Constructing simple XML?

[Brett Romero]

What is the best way to construct this type of XML

<ROOT><Person PersonId=\"1000\" /><Person PersonId=\"1001\" /><Person
PersonId=\"1002\" /></ROOT>

when I'm getting back only the three numerical values? The is occuring
in a loop and I'm getting a variable such as PersonID, which holds the
values. The XML needs to be a string but I'd like to see what a more
complex type would look like (XMLdoc type?).

Thanks,
Brett

 

[Brett Romero]

This pretty much covers the creation part:
http://www.csharphelp.com/archives/archive199.html. Still not sure how
to get the two backslashes in there.

Brett

 

[Kai Brinkmann [MSFT]]

Brett, you create attribute nodes just like element nodes. Here's the code
from the link you posted, modified to create the XML in your original post:

XmlDocument xmldoc;
XmlAttribute attr;
XmlElement elem;
XmlElement root;

xmldoc=new XmlDocument();

//let's add the XML declaration section
XmlNode xmlnode=xmldoc.CreateNode(XmlNodeType.XmlDeclaration,"","");
xmldoc.AppendChild(xmlnode);

//let's add the root element
elem=xmldoc.CreateElement("ROOT");
xmldoc.AppendChild(elem);
root = elem;

//add person elements. assume you have some data structure called ListOfIds
containing the numbers
foreach (int personId in ListOfIds)
{
elem = xmldoc.CreateElement("Person");
attr = xmldoc.CreateAttribute("PersonId");
attr.InnerText = personId.ToString();
elem.Attributes.Append(attr)
root.AppendChild(elem);
}

--
Kai Brinkmann [MSFT]

Please do not send e-mail directly to this alias. This alias is for
newsgroup purposes only.
This posting is provided "AS IS" with no warranties, and confers no rights.

 

[Brett Romero]

Thanks.

If I try to read my file, which looks like this:
<?xml version="1.0"?>
<ROOT>
<Person PersonID="83615">
</Person>
</ROOT>

I get this error:
"The data at the root level is invalid. Line 1, position 1."

I'm using:
xmlDoc.LoadXml(@"C:\Documents and Settings\BROMERO\My
Documents\testing.spec");

I'm not sure if it is referring to the LoadXML method or the file. Any
ideas?

The <?xml version="1.0"?> part was generated with the XMLDeclaration
enumerator.

Thanks,
Brett

 

[Guest]

Hi Brett,
the LoadXML method assumes the string you are passing as a parameter is
XML, not a file path. If you want to load XML from a file use the
XmlDocument.Load method and pass in the file name.

Hope that helps
Mark Dawson
http://www.markdawson.org

 

[Anders Borum]

For really quick XML construction, I suggest looking at the XmlWriter class
(and its descending implementations).