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Home/ Questions/Q 4254206
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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T05:03:06+00:00 2026-05-21T05:03:06+00:00

Suppose, I have a simple type: public class Report { public Report() { BirthDate

  • 0

Suppose, I have a simple type:

public class Report
{
    public Report()
    {
        BirthDate = new Element();
        BirthPlace = new Element();
    }

    public Element BirthDate { get; set; }
    public Element BirthPlace { get; set; }
}

public class Element
{
    [XmlAttribute("published")]
    public bool Published { get; set; }
    [XmlText]
    public string Value { get; set; }
}

I defined simple extension method for serialization purposes:

public static class TheHelper
{
    public static string Serialize<T>(this T source, Encoding encoding)
    {
        MemoryStream memoryStream = new MemoryStream();
        XmlSerializer xs = new XmlSerializer(typeof(T));
        XmlTextWriter xmlTextWriter = new XmlTextWriter(memoryStream, encoding);

        xs.Serialize(xmlTextWriter, source);
        memoryStream = (MemoryStream)xmlTextWriter.BaseStream;
        return encoding.GetString(memoryStream.ToArray());
    }
}

When sample Report object is created, it is next serialized to xml format:

Report r = new Report();
r.BirthDate.Published = true;
r.BirthDate.Value = DateTime.Now.AddYears(-1000).ToString("yyyy-MM-dd");
r.BirthPlace.Published = false;
r.BirthPlace.Value = "K-PAX";

string xml = r.Serialize(Encoding.UTF8);

Output document shown below is created:

<?xml version="1.0" encoding="utf-8"?>
<Report xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <BirthDate published="true">1011-04-07</BirthDate>
    <BirthPlace published="false">K-PAX</BirthPlace>
</Report>

But I would like to add language identification using special attribute named xml:lang:

<?xml version="1.0" encoding="utf-8"?>
<Report xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <BirthDate published="true" xml:lang="en-GB">1011-04-07</BirthDate>
    <BirthDate published="true" xml:lang="kp-AX">07.04.1011</BirthDate>
    <BirthPlace published="false" xml:lang="en-GB">K-PAX</BirthPlace>
    <BirthPlace published="false" xml:lang="kp-AX">k_p4x</BirthPlace>
</Report>

What can be the smart way to achieve this ? I have resources defined for en-GB and kp-AX.. languages. How to modify and create Report object to have multiple tags with different xml-lang attributes to be serializable using XmlSerializer ?

Regards.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-21T05:03:07+00:00Added an answer on May 21, 2026 at 5:03 am

    This should work:

    public class Element
    {
        [XmlAttribute("published")]
        public bool Published { get; set; }
        [XmlAttribute("xml:lang", DataType = "language")]
        public string Language { get; set; }
        [XmlText]
        public string Value { get; set; }
    }
    

    If you wanted to use the CultureInfo class, you could create a property and use the XmlIgnore attribute on the CultureInfo property and have an addition string property that converts it, like in this LocalizableString example:

    /// <summary>
    /// The language of the <see cref="Value"/>
    /// </summary>
    [XmlIgnore]
    public CultureInfo Language { get; set; }
    
    /// <summary>Used for XML serialization.</summary>
    /// <seealso cref="Language"/>
    [XmlAttribute("xml:lang", DataType = "language")]
    public string LanguageString
    {
        get { return (Language == null || string.IsNullOrEmpty(Language.ToString())) ?
                 null : Language.ToString(); }
        set { Language = string.IsNullOrEmpty(value) ?
                 CultureInfo.InvariantCulture : new CultureInfo(value); }
    }
    
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