在我來說,我只是想序列_recordsets所以我的解決辦法是:
1)獲取所有_Recordset類型字段:
function Test.GetRecordsetFieldsFromObject(
AObject: TObject): TStringList;
var
Obj: TRttiType;
Rtti: TRTTIContext;
ObjField: TRttiField;
IntfObj: IInterface;
rsOut: _recordset;
begin
Result := TStringList.Create;
Obj := Rtti.GetType(AObject.ClassType);
for ObjField in Obj.GetFields do
if ObjField.FieldType.TypeKind = tkInterface then
begin
IntfObj := ObjField.GetValue(AObject).AsInterface;
if IntfObj.QueryInterface(_Recordset, rsOut) = 0 then
begin
Result.Add(ObjField.Name);
rsOut := nil;
end;
end;
end;
2)註冊轉換器和返還的每個字段創建
for FieldName in FieldNameList do
begin
JsonMarshal.RegisterConverter(TFoo, FieldName, function(Data: TObject; Field: String): TListOfStrings
var
Obj: TRttiType;
ObjField: TRttiField;
rsProp: _Recordset;
strStream: TStringStream;
begin
SetLength(Result, 1);
strStream := TStringStream.Create;
try
Obj := Rtti.GetType(data.ClassType);
ObjField := Obj.GetField(Field);
rsProp := ObjField.GetValue(Data).AsInterface as _Recordset;
rsProp.Save(TStreamAdapter.Create(strStream) as IUnknown, adPersistXML);
Result[0] := strStream.DataString;
finally
rsProp := nil;
strStream.Free;
end;
end);
JsonUnMarshal.RegisterReverter(TFoo, FieldName, procedure(Data: TObject; Field: String; Args: TListOfStrings)
var
Obj: TRttiType;
ObjField: TRttiField;
rsProp: _Recordset;
strStream: TStringStream;
begin
rsProp := coRecordset.Create;
strStream := TStringStream.Create(Args[0]);
try
Obj := Rtti.GetType(data.ClassType);
ObjField := Obj.GetField(Field);
strStream.Position := 0;
rsProp.Open(TStreamAdapter.Create(strStream) as IUnknown, EmptyParam, adOpenUnspecified, adLockUnspecified, 0);
ObjField.SetValue(Data, TValue.From<_Recordset>(rsProp));
finally
rsProp := nil;
strStream.Free;
end;
end);
end;
因爲沒有實現,所以接口不能很容易地序列化,它們只是接口。您如何期望序列化工作?您可能需要獲得實現來支持提供序列化服務的另一個接口。 –
是的你是對的!但我有一個使用rtti將serialize對象轉換爲XML的實現,當出現「_recordset」時,我將它保存到一個流(TStringStream)中,這就是我如何獲得它的「序列化」。我想也許我可以用TJSONMarshal做那樣的事情,但「TJSONMarshal.RegisterConverter」只接受TClass –