It has taken a while, but I have finally been able to modify an XML document based on user input for the namespace and node name:
string nodeName = "DefinitionName"; // this is really provided by the user
string namespace = "http://schemas.datacontract.org/2004/07/Xxx.Session"; // also user-provided
XmlDocument xmlDocument = new XmlDocument();
xmlDocument.Load(taskResolved.XmlPathAndFileName);
XmlElement rootElement = xmlDocument.DocumentElement;
XmlNamespaceManager namespaceManager = new XmlNamespaceManager(xmlDocument.NameTable);
namespaceManager.AddNamespace("snuh", namespace); // hard-coded prefix, grrr...
XmlNodeList xmlNodes;
xmlNodes = rootElement.SelectNodes("//snuh:" + nodeName, namespaceManager);
I feel like I'm doing something wrong because I have to hard-code a prefix (snuh). I could try and choose a prefix, like snuh, that I can hope will never appear in a document, but that isn't foolproof. Another option is to use a GUID for a prefix, but this just seems like a hack work-around. Am I missing something? Is there a better way?
The top of the XML doc looks like this:
<?xml version="1.0" encoding="utf-8"?>
<SessionStateInfo xmlns:i="http://www.w3.org/2001/XMLSchema-instance" z:Id="1"
z:Type="Xxx.SessionStateInfo"
z:Assembly="Xxx.Common, Version=6.2.0.0, Culture=neutral, PublicKeyToken=null"
xmlns:z="http://schemas.microsoft.com/2003/10/Serialization/"
xmlns="http://schemas.datacontract.org/2004/07/Xxx.Session">
<CoaterNumber>25</CoaterNumber>
<DefinitionName z:Id="2">TwoLineMarkerDefinition</DefinitionName>
<EnableManualMode>true</EnableManualMode>
If you want to simply select the first DefinitionName node.
You may write
XmlNode node = rootElement[nodeName, namespace];
and if you want the whole list:
XmlNodeList nodeList = rootElement.GetElementsByTagName(nodeName, namespace);
What about using the XPath local-name() and namespace-uri() functions?
string xpath = string.Format("//*[local-name()='{0}' and namespace-uri()='{1}']", nodeName, namespace);
xmlNodes = rootElement.SelectNodes(xpath);
Don't know if those functions are supported in XmlDocument though, haven't tested it.
Related
i have to fill up an XML file from a DATA TABLE ,my problem is that i have to get the schemaLocation in the root node ,for this i use the code below ,then i have this result,and i dont know where is p1 coming from
In your resulting XML, p1 is a namespace. The code you have posted (in a screenshot) is defining the namespace "xsi", I'm not sure why your result is generating p1 unless you are renaming xsi somewhere that is not shown.
XmlDocument doc = new XmlDocument();
XmlDeclaration declaire = doc.CreateXmlDeclaration("1.0", "utf-8", null);
XmlElement rootnode = doc.CreateElement("BMECAT");
doc.InsertBefore(declaire, doc.DocumentElement);
doc.AppendChild(rootnode);
rootnode.SetAttribute("version", "2005");
XmlAttribute atr = doc.CreateAttribute("xsi", "schemaLocation", "http://www.w3.org/2001/XMLSchema-instance");
atr.Value = "http://www.adlnet.org/xsd/adlcp.vlp3";
rootnode.SetAttributeNode(atr);
rootnode.Attributes.Append(atr);
In your code:
XmlAttribute atr = doc.CreateAttribute("xsi", "schemaLocation", "http://www.w3.org/2001/XMLSchema-instance");
"xsi" is the name of the namespace it generates, you can control it there. This results in:
<?xml version="1.0" encoding="utf-8"?>
<BMECAT version="2005" xsi:schemaLocation="http://www.adlnet.org/xsd/adlcp.vlp3" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" />
I'm not sure your code matches the result file you provided. When I ran the code, I get "xsi" as expected. If I set "xsi" to null there, it uses a default name, which in my case was d1p1. All instances of "xsi" were replaced with "d1p1". This makes me believe that code might be slightly different from what generated your result. I don't know where "d1p1" came from, it's likely a generated default namespace. This seems like a common default (Remove "d1p1" namespace prefix in DataContractSerializer XML output). In your provided code, if you change "xsi" to "p1" you would get your result.
