ryu*_*ice 21 .net c# xml-serialization
我收到此消息,"字符串'7/22/2006 12:00:00 AM'不是有效的AllXsd值.",在反序列化XML时,该元素包含一个日期,这是所谓的属性要映射到元素:
[XmlElement("FEC_INICIO_REL",typeof(DateTime))]
public DateTime? FechaInicioRelacion { get; set; }
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难道我做错了什么?
更新:这是XML:
<Detalle>
<NOM_ASOC>Financiera Panameña</NOM_ASOC>
<DESCR_CORTA_RELA>PREST. PERSONAL</DESCR_CORTA_RELA>
<FEC_INICIO_REL>7/22/2006 12:00:00 AM</FEC_INICIO_REL>
<FEC_FIN_REL>9/22/2008 12:00:00 AM</FEC_FIN_REL>
<MONTO_ORIGINAL>1160.0000</MONTO_ORIGINAL>
<NUM_PAGOS>2</NUM_PAGOS>
<DESCR_FORMA_PAGO>PAGOS VOLUNTARIOS</DESCR_FORMA_PAGO>
<IMPORTE_PAGO>59.9400</IMPORTE_PAGO>
<FEC_ULTIMO_PAGO>11/15/2006 12:00:00 AM</FEC_ULTIMO_PAGO>
<MONTO_ULTIMO_PAGO>0.0000</MONTO_ULTIMO_PAGO>
<DESCR_OBS_CORTA />
<SALDO_ACTUAL>1078.3900</SALDO_ACTUAL>
<NUM_DIAS_ATRASO>0</NUM_DIAS_ATRASO>
<HISTORIA>1</HISTORIA>
<MONTO_CODIFICADO />
<FEC_ACTUALIZACION>10/17/2008 12:00:00 AM</FEC_ACTUALIZACION>
<COD_GRUPO_ECON> </COD_GRUPO_ECON>
<TIPO_ASOC> </TIPO_ASOC>
<NUM_REFER>2008628116</NUM_REFER>
</Detalle>
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ryu*_*ice 20
我通过将日期存储在字符串中然后创建一个解析日期并将其作为DateTime返回的getter解决了这个问题.
示例代码:
[XmlElement("Valid")]
public string _Valid
{
get;
set;
}
[XmlIgnore]
public bool? Valid
{
get
{
if (!string.IsNullOrWhiteSpace(_Valid))
{
return bool.Parse(_Valid);
}
return null;
}
}
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小智 12
AllocationDate是必填字段,但可以作为空白提供,通过AllocationDateString表示它来处理:
private DateTime? _allocationDate;
[XmlIgnore]
public DateTime? AllocationDate
{
get { return _allocationDate; }
set { _allocationDate = value; }
}
[XmlAttribute("AllocationDateTime")]
public string AllocationDateTimeString
{
get
{
return _allocationDate.HasValue ? XmlConvert.ToString(_allocationDate.Value, XmlDateTimeSerializationMode.Unspecified)
: string.Empty;
}
set
{
_allocationDate = !string.IsNullOrEmpty(value) ? XmlConvert.ToDateTime(value, XmlDateTimeSerializationMode.Unspecified) : (DateTime?)null;
}
}
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