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copied!<p>xml-rpc.net certainly takes a System.Datetime and converts to dateTime.iso8601. A gotcha I had was with the various interpretations of dateTime.iso8601. From your xml template, your datetime output is supported (20101117T09:42:00) and you won't have the same problem. I had a very strict upstream server that wanted some offset (e.g +001) and I had to extend and pass a datetime format to xml-rpc.net.</p> <p><strong>Edit</strong></p> <p>Yea I modified the code to achieve the new DateTimeFormat. I downloaded the source and made the following modifications (I wish it was better hosted and one could offer a modification).</p> <p>There is an XmlRpcSerializer.cs, which is where the serialization of the type to xml-rpc format happens. The method you need is naturally...</p> <pre><code>void Serialize( XmlWriter xtw, Object o, MappingActions mappingActions, List<object> nestedObjs) </code></pre> <p>Somewhere in this method, is a check for XmlRpcType.tDateTime:</p> <pre><code>else if (xType == XmlRpcType.tDateTime) { DateTime dt = (DateTime)o; //The following line is what you need, if a custom DateTime format was supplied, to override the default string sdt = dt.ToString((string.IsNullOrEmpty(DateTimeFormat) ? "yyyyMMdd'T'HH':'mm':'ss" : DateTimeFormat), DateTimeFormatInfo.InvariantInfo); WriteFullElementString(xtw, "dateTime.iso8601", sdt); } </code></pre> <p>The rest of the story is how to pass the custom datetime format to the XmlRpcSerializer class. You could just add a property to this class to make it happen. There is an XmlRpcFormatSettings that could be used but that's up to you.</p>

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