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[JsonPropertyName("xyz")] how to ignore this so serialization to json doesnt use these names, but insted use real names of parameters (vars)

Colorful Copperhead answered on April 15, 2024 Popularity 1/10 Helpfulness 1/10

answer [JsonPropertyName("xyz")] how to ignore this so serialization to json doesnt use these names, but insted use real names of parameters (vars)

[JsonPropertyName("xyz")] how to ignore this so serialization to json doesnt use these names, but insted use real names of parameters (vars)

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var options = new JsonSerializerOptions
{
    PropertyNamingPolicy = null //null for exact property names (other option is JsonNamingPolicy.CamelCase)
};
string jsonString = JsonSerializer.Serialize(yourObject, options);

Popularity 1/10 Helpfulness 1/10 Language csharp

Source: Grepper

Tags: c# ignore names parