Deserialize json that has some property name starting with a number

Question

JSON data looks like this

[
    {
        "market_id": "21",
        "coin": "DarkCoin",
        "code": "DRK",
        "exchange": "BTC",
        "last_price": "0.01777975",
        "yesterday_price": "0.01770278",
        "change": "+0.43",
        "24hhigh": "0.01800280",
        "24hlow": "0.01752015",
        "24hvol": "404.202",
        "top_bid": "0.01777975",
        "top_ask": "0.01790000"
    }
]

Notice these 3 properties here 24high, 24hhlow, and 24hvol how do you make a class for that. I need all those properties by the way, not just those 3 properties I mentioned.

score 26 · Accepted Answer · answered Jun 14 '14 at 09:51

26

You should use JSON.NET or similar library that offers some more advanced options of deserialization. With JSON.NET all you need is adding JsonProperty attribute and specify its custom name that appears in resulting JSON. Here is the example:

   public class MyClass
   {
        [JsonProperty(PropertyName = "24hhigh")]
        public string Highest { get; set; }
        ...

Now to deserialize:

    string jsonData = ...    
    MyClass deserializedMyClass = JsonConvert.DeserializeObject<MyClass>(jsonData);

answered Jun 14 '14 at 09:51

Borys Generalov

2,255
17
19

Thanks for that "JsonProperty" – jaysonragasa Jun 14 '14 at 10:55
2

To deserialize he would need `JsonConvert.Deserialize>(jsonData)` because the JSON represents an array, not a single object. – Brian Rogers Jun 14 '14 at 15:43
You save my lots of time. – Durgesh Pandey Dec 12 '17 at 18:53
in my case, the property represents some year, so the name isn't fixed, example: `"serie": { "2021" : "85541" }`. In this json 2021 is the year, how to deserialize in this case? – Ewerton Oct 29 '21 at 15:19

score 3 · Answer 2 · edited May 06 '20 at 10:21

3

For .NET Core 3.0 and beyond, you can now use the System.Text.Json namespace. If you are using this:

public class MyClass
{
    ...
    [JsonPropertyName("24hhigh")]
    public string twentyFourhhigh { get; set; }
    ...
}

You can use JsonPropertyName Attribute.

edited May 06 '20 at 10:21

Jeremy Caney

7,102
69
48
77

answered May 06 '20 at 09:32

Pranav

105
5

Deserialize json that has some property name starting with a number

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