I might suggest using this method instead:
How to Add schemaLocation attribute to an XML document
Here you would use the accepted answer against your XmlElement rootnode.
XmlElement.SetAttributeValue (localname, prefix, namespace, value)
Please try this code and let me know whether this helped you or not.
Particularly in this code i parse XML file and get the root element:
Then use it to select all attributes named schemaLocation. There is only one, so you can use SelectSingleNode:
The variable
schemaLocationAttribute
Contanins Value attribute through which you can get actual value.
XmlReader xmlReader = XmlReader.Create("MyXML.xml");
XmlDocument xmlDocument = new XmlDocument();
xmlDocument.Load(xmlReader);
XmlElement root = xmlDocument.DocumentElement;
XmlNode schemaLocationAttribute = root.SelectSingleNode("//#*[local-name()='schemaLocation']");
//Single schema value
string schemaValue = schemaLocationAttribute.Value;
//If you have multiple values in your schema
//you have to store it inside of array
string[] multipleShcemavalues = schemaLocationAttribute.Value.Split(null);
//And you have to choose whuickelement you want to use
string chooosendShcema = multipleShcemavalues[1]; //For example
Using C# I have an XML file like
<?xml version="1.0" encoding="utf-8"?>
<root>
<Account>
<name>Jani</name>
</Account>
</root>
and I also have a function to read the name node as:
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load("lib//user.xml");
XmlNode node;
node = xmlDoc.DocumentElement;
string name = node.Attributes[0].Value;
label1.Text = name.ToString();
but I am getting index out of range error as:
Why is this happening?
node = xmlDoc.DocumentElement;
string name = node.Attributes[0].Value;
node is your root node. Which looks like this:
<root>
How many attributes does it have? None, as it turns out. An attribute in XML is one of these bar="baz" things:
<foo bar="baz">
node.Attributes[0] refers to the first attribute. There is no first attribute, there's no second attribute -- you didn't use attributes in this XML at all. Hence, that's out of range. There's no first item in an empty collection.
What you want is an element named name, which is farther down inside your XML tree.
Probably this:
var node = xmlDoc.DocumentElement.SelectSingleNode("/root/Account/name");
And then you'll want to look at node.InnerText to get "Jani" out of it.
You are trying to read node.Attributes[0].Value but there is no attribtues in your sample XML file. Not sure of the exact syntax but it should probably be closer to node.Value
As mentioned by other answers, your current XML does not have attributes.
private void DoIt()
{
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load(#"M:\StackOverflowQuestionsAndAnswers\38924171\38924171\data.xml");
XmlNode node;
node = xmlDoc.DocumentElement;
//string name = node.Attributes[0].Value;
string name = node["Account"].InnerText;
}
If your XML did have attributes
<?xml version="1.0" encoding="utf-8"?>
<root>
<Account name="Jani" />
</root>
Then you could do this:
private void DoItAgain()
{
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load(#"M:\StackOverflowQuestionsAndAnswers\38924171\38924171\data2.xml");
XmlNode node;
node = xmlDoc.DocumentElement;
string name = node["Account"].Attributes[0].Value;
}
I have an XML file that looks like:
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="https://www.someurl.com/somefile.xslt"?>
<AutoInsuranceClaim xmlns="http://www.someurl.com/schemas/AutoInsuranceClaim">
<Identification>
<BaseOwner>3</BaseOwner>
<BaseType>ABC123</BaseType>
<BaseTypeRef>471038341757</BaseTypeRef>
</Identification>
</AutoInsuranceClaim>
and I'm trying to read the Identification node. Here's my code:
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load(#"..\..\Data.xml");
XmlNamespaceManager nsmgr = new XmlNamespaceManager(xmlDoc.NameTable);
nsmgr.AddNamespace("ns", "http://www.someurl.com/schemas/AutoInsuranceClaim");
XmlNodeList nodeList = xmlDoc.SelectNodes(#"/ns:AutoInsuranceClaim/Identification", nsmgr);
Console.WriteLine("There are {0} nodes...", nodeList.Count);
I know I should get a least 1 value. My understanding of .NET XML parsing is that if you have a default namespace with no prefix, you have to create your own namespace. But this should have returned 1.
