Can't deserialize valid JSON using system.json - json

Question

I write some code (rest server) that produce for me data in JSON format. When I use it in PHP it works fine, JSON is valid, everything is ok. When I use it in Delphi nothing works.
When I search internet I found:
desearilizing JSON using SuperObject
but that method returns empty strings for me.
I want to use that JSON elements as array (eg. JSONValue.items[i]).
I'm using Delphi XE7 System.JSON and don't want to use superobject or any others libraries.
How use it as array?
I paste my code that generates JSON:
var
qry: TfdQuery;
FieldsObj: TJSONObject;
FieldNameArray: TJSONArray;
I: Integer;
DataObj: TJSONObject;
DataRows: TJSONArray;
RowFields: TJSONArray;
tablename:string;
begin
tablename:='produkt';
qry := TfdQuery.Create(Self);
qry.SQL.Text := 'select * from produkt where (id ='''+ProductID+''')';
qry.Connection := FDConnection1;
qry.Open;
FieldsObj := TJSONObject.Create;
FieldNameArray := TJSONArray.Create;
for I := 0 to qry.FieldCount - 1 do
FieldNameArray.Add(qry.Fields[I].FieldName);
FieldsObj.AddPair(TableName, FieldNameArray);
DataObj := TJSONObject.Create;
DataRows := TJSONArray.Create;
qry.First;
while not qry.Eof do
begin
RowFields := TJSONArray.Create;
for I := 0 to qry.FieldCount - 1 do
RowFields.Add(qry.Fields[I].AsString);
DataRows.Add(RowFields);
qry.Next;
end;
DataObj.AddPair('data', DataRows);
Result := TJSONArray.Create(FieldsObj, DataObj);
qry.Free;
And this is the result:
{
"ProductID": "1",
"result": [{
"produkt": ["id", "parent_id", "full_name", "opcja_1", "opcja_2", "opcja_3", "opcja_4", "opcja_5", "opcja_6", "opcja_7", "opcja_8", "opcja_9", "opcja_10", "opcja_11", "opcja_12", "field_address1", "field_address2", "quantity", "opis", "zdjecie1", "zdjecie2", "zdjecie3", "samples", "link_stable0", "link_stable1", "link_stable2", "price1", "price2", "price3"]
}, {
"data": [
["1", "1", "name", "1", "1", "1", "1", "0", "0", "0", "0", "0", "0", "0", "12", "10", "20", "1,2", "description of product", "http://www.vphosted.com/e6=0", "photo link2", "photo link 3", "sample project file link", "link option", "10", "link", "10", "link", "10"]
]
}]
}

Answer 1

This would produce JSON more in the format that I would expect:
var
qry: TfdQuery;
FieldsObj: TJSONObject;
//FieldNameArray: TJSONArray;
I: Integer;
DataObj: TJSONObject;
FieldObj: TJSONObject;
DataRows: TJSONArray;
RowFields: TJSONArray;
tablename:string;
begin
tablename:='produkt';
qry := TfdQuery.Create(Self);
qry.SQL.Text := 'select * from produkt where (id ='''+ProductID+''')';
qry.Connection := FDConnection1;
qry.Open;
FieldsObj := TJSONObject.Create;
//FieldNameArray := TJSONArray.Create;
//for I := 0 to qry.FieldCount - 1 do
// FieldNameArray.Add(qry.Fields[I].FieldName);
//FieldsObj.AddPair(TableName, FieldNameArray);
DataObj := TJSONObject.Create;
DataRows := TJSONArray.Create;
qry.First;
while not qry.Eof do
begin
RowFields := TJSONArray.Create;
for I := 0 to qry.FieldCount - 1 do
begin
FieldObj := TJSONObject.Create;
FieldObject.AddPair(qry.Fields[I].FieldName, qry.Fields[I].AsString));
RowFields.Add( FieldObj );
end;
DataRows.Add(RowFields);
qry.Next;
end;
DataObj.AddPair('data', DataRows);
Result := TJSONArray.Create(FieldsObj, DataObj);
qry.Free;
If you know the record structure, though, I would prefer to use REST.JSON, which I am pretty sure ships with XE7 and is much simpler to use. You just create your object structure, create an instance of that structure, populate it with your field values and use
TJSON.ObjectToJsonString( fObject )
to create your string and
iObject := TJSON.JsonToObject<TMyObject>( pTransferString );
to get your object back.
If you want a more complete example using this method, let me know and I will post.