how Deserializing this type of xml file?

Question

<row>
    <id>1</id>
    <code></code>
    <name></name>
    <address></address>
    <state></state>
    <zone>?</zone>
</row>
<row>
    <id>2</id>
    <code>AA</code>
    <name>Ataria</name>
    <address>Sitapur National Highway 24, Uttar Pradesh</address>
    <state>Uttar Pradesh</state>
    <zone>NER</zone>
</row>

i have no root element in this xml file only row element start and end this xml file. how Deserializing this type of data ? in c#

Answer 1

If you sure that missing root is only the one issue with your XML - just add it manually:

string fileContent = File.ReadAllText(path);
string rawXml = "<root>" + fileContent + "</root>";

// now you can use LINQ-to-XML or whatever
XDocument xdoc = XDocument.Load(rawXml);

Answer 2

You can also load an XML Fragment directly, via

XmlReaderSettings settings = new XmlReaderSettings();
settings.ConformanceLevel = ConformanceLevel.Fragment;

using (XmlReader reader = XmlReader.Create("tracelog.xml", settings))
{
    while (reader.Read())
    {
        // Process each node of the fragment,
        // possibly using reader.ReadSubtree()
    }
}

You would create XElement s by passing the results of reader.ReadSubTree() to XElement.Load(...) .

Answer 3

Well to start with, it's not an XML file - or at least, it doesn't represent an XML document.

One option would be to copy the file into a new file which does have document start/end tags... then you can load it as a normal document. Just create a file, write a document start tag, copy the contents of this file, then write a document end tag, and close the file handle. You could even do this in memory.

Alternatively, it may be possible to read it as it is, in fragments - possibly via XmlReader . I can't say it's something I've done, and I'd generally encourage you to create a full XML file instead, as then you'll be on more familiar territory.

Answer 4

its not an XML file if it doesn't have the root. parser will throw an error if you try to parse it. you can do this way

<?xml version="1.0"?>
<Root>
--- add your file content here
</Root>  

then give this file path to the parser.