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c#xml-deserializationxmltypexmlroot
Evgeniy
Evgeniy

Reputation: 555

xmlconvert string to date

I have model:

    [XmlRoot(ElementName = "event", IsNullable=true)]
public class Event
{
    public int id { get; set; }
    public string title { get; set; }
    public eventArtists artists { get; set; }
    public venue venue { get; set; }
    public string startDate { get;set;}
    public string description { get; set; }
    [XmlElement("image")]
    public List<string> image { get; set; }
    public int attendance { get; set; }
    public int reviews { get; set; }
    public string url { get; set; }
    public string website { get; set; }
    public string tickets { get; set; }
    public int cancelled { get; set; }
    [XmlArray(ElementName="tags")]
    [XmlArrayItem(ElementName="tag")]
    public List<string> tags { get; set; }
}

now I want to convert public string startDate { get;set;} to DatiTime:

public DateTime startDate { get{return startDate;} set{startDate. = DateTime.Parse(startDate);}}

How can I do that?

Upvotes: 0

Views: 931

Answers (1)

Thomas Levesque
Thomas Levesque

Reputation: 292765

You don't have anything special to do, just declare the property as a DateTime. The XmlSerializer will convert it automatically to a string like 2012-03-27T16:21:12.8135895+02:00

If you need to use a specific format, you have to use a small trick... Put a [XmlIgnore] attribute on the DateTime property, and add a new string property that handles the formatting:

[XmlIgnore]
public DateTime startDate { get;set;}

private const string DateTimeFormat = "ddd, dd MMM yyyy HH:mm:ss";

[XmlElement("startDate")]
[EditorBrowsable(EditorBrowsableState.Never)]
public string startDateXml
{
    get { return startDate.ToString(DateTimeFormat, CultureInfo.InvariantCulture); }
    set { startDate = DateTime.ParseExact(value, DateTimeFormat, CultureInfo.InvariantCulture); }
}

(the [EditorBrowsable] attribute is there to avoid showing the property in intellisense, since it's only useful for serialization)

Upvotes: 3

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