How to deserialize JSON array with "root" element for each object in array using Json.NET?

Question

I have following JSON string:

[
  { "Person" : { "Name" : "John", "Gender" : "male" } },
  { "Person" : { "Name" : "John", "Gender" : "male" } }
]

(As you may notice unfortunately I have a sort of "root" element for each object in the array. Without this "root" element the task becomes quite trivial.)

I have to deserialize it into a list of Person class:

class Person {
    public string Name { get; set; }
    public string Gender { get; set; }
}
...
List<Person> ListPersons() {
    return JsonConvert.DeserializeObject<List<Person>>(jsonString);
}

Is it possible to do with Json.NET without creating wrapper class like PersonResult?

class PersonResult {
    public Person Person { get; set; }
}
...
List<Person> ListPersons() {
    return JsonConvert.DeserializeObject<List<PersonResult>>(jsonString)
                      .Select(p => p.Person)
                      .ToList();
}

The perfect solution for me is to be able somehow explicitly specify this "root" (e.g. via attribute) and do not create any wrappers, helpers, etc.

k.m · Accepted Answer · 2012-01-06 09:36:23Z

6

Unfortunatelly, there's not much you can do about this issue. That's how JSON format looks like and there's no way around that. As a result, Json.Net "sees" your string as more or less:

an array of objects with Person property, which is another object with Name and Gender properties

You could possibly play some with custom ContractResolvers to force serializer to work differently... but that's quite a bit of work. Wrapper class like you suggested is how those problems are dealt with, and I suggest sticking to it.

answered Jan 6, 2012 at 9:36

k.m

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1 Comment

Anton Moiseev Over a year ago

Yeah, unfortunately I had to end up with wrapper approach. Thanks.

Ken Tseng · Accepted Answer · 2022-06-16 05:58:21Z

0

class PersonResult : List<Person> {}

class Person 
{
    public string Name { get; set; }
    public string Gender { get; set; }
}

main() {
    return JsonConvert.DeserializeObject<PersonResult>(jsonString);
}

answered Jun 16, 2022 at 5:58

Ken Tseng

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