If not, what am I missing?
I might be grasping at straws here, but shouldn't you be namespacing both entities in your xpath expression?
XmlNodeList nodeList = xmlDoc.SelectNodes(#"/ns:AutoInsuranceClaim/ns:Identification", nsmgr);
XElement root = XElement.Load("Data.xml");
var identifications = root.Descendants()
.Where(x => x.Name.LocalName == "Identification")
.ToList()
The problem is that you're trying to find an Identification node without a namespace, but it will have defaulted to the same namespace as the parent due to the xmlns=... part. Try this:
var nodeList = xmlDoc.SelectNodes("/ns:AutoInsuranceClaim/ns:Identification",
nsmgr);
Having tried it myself, it printed a count of 1.
Personally I'd use LINQ to XML instead though, which makes namespace easier handling:
XDocument doc = XDocument.Load(#"..\..\Data.xml");
XNamespace ns = "http://www.someurl.com/schemas/AutoInsuranceClaim";
var nodes = doc.Root.Elements(ns + "Identification");
How to parse the xml file?
<?xml version="1.0" encoding="UTF-8"?>
<sitemapindex xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">
<sitemap>
<loc>link</loc>
<lastmod>2011-08-17T08:23:17+00:00</lastmod>
</sitemap>
<sitemap>
<loc>link</loc>
<lastmod>2011-08-18T08:23:17+00:00</lastmod>
</sitemap>
</sitemapindex>
I am new to XML, I tried this, but it seems to be not working :
XmlDocument xml = new XmlDocument(); //* create an xml document object.
xml.Load("sitemap.xml");
XmlNodeList xnList = xml.SelectNodes("/sitemapindex/sitemap");
foreach (XmlNode xn in xnList)
{
String loc= xn["loc"].InnerText;
String lastmod= xn["lastmod"].InnerText;
}
The problem is that the sitemapindex element defines a default namespace. You need to specify the namespace when you select the nodes, otherwise it will not find them. For instance:
XmlDocument xml = new XmlDocument();
xml.Load("sitemap.xml");
XmlNamespaceManager manager = new XmlNamespaceManager(xml.NameTable);
manager.AddNamespace("s", "http://www.sitemaps.org/schemas/sitemap/0.9");
XmlNodeList xnList = xml.SelectNodes("/s:sitemapindex/s:sitemap", manager);
Normally speaking, when using the XmlNameSpaceManager, you could leave the prefix as an empty string to specify that you want that namespace to be the default namespace. So you would think you'd be able to do something like this:
// WON'T WORK
XmlDocument xml = new XmlDocument();
xml.Load("sitemap.xml");
XmlNamespaceManager manager = new XmlNamespaceManager(xml.NameTable);
manager.AddNamespace("", "http://www.sitemaps.org/schemas/sitemap/0.9"); //Empty prefix
XmlNodeList xnList = xml.SelectNodes("/sitemapindex/sitemap", manager); //No prefixes in XPath
However, if you try that code, you'll find that it won't find any matching nodes. The reason for this is that in XPath 1.0 (which is what XmlDocument implements), when no namespace is provided, it always uses the null namespace, not the default namespace. So, it doesn't matter if you specify a default namespace in the XmlNamespaceManager, it's not going to be used by XPath, anyway. To quote the relevant paragraph from the Official XPath Specification:
A QName in the node test is expanded into an expanded-name using the
namespace declarations from the expression context. This is the same
way expansion is done for element type names in start and end-tags
except that the default namespace declared with xmlns is not used: if
the QName does not have a prefix, then the namespace URI is null (this
is the same way attribute names are expanded). It is an error if the
QName has a prefix for which there is no namespace declaration in the
expression context.
