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creating class from JSON with different types of a node


By : Kishore Brahma
Date : September 29 2020, 09:00 AM
With these it helps You need a custom converter.
You are trying to convert a complex type, as a result Newtonsoft doesn't know what to do. The fix is rather trivial, and is documented here: Using a Custom JsonConverter to fix bad JSON results.
code :
class SingleValueArrayConverter<T> : JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        object retVal = new Object();
        if (reader.TokenType == JsonToken.StartObject)
        {
            T instance = (T)serializer.Deserialize(reader, typeof(T));
            retVal = new [] { instance };
        }
        else if (reader.TokenType == JsonToken.StartArray)
        {
            retVal = serializer.Deserialize(reader, objectType);
        }
        return retVal;
    }

    public override bool CanConvert(Type objectType) => objectType == typeof(T);
}
class SamPointOfContact
{
    public string firstName { get; set; }
    public string lastName { get; set; }
    public string title { get; set; }
}

class Naics
{
    public string naicsName { get; set; }
    public string hasSizeChanged { get; set; }
    public bool isPrimary { get; set; }
    public int ExcpCounter { get; set; }
    public string isSmallBusiness { get; set; }
    public int naicsCode { get; set; }
}

class Answer
{
    public string answerText { get; set; }
    public string section { get; set; }
    public SamPointOfContact samPointOfContact { get; set; }
    public Naics[] naics { get; set; }
}

class AnswerNode
{
    [JsonConverter(typeof(SingleValueArrayConverter<Answer>))]
    public Answer[] answers { get; set; }
    public string id { get; set; }
}

class Root
{
    public AnswerNode[] listOfAnswers { get; set; }
}
Root result =
    JsonConvert.DeserializeObject<Root>(
        System.IO.File.ReadAllText("test.json"));


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Creating a view linking three different node types with two node references


By : Heshitha Chathuranga
Date : March 29 2020, 07:55 AM
may help you . Start at the bottom, with a node type filter of Medical Release Form. Create a relationship on the Camper field. Now you have Medical Release Form -> Registration Information (Camper).
Create another relationship on the Camp field, this time with the "Relationship:" set to Camper. That gets you all the nodes linked together, Medical Release Form -> Registration Information (Camper) -> Camp.

Creating a Child Binary Tree Node class from a template Graph Node class


By : Benny Reiner
Date : March 29 2020, 07:55 AM
Does that help The source of your problem is described here
This is my solution:
code :
#include <iostream>
#include <vector>

using namespace std;

template <typename T>
class GraphNode {

public:
    T data;
    vector<GraphNode*> adj;

    void Print(GraphNode<T>* node) {

        if(!node) {
            std::cout << "*";
            return;
        }

        std::cout << node->data << ":";

        for(typename vector<GraphNode<T>* >::iterator iter = adj.begin();
                iter != adj.end();
                iter++)
        {
            Print(iter);
        }
    }
};

template <typename T>
class BinaryTreeNode : public GraphNode<T> {
    using GraphNode<T>::data;
    using GraphNode<T>::adj;

public:
    BinaryTreeNode<T>* lhs;
    BinaryTreeNode<T>* rhs;

    BinaryTreeNode() {
        adj.push_back(NULL);
        adj.push_back(NULL);

        lhs = NULL;
        rhs = NULL;
    }

    BinaryTreeNode(T in_data) {
        data = in_data;

        adj.push_back(NULL);
        adj.push_back(NULL);

        lhs = NULL;
        rhs = NULL;
    }


    BinaryTreeNode& operator=(const BinaryTreeNode& other) {

        // if the other item is this, then return itself
        if(&other != this) {
            data = other.data;
            // copy the vector
            lhs = other.lhs;
            rhs = other.rhs;
        }
        return *this;
    }

};

int main() {
    BinaryTreeNode<int> node;
    cout << 42;
    return 0;
}

Creating new types using Typedef on Vibe.data.json.Json, implicit conversion error


By : sohini gudapati
Date : March 29 2020, 07:55 AM
this will help It is an error in Typedef it seems it is not able to alias a type which use array of type void, Same problem exist with JSONValue from std_data_json. In this case I change this line: https://github.com/s-ludwig/taggedalgebraic/blob/master/source/taggedalgebraic.d#L58 to ubyte type and it is working now. I will try look at it further and fix it.
Issue link

Creating JSON Objects in Node.js loop - Node stops on undefined object


By : user2366978
Date : March 29 2020, 07:55 AM
I wish this helpful for you In JavaScript, you can use the ternary operator on object.capacity, like so: object.capacity ? object.capacity[0] : null. This checks whether object.capacity exists, and executes the line after ? if it does, and executes the line after : if it does not.
More generally, this is how you use the ternary operator, {condition} ? {if true} : {otherwise}
code :
JSONObject = {
    "manufacturer": JSON.parse(object.deviceSpecificationJson[0]).manufacturer,
    "model": JSON.parse(object.deviceSpecificationJson[0]).model,
    "capacity": object.capacity ? object.capacity[0] : null,
    // null can be replaced with the value you want it to show if the property does not exist
    // like 0, or an empty string ''
 }

types of creating a new node of a linked list, when to use what


By : Pramath S.B
Date : March 29 2020, 07:55 AM
around this issue node *x = NULL; node *x = new node();
code :
node *x = NULL;       // 1
node *x = new node(); // 2
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