How to Ignore JSONproperty(Propertyname = "Somename") When Serializing JSON

How can I ignore the property name that I use in the jsonproperty attribute, provided that it is only once

you can add a constructor to your class

public class DocumentDetail
{
    public string qrCode { get; set; }

    public string gtinNumber { get; set; }

    [JsonProperty("lotNumber")] // optional, you can assign any name for serialization
    public string lotNumber { get; set; }
    
    [Newtonsoft.Json.JsonConstructor]
    public  DocumentDetail( string KAREKOD,string GTIN, string LOTNUMBER)
    {
        qrCode=KAREKOD;
        gtinNumber=GTIN;
        lotNumber=LOTNUMBER;
    }
    public DocumentDetail() {}
}

and you don't need to include all properties in the constructor, just include the properties that need different names for a serialization and a deserialiazation.

Ignore some of the properties during Json.NET serializing but required on others

Try the below code

    [JsonProperty(PropertyName = "componentMainVersion", DefaultValueHandling = DefaultValueHandling.Include)] 
    public ushort Version { get; set; }

Required is a different property which makes sure that value for the property is required always

How to ignore @JsonProperty when Serialization in Java

Use simple access methods annotated with @JsonGetter or @JsonSetter respectively, each configured with required name of json property.
In your case the code could be something like this:

public class Account {

    private String reasonCode;

    @JsonGetter("reasonCode")
    public String getReasonCode() {
        return reasonCode;
    }

    @JsonSetter("rCode")
    public void setReasonCode(String reasonCode) {
        this.reasonCode = reasonCode;
    }
}

Using JsonProperty for Deserializing but NOT serializing

You could use a set-only property that points to the 'good' property.

public class Item
{
    public int Id { get; set; }

    [JsonProperty("pkID")]
    public int BackwardCompatibleId { set => Id = value; }
}

// test
var x = new Item { Id = 88 };
var json = JsonConvert.SerializeObject(x); // {"Id":88}
var clone = JsonConvert.DeserializeObject<Item>("{\"pkId\":99}");

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