Serializing a DataType="time" field using XmlSerializer
I'm getting an odd result when serializing a DateTime field using XmlSerializer.
I have the following class:
public class RecordExample
{
[XmlElement("TheTime", DataType = "time")]
public DateTime TheTime { get; set; }
[XmlElement("TheDate", DataType = "date")]
public DateTime TheDate { get; set; }
public static bool Serialize(
Stream stream, object obj, Type objType, Encoding encoding)
{
try
{
var settings = new XmlWriterSettings { Encoding = encoding };
using (var writer = XmlWriter.Create(stream, settings))
{
var xmlSerializer = new XmlSerializer(objType);
if (writer != null) xmlSerializer.Serialize(writer, obj);
}
return true;
}
catch (Exception)
{
return false;
}
}
}
When i call the use the XmlSerializer with the following testing code:
var obj = new RecordExample
{
TheDate = DateTime.Now.Date,
TheTime = new DateTime(0001, 1, 1, 12, 00, 00)
};
var ms = new MemoryStream();
RecordExample.Serialize(ms, obj, typeof(RecordExample), Encoding.UTF8);
txtSource2.Text = Encoding.UTF8.GetString(ms.ToArray());
I get some strange results, here's the xml that is produced:
<?xml version="1.0" encoding="utf-8"?>
<RecordExample
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://w开发者_StackOverflow中文版ww.w3.org/2001/XMLSchema">
<TheTime>12:00:00.0000000+00:00</TheTime>
<TheDate>2010-03-08</TheDate>
</RecordExample>
Any idea's how i can get the "TheTime" element to contain a time which looks more like this:
<TheTime>12:00:00.0Z</TheTime>
...as that's what i was expecting?
Thanks
Dave
I've had different issues with this myself... however I was attempting to serialize a TimeSpan object. The solution was to have two properties, one that held the TimeSpan, and one that was a string representation of the TimeSpan which got Serialized. Here was the pattern:
[XmlIgnore]
public TimeSpan ScheduledTime
{
get;
set;
}
[XmlElement("ScheduledTime", DataType="duration")]
public string XmlScheduledTime
{
get { return XmlConvert.ToString(ScheduledTime); }
set { ScheduledTime = XmlConvert.ToTimeSpan(value); }
}
However, with this code, the time is printed out in the following format:
<ScheduledTime>PT23H30M</ScheduledTime>
The W3C definition of duration is here which explains it.
take a look at this question Serializing DateTime to time without milliseconds and gmt
Expanding on the comment I made on one of the others answers.
public class RecordExample : IXmlSerializable
{
public DateTime TheTime { get; set; }
public DateTime TheDate { get; set; }
public XmlSchema GetSchema()
{
return null;
}
public void ReadXml(XmlReader reader)
{
// TODO : Deserialization logic here
}
public void WriteXml(XmlWriter writer)
{
writer.WriteElementString(
"date",
this.TheDate.ToString("yyyy-MM-dd"));
writer.WriteElementString(
"time",
this.TheTime.ToString("hh:mm:ss.fK"));
}
}
Serializing like this:
var rc = new RecordExample()
{
TheDate = DateTime.Today,
TheTime = DateTime.UtcNow
};
var serializer = new XmlSerializer(typeof(RecordExample));
var ms = new MemoryStream();
serializer.Serialize(ms, rc);
ms.Seek(0, SeekOrigin.Begin);
Console.WriteLine(new StreamReader(ms).ReadToEnd());
Output example:
<?xml version="1.0"?>
<RecordExample>
<date>2010-03-08</date>
<time>04:26:16.1Z</time>
</RecordExample>
I concur with the other answers (I was not done writing when they popped up). it does not look like it is possible, in a direct way. A look at the source with Reflector shows that a time value ends up being converted to a string with the System.Xml.XmlConvert.ToString, that has a hard-coded format of:
HH:mm:ss.fffffffzzzzzz
So having two properties, the real one being [XmlIgnore] and a string that you build yourself is a good way to go.
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