Therefore, when the elements you are reading belong to a namespace, you can't avoid putting the namespace prefix in your XPath statements. However, if you don't want to bother putting the namespace URI in your code, you can just use the XmlDocument object to return the URI of the root element, which in this case, is what you want. For instance:
XmlDocument xml = new XmlDocument();
xml.Load("sitemap.xml");
XmlNamespaceManager manager = new XmlNamespaceManager(xml.NameTable);
manager.AddNamespace("s", xml.DocumentElement.NamespaceURI); //Using xml's properties instead of hard-coded URI
XmlNodeList xnList = xml.SelectNodes("/s:sitemapindex/s:sitemap", manager);
Sitemap has 2 sub nodes "loc" and "lastmod". The nodes that you are accessing are "name" and "url". that is why you are not getting any result. Also in your XML file the last sitemap tag is not closed properly with a corresponding Kindly try xn["loc"].InnerText and see if you get the desired result.
I would definitely use LINQ to XML instead of the older XmlDocument based XML API. You can accomplish what you are looking to do using the following code. Notice, I changed the name of the element that I am trying to get the value of to 'loc' and 'lastmod', because this is what is in your sample XML ('name' and 'url' did not exist):
XElement element = XElement.Parse(XMLFILE);
IEnumerable<XElement> list = element.Elements("sitemap");
foreach (XElement e in list)
{
String LOC= e.Element("loc").Value;
String LASTMOD = e.Element("lastmod").Value;
}
Can someone help me read attribute ows_AZPersonnummer with asp.net using c# from this xml structure
<listitems
xmlns:s="uuid:BDC6E3F0-6DA3-11d1-A2A3-00AA00C14882"
xmlns:dt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882"
xmlns:rs="urn:schemas-microsoft-com:rowset"
xmlns:z="#RowsetSchema"
xmlns="http://schemas.microsoft.com/sharepoint/soap/">
<rs:data ItemCount="1">
<z:row
ows_AZNamnUppdragsansvarig="Peter"
ows_AZTypAvUtbetalning="Arvode till privatperson"
ows_AZPersonnummer="196202081276"
ows_AZPlusgiro="5456436534"
ows_MetaInfo="1;#"
ows__ModerationStatus="0"
ows__Level="1" ows_ID="1"
ows_owshiddenversion="6"
ows_UniqueId="1;#{11E4AD4C-7931-46D8-80BB-7E482C605990}"
ows_FSObjType="1;#0"
ows_Created="2009-04-15T08:29:32Z"
ows_FileRef="1;#uppdragsavtal/Lists/Uppdragsavtal/1_.000"
/>
</rs:data>
</listitems>
And get value 196202081276.
Open this up in an XmlDocument object, then use the SelectNode function with the following XPath:
//*[local-name() = 'row']/#ows_AZPersonnummer
Basically, this looks for every element named "row", regardless of depth and namespace, and returns the ows_AZPersonnummer attribute of it. Should help avoid any namespace issues you might be having.
The XmlNamespaceManager is your friend:
string xml = "..."; //your xml here
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
XmlNamespaceManager nsm = new XmlNamespaceManager(new NameTable());
nsm.AddNamespace("z", "#RowsetSchema");
XmlNode n = doc.DocumentElement
.SelectSingleNode("//#ows_AZPersonnummer", nsm);
Console.WriteLine(n.Value);
You can also use LINQ to XML:
XDocument xd = XDocument.Parse(xml);
XNamespace xns = "#RowsetSchema";
string result1 = xd.Descendants(xns + "row")
.First()
.Attribute("ows_AZPersonnummer")
.Value;
// Or...
string result2 =
(from el in xd.Descendants(xns + "row")
select el).First().Attribute("ows_AZPersonnummer").Value;
I'd say you need an XML parser, which I believe are common. This looks like a simple XML structure, so the handling code shouldn't be too hard.
Use <%# Eval("path to attribute") %> but you need to load the xml has a DataSource.
Otherwise you can load it using XmlTextReader. Here's an example.