 3 would fail and case 2 succeed, it would have made some sense to me, but now I don't get it...
The problem is that the XPath expression is starting with the children of the specified node. If you start with an XDocument, the root element is the child node. If you start with an XElement representing your exampleRoot node, then the children are the two exampleSection nodes.
If you change your XPath expression to "/exampleSection[#name='test']/#value", it will work from the element. If you change it to "//exampleSection[#name='test']/#value", it will work from both the XElement and the XDocument.
I want to merge two XmlDocuments by inserting a second XML doc to the end of an existing Xmldocument in C#. How is this done?
Something like this:
foreach (XmlNode node in documentB.DocumentElement.ChildNodes)
{
XmlNode imported = documentA.ImportNode(node, true);
documentA.DocumentElement.AppendChild(imported);
}
Note that this ignores the document element itself of document B - so if that has a different element name, or attributes you want to copy over, you'll need to work out exactly what you want to do.
EDIT: If, as per your comment, you want to embed the whole of document B within document A, that's relatively easy:
XmlNode importedDocument = documentA.ImportNode(documentB.DocumentElement, true);
documentA.DocumentElement.AppendChild(importedDocument);
This will still ignore things like the XML declaration of document B if there is one - I don't know what would happen if you tried to import the document itself as a node of a different document, and it included an XML declaration... but I suspect this will do what you want.
Inserting an entire XML document at the end of another XML document is actually guaranteed to produce invalid XML. XML requires that there be one, and only one "document" element. So, assuming that your files were as follows:
A.xml
<document>
<element>value1</element>
<element>value2</element>
</document>
B.xml
<document>
<element>value3</element>
<element>value4</element>
</document>
The resultant document by just appending one at the end of the other:
<document>
<element>value1</element>
<element>value2</element>
</document>
<document>
<element>value3</element>
<element>value4</element>
</document>
Is invalid XML.
Assuming, instead, that the two documents share a common document element, and you want to insert the children of the document element from B into A's document element, you could use the following:
var docA = new XmlDocument();
var docB = new XmlDocument();
foreach (var childEl in docB.DocumentElement.ChildNodes) {
var newNode = docA.ImportNode(childEl, true);
docA.DocumentElement.AppendChild(newNode);
}
This will produce the following document given my examples above:
<document>
<element>value1</element>
<element>value2</element>
<element>value3</element>
<element>value4</element>
</document>
This is the fastest cleanest way to merge xml documents.
XElement xFileRoot = XElement.Load(file1.xml);
XElement xFileChild = XElement.Load(file2.xml);
xFileRoot.Add(xFileChild);
xFileRoot.Save(file1.xml);
Bad news. As long as the xml documents can have only one root element you cannot just put content of one document at the end of the second. Maybe this is what you are looking for? It shows how easily you can merge xml files using Linq-to-XML
Alternatively if you are using XmlDocuments you can try make it like this:
XmlDocument documentA;
XmlDocument documentB;
foreach(var childNode in documentA.DocumentElement.ChildNodes)
documentB.DocumentElement.AppendChild(childNode);