How to deserialize JSON object with Newtonsoft when it's fields names are reserved keywords like "short"?

user_1154283 0
Viewed 887

I have some JSON object:

"opf": {
                "type": "2014",
                "code": "12247",
                "full": "Публичное акционерное общество",
                "short": "ПАО"
            }

I want it to deserialize it into my class:

class SuggestionInfoDataOpf
{
    public string code;
    public string full;
    public string short; //ERROR. Of course I can't declare this field
    public string type;
}

What to do?.. I want to deserialize it like that: Newtonsoft.Json.JsonConvert.DeserializeObject<SuggestionInfoDataOpf>(json_str);, but fields names should match.

json-deserialization reserved-words json.net .net c#
2 Answers

By using the JsonProperty attribute

class SuggestionInfoDataOpf
{
    [JsonProperty("short")]
    public string Something {get; set;}
}

Or using the prefix "@" before the name of the property. Using it you can name a member the same as a keyword

class SuggestionInfoDataOpf
{
    public string @short;
}

But IMO the JsonProperty is better, as it allows you to keep to the C# naming guidelines as well as visually separating members from keywords

user_93639730

You should use keywords with @ like this:

class SuggestionInfoDataOpf
{
    public string code;
    public string full;
    public string @short;
    public string type;
}
user_108187300
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