I have an XML file with some data on it..for example
<?xml version="1.0" encoding="utf-8"?>
<CreateAndSendMessageRequest xmlns:xsi="http://www.w3.org/2001/XMLSchema-
instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns="http://schemas.communisis.com/lv">
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>250028766505DN</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
Now i want to replace POLICY REFERENCE value with some dummy data. I am able to do that but now i have to save it as a new file with a different file name..
How can i achieve that..
For reference i am giving my code.
XmlDocument doc = new XmlDocument();
doc.Load(filepath);
XmlNodeList node = doc.GetElementsByTagName("PolicyReference");
var item =node.Item(0);
string value = item.FirstChild.Value;
string nevalue = value.Replace(value, "Test123");
doc.DocumentElement.;
doc.Save(#"C:\Test\file.xml");
The xml wasn't valid.
Anyway, this is how to make it work:
Xml file:
<?xml version="1.0" encoding="UTF-8"?>
<CreateAndSendMessageRequest>
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>250028766505DN</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
</CompositionRequest>
</CreateAndSendMessageRequest>
Code:
XmlDocument doc = new XmlDocument();
doc.Load("bho.xml");
XmlNodeList node = doc.GetElementsByTagName("PolicyReference");
node[0].InnerText = "Test123";
doc.Save("file.xml");
Result xml:
<?xml version="1.0" encoding="UTF-8"?>
<CreateAndSendMessageRequest>
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>Test123</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
</CompositionRequest>
</CreateAndSendMessageRequest>
Hope it helps you ^^
Related
I build an XML Document which needs to be validated against a xsd file. Thus I need a reference to the xsd file in the root element of the xml. So far I use this C# Code:
var ser = new XmlSerializer(typeof(myspecialtype));
XmlSerializerNamespaces MainNamespace = new XmlSerializerNamespaces();
MainNamespace.Add("xlink", "http://www.w3.org/1999/xlink");
MainNamespace.Add("xsi", "http://www.w3.org/2001/XMLSchema-instance");
using (XmlWriter w = XmlWriter.Create(#"C:\myxmlfile.xml"))
{
w.WriteProcessingInstruction("xml-stylesheet", "type=\"text/xsl\" href=\"utils/somexsl.xsl\"");
ser.Serialize(w, LeBigObject, HauptNs);
}
The resulting Xml begins like this:
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="utils/somexsl.xsl"?>
<caddy-xml xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlVersion="03.07.00">
but I need this:
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="utils/somexsl.xsl"?>
<caddy-xml xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlVersion="03.07.00" xsi:noNamespaceSchemaLocation="utils/theveryimportant.xsd">
I came across "CreateAttribute" here: Add Namespace to an xml root node c# but I can't put it together with the Serializer. Thank you!
I was pointed to the solution here:
https://social.msdn.microsoft.com/Forums/en-US/e43585c6-181b-4449-8806-b07f82681a2a/how-to-include-xsinonamespaceschemalocation-in-the-xml?forum=asmxandxml
I added this to my class:
[XmlAttribute("noNamespaceSchemaLocation", Namespace = XmlSchema.InstanceNamespace)]
public string attr = "utils/theveryimportant.xsd";
and it works.
Example
<?xml version="1.0" encoding="UTF-8"?>
<Settings>
<Tag1>XXXX</Tag1>
<Tag2>YYYY</Tag2>
<Tag3>true</Tag3>
<Tag4>ZZZZ</Tag4>
</Settings>
I want to edit only the contents of Tag3 without having to create another .xml file
You may edit your XML file like this:
XmlDocument doc = new XmlDocument();
doc.Load("D:\\somefile.xml");
XmlNode root = doc.DocumentElement;
XmlNode myNode = root.SelectSingleNode("Settings::Tag3");
myNode.Value = "blabla";
doc.Save("D:\\somefile.xml");
The code below works in C#, no error, no exception. The problem is that the numconfig.xml file won't change after running the code.
C# code:
XmlNodeList xm = new XmlManager(System.Web.HttpContext.Current.Server.MapPath("~/Xml/numconfig.xml")).ReadAllChild(#"//number");
xm[0].SelectSingleNode("abc[#name='upper']").Attributes["value"].Value = "201";
Xml file:
<?xml version="1.0" encoding="utf-8" ?>
<number>
<aaa>
<abc value="200" text="xxxx" name="upper"/>
</aaa>
</number>
How are you trying to commit the changes back to the file? Your code does not show an example eg
using (var streamWriter = new StreamWriter(location))
{
foreach (XmlNode xmlNode in xm)
{
streamWriter.WriteLine(xmlNode.OuterXml);
}
}
The Application serializes a List to xml:
<?xml version="1.0"?>
<Tools xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Tool>
<Name>test1</Name>
<Path>C:\Program Files\FreePDF_XP\fpmailer.exe</Path>
</Tool>
<Tool>
<Name>test2</Name>
<Path>C:\Program Files\FreePDF_XP\fpassist.exe</Path>
</Tool>
<Tool>
<Name>test3</Name>
<Path>C:\Program Files\FreePDF_XP\ShellMail.exe</Path>
</Tool>
</Tools>
If I'm don't on the wrong way I must create new Objects and add them to a list:
Tool tool = new Tool();
XmlSerializer deserializer = new XmlSerializer(typeof(List<Tool>));
using (var reader = new StreamReader(#Start.userConfigurePath + "\\config.xml"))
{
tool = (Tool)deserializer.Deserialize(reader);
reader.Close();
}
toolList.Add(tool);
}
In the result there isn't a Object in the List. How can I deserialize the serialized objects in the xml in a List? Could it be that there is something wrong at the code for derisalization?
Edit:
Something seems to be wrong with my xml declaration (XML File Error 2,2). That I don't understand because i create the xml document on this way:
XmlDocument toolConfig = new XmlDocument();
XmlNode myRoot;
myRoot = toolConfig.CreateElement("Tool");
toolConfig.AppendChild(myRoot);
It should be
toolList = (List<Tool>)deserializer.Deserialize(reader);
I have a xml file, that looks something like this:
<?xml version="1.0" encoding="utf-8"?>
<Settings>
<WhereDoCopy Path="C:\Users\USER\Desktop\jobs" />
</Settings>
What can I do, to add an Element to that file?
XDocument doc = XDocument.Load("your.xml");
XElement item = new XElement(WhereDoCopy ,
new XAttribute("Path", "C:\Users\USER\Desktop\jobs2"));
doc.Root.Add(item);
http://msdn.microsoft.com/en-us/library/system.xml.linq.xdocument.aspx