C# JSON list of list omittings tags

Question

I would like to create a JSON proto3 compliant string including a section like this:

"foo": [[1.0, 2.0], [3.0, 4.0], [5.0, 6.0]]

I can't figure out how to do it. This seems to be a List<List<T>> problem (that has been discussed) but when I feed such a solution to Newtonsoft Json.NET I get:

"foo": [ "bar": [1.0, 2.0], "bar": [3.0, 4.0], "bar": [5.0, 6.0]]

I tried only "Newtonsoft Json.NET" but the question is open to any other JSON libraries or custom coding methods. How can this be done?

EDIT: This is a serialization problem. I have no control over the deserializer code.

J. Tom · Accepted Answer · 2019-02-15 00:01:56Z

2

I'm not sure where you're going wrong but the following code works perfectly fine:

var lists = new List<List<double>>
{
    new List<double> {1.0, 2.0},
    new List<double> {3.0, 4.0}
};
var test = (lists, "test");
Console.Write(JsonConvert.SerializeObject(test));

Output:

{"Item1":[[1.0,2.0],[3.0,4.0]],"Item2":"test"}

Could you edit your post with some sample code?

answered Feb 15, 2019 at 0:01

J. Tom

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1 Comment

Manngo Over a year ago

Actually this looks very good. Give me the rest of Valentine's Day and I will confirm ASAP.

bdn02 · Accepted Answer · 2019-02-14 23:44:23Z

1

Try this:

string str = "{ \"foo\": [[1.0, 2.0], [3.0, 4.0], [5.0, 6.0]] }";
var mytype = new { foo = new List<List<decimal>>() };
var mydeserializedtype = JsonConvert.DeserializeAnonymousType(str, mytype);
string str2 = JsonConvert.SerializeObject(mydeserializedtype);

or more classic

class myclass
{
 public List<List<decimal>> foo { get; set; }
}
….
string str = "{ \"foo\": [[1.0, 2.0], [3.0, 4.0], [5.0, 6.0]] }";
myclass my = JsonConvert.DeserializeObject<myclass>(str);
string str2 = JsonConvert.SerializeObject(my);

answered Feb 14, 2019 at 23:44

bdn02

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2 Comments

Manngo Over a year ago

Ehm, actually I would like to serialize and not to deserialize. I thought I made it clear, nevertheless I will edit.

xdtTransform Over a year ago

deserializeis the exact opoposite of deserialize. It means that if you find the class that will result the deserialize, you know what object to instancite in order to produce the right serialize. This answer show you this concept by deserialize str into my then deserialize my into str2. Poor choice or var name but still a clear path to the solution. The only answer of how to serialize that is to learn how to deserialize, this make all question about serialisation duplicate of how to deserialize in my eyes.

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