TJSON.JsonToObject not working in Delphi

Question

I am trying to convert Json trying to class object, but the values are not appearing in the new object variable. The resulting object has blank value in string and 0 in integer. Thanks in advance.

Code:

type
Student =class
    public
    Name : string;
    Age : Integer;
end;

procedure TForm2.Button5Click(Sender: TObject);
var
  Student1, Student2: Student;
  STR: string;
begin
  Student1 := Student.Create;
  Student2 := Student.Create;

  try
  Student1.Name := 'Sam';
  Student1.Age := 24;
  str := TJson.ObjectToJsonString(Student1);
  Form2.outputMemo.Lines.Add(str);
  Student2 := TJSON.JsonToObject<Student>(str);
  Form2.outputMemo.Lines.Add(Student2.Name);
  Form2.outputMemo.Lines.Add(Student2.Age.ToString);
  finally
  Student1.Free;
  Student2.Free;
  end;
//Form2.outputMemo.Lines.Text :=TJson.ObjectToJsonObject(Student1);

end;

Output:

{"name":"Sam","age":24}

0

Edit: I just saw this, and it worked when I changed the names to FName and FAge... what a sorcery!, can anyone please explain the logic behind this? Delphi Rest.JSON JsonToObject only works with f variables

Why are you using classes rather than records for your data persistence? — David Heffernan, Mar 27 '22 at 08:44
@DavidHeffernan, perhaps because there is no built-in conversion for records as there is with TJSON for classes. — Uwe Raabe, Mar 27 '22 at 08:54
BTW, there is a memory leak from `Student2 := Student.Create`, as `Student2` is assigned later with `TJSON.JsonToObject`, leaving the former instance in memory. — Uwe Raabe, Mar 27 '22 at 08:57
@DavidHeffernan, I am new to Delphi, in C# we usually use data classes, if there is any other way to do it efficiently, please suggest the same. But I do need to convert the objects to JSON and vice versa here. — S. B., Mar 27 '22 at 09:14
@UweRaabe, what can be done to avoid this? Shall I use `Student1.Free` before assigning Student2? As I said I am very new to Delphi, please suggest if I am doing it wrong. Also this is just a practice example to understand. In my real code I will be calling `Post().ContentAsString()` and convert that to class object. — S. B., Mar 27 '22 at 09:17
There are lots of ways to work with json. But you look like you have data only structures here and surely they are a better fit for records. But we don't have much context to advise you. — David Heffernan, Mar 27 '22 at 09:44
Don't create `Student2` in the first place. It would be sufficient to initialize it with `nil`. — Uwe Raabe, Mar 27 '22 at 13:24
@UweRaabe, Okay thanks! so removing the line `Student2 := Student.Create;` will be enough right? or do I need to specify `Student2 := nil;`? — S. B., Mar 27 '22 at 13:28
As long as `Student2` is a local variable of the current method an initialization with `nil` is recommended. Otherwise the content is arbitrary and in case of an error before the assignment with `TJSON.JsonToObject` (which can happen during the conversion), the call to `Student2.Free` may use uninitialized memory. — Uwe Raabe, Mar 27 '22 at 13:33
@DavidHeffernan, in theory I would agree with you, but while records may serve well for flat structures like shown in the question, JSON is usually made of hierarchies of objects (the "O" in JSON). I have found that it is easier to use classes in all cases than to have a different record approach in only some cases where it may accidentally fit. It happens all too often that a flat record-like structure soon evolves to a complex hierarchy. Starting with classes right away significantly reduces the stress when this happens. — Uwe Raabe, Mar 27 '22 at 13:43
@uwe json has scalars, arrays and objects. Arrays can map to dynamic arrays and objects to records. — David Heffernan, Mar 27 '22 at 15:35

score 4 · Accepted Answer · answered Mar 26 '22 at 12:11

4

The internal mapping of JSON fields to Delphi fields is prefixing them with F and changing the following character to upper case. If you want complete control over this you can specify the JSON name with an attribute:

type
  Student =class
  public
    [JSONName('name')]
    Name : string;
    [JSONName('age')] 
    Age : Integer;
  end;

Note that the JSON names given are case sensitive.

You need to include REST.Json.Types to the uses so that the attribute declaration can be found.

answered Mar 26 '22 at 12:11

Uwe Raabe

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Thanks for your answer, but currently I'll keep them as FFieldName, I have huge class, it will be very time consuming. – S. B. Mar 27 '22 at 09:19
1

If you aren't yet, definitely try out MMX (https://www.mmx-delphi.de/). It makes adding properties with private fields very easy and fast – RaelB Mar 27 '22 at 18:45

TJSON.JsonToObject not working in Delphi

1 Answers1