需要序列化為 JSON 字符串并將 JSON 字符串反序列化為對象。這里是我的課堂流程。[DataContract]public class BatchItem_NF : INF{ [DataMember] public IDebtor Debtor { get; set; } [DataMember] public IList<IDebtor> AllDebtors { get { return _debtors; } }}[DataContract]public class Debtor : IDebtor{ [DataMember] public IAddress Address { get; set; } [DataMember] public string Name { get; set; }}[DataContract]public class Address : IAddress{ [DataMember] public string Street { get; set; } [DataMember] public string City { get; set; }}這是我試圖序列化但不起作用的內容private string BatchItem_NFSerialization(INF Alldebtors){ Type t = Alldebtors.GetType(); extraTypes1 = t.GetProperties() .Where(p => p.PropertyType.IsInterface) .Select(p => p.GetValue(Alldebtors, null).GetType()) .ToArray(); using (MemoryStream ms = new MemoryStream()) { DataContractJsonSerializer serializer = new DataContractJsonSerializer(t, extraTypes1); serializer.WriteObject(ms, ((BatchItem_NF)Alldebtors)); retVal = Encoding.Default.GetString(ms.ToArray()); }}它拋出以下錯誤：輸入“地址”，數據合同名稱為“地址：http : //schemas.datacontract.org/2004/07/ ”是不期望的。如果您正在使用 DataContractSerializer 或將任何靜態未知的類型添加到已知類型列表中，請考慮使用 DataContractResolver - 例如，通過使用 knownTypeAttribute 屬性或將它們添加到傳遞給序列化程序的已知類型列表中
1 回答
白衣染霜花
TA貢獻1796條經驗 獲得超10個贊
您只能獲得第一級屬性的已知類型。在您的代碼中，您只能獲取Debtor類型，但您還應該處理此屬性以獲取Address類型，例如通過遞歸。
private static string BatchItem_NFSerialization(INF Alldebtors)
{
Type t = Alldebtors.GetType();
var extraTypes1 = GetKnownTypes(Alldebtors);
using (MemoryStream ms = new MemoryStream())
{
DataContractJsonSerializer serializer = new DataContractJsonSerializer(t, extraTypes1);
serializer.WriteObject(ms, ((BatchItem_NF)Alldebtors));
var retVal = Encoding.Default.GetString(ms.ToArray());
return retVal;
}
}
private static IEnumerable<Type> GetKnownTypes(object property)
{
Type t = property.GetType();
var extraValues = t.GetProperties()
.Where(p => p.PropertyType.IsInterface && !p.PropertyType.IsGenericType)
.Select(p => p.GetValue(property, null))
.ToArray();
var extraTypes = extraValues.SelectMany(GetKnownTypes).ToArray();
return extraValues.Select(v => v.GetType()).Concat(extraTypes).ToArray();
}
- 1 回答
- 0 關注
- 235 瀏覽
添加回答
舉報
0/150
提交
取消