Combine json objects into one?

[Merethiel]

The title might not correctly describe what I'm attempting but essentially I have one structure that I can convert to a json string but I want to append a list of objects also to be converted into json strings and send that all together as one packet.

Something like this.
`json Packet;

Packet["PacketType"] = config::packet::type::TYPE;

Packet["ArrayOfObjects"] = ArrayOfObjects;`

Or make two separate json objects and combine them.

Another question I have is, does an array of objects have to be set at initialisation?

[pboettch]

What have you tried? Could you show us some more elaborate code-examples of what you do and how you do it?

To push an object into a list you can do:

json a = json::array();
a.push_back({{"key", "value"}}); // two braces!

// or
json d;
d["key"] = value;
a.push_back(d);

[Merethiel]

Hi thank you for the response!
This is what I tried as an example:

    Packet["Id"] = 10;
    typedef struct Movement_{
        int XPosition;
        int YPosition;
    }Movement;
    std::set<Movement> v;

    for(int x = 0; x < 10; ++x)
    {
        Movement Mov = {0};
        Mov.XPosition = x;
        Mov.YPosition = x+1;
        v.insert(Mov);
    }
    Packet["MovementPackets"] = v;```



Essentially I wanted one "Id" while the rest was a set of objects following it.
It seems like the code tags are being buggy on here.

[pboettch]

Before proposing something, could you write (by hand if needed) the json you would expect? Where do you see the array, where the object(s)?

[Merethiel]

It was more of a question than a proposal if it already existed, but the JSON results I would have been expecting from the code above, to look something like this:

    /*
     {
     "Id" : 10,
     "Xposition[0]": 0,
     "YPosition[0]": 1,
     "XPosition[1]" : 1,
     "YPosition[1]" : 2,
     "XPosition[2]" : 2,
     "YPosition[2]" : 3,
     */

And so on until XPosition[9] and YPosition[9]

But I'm thinking a way I can do this wiht the existing code is simply appending the string value of [0],[1] as I loop through?

[pboettch]

I meant before I can propose something.

However, are you sure your JSON is efficient? Why not:

{"Id": 10,
 "Position": [ 
     { "x": 0, "y": 1 },
     { "x": 1, "y": 2 }
 ]
}

[Merethiel]

I think the way you wrote it is far more efficient and better looking than I how I wrote it!

[pboettch]

Hmm, what do you mean with unique keys? Here is how you would access this with nlohmann-json:

d["Id"] == 10

d["Position"][0]["x"] == 0

d["Position"][1]["y"] == 2

[Merethiel]

Would that need to be implemented or is there a workaround I can use for it? The other concern I have is, I'm sending this over through a Websocket to a javascript client and I'm not certain if the JSON implementation in Javascript would be compatible if the feature was implemented here.

[pboettch]

This is standard-compliant JSON:

{"Id": 10,
 "Position": [ 
     { "x": 0, "y": 1 },
     { "x": 1, "y": 2 }
 ]
}

[Merethiel]

Ah okay. Thank you so much! I'll see if I can find a work around until then.

[pboettch]

Workaround for what?

[Merethiel]

What you posted:

{"Id": 10,
 "Position": [ 
     { "x": 0, "y": 1 },
     { "x": 1, "y": 2 }
 ]
}

Oh I think I just realised what you did. You pushed back the previous json into the original. I get it now. I'm sorry for the confusion!

[pboettch]

Of course.

#include <nlohmann/json.hpp>

#include <iostream>

int main(void)
{
	nlohmann::json pos;

	pos["Id"] = 10;
	pos["Position"] = nlohmann::json::array();

	for (int i = 0; i < 10; i++)
		pos["Position"].push_back({{"x", i}, {"y", i + 1}});

	std::cout << std::setw(2) << pos << "\n";
	return 0;
}

Prints

{
  "Id": 10,
  "Position": [
    {
      "x": 0,
      "y": 1
    },
    {
      "x": 1,
      "y": 2
    },
    {
      "x": 2,
      "y": 3
    },
    {
      "x": 3,
      "y": 4
    },
    {
      "x": 4,
      "y": 5
    },
    {
      "x": 5,
      "y": 6
    },
    {
      "x": 6,
      "y": 7
    },
    {
      "x": 7,
      "y": 8
    },
    {
      "x": 8,
      "y": 9
    },
    {
      "x": 9,
      "y": 10
    }
  ]
}

[Merethiel]

Thank you! That's exactly what I was trying to do!

[pboettch]

You're welcome. Don't forget to close this issue if you have no further problems regarding your initial questions.