Question

<results>
    <result>
        <egov_ref_no>20121203001</egov_ref_no>
        <status>OK</status>
        <err_code>01</err_code>
    </result>
    <result>
        <egov_ref_no>20121203002</egov_ref_no>
        <status>OK</status>
        <err_code>02</err_code>
    </result>
    <result>
        <egov_ref_no>20121203003</egov_ref_no>
        <status>OK</status>
        <err_code>03</err_code>
    </result>
</results> 

The code above shows that the root node is already an IEnumerable where as in online examples they are only elements.

Answer 1

If you have a results object you can just pass in a XmlRootAttribute into the XmlSerializer constructor. In this case its “results”

Example:

List<Result> results = new List<Result>();

XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<Result>), new XmlRootAttribute("results"));

using (FileStream stream = new FileStream(@"C:\Test.xml", FileMode.Open))
{
    results = (List<Result>)xmlSerializer.Deserialize(stream);
}

My result object:

[XmlType(TypeName = "result")]
public class Result
{
    [XmlElement(ElementName = "egov_ref_no")]
    public long EgovRefNo { get; set; }

    [XmlElement(ElementName = "status")]
    public string Status { get; set; }

    [XmlElement(ElementName = "err_code")]
    public int ErrorCode { get; set; }
}

This will return as List<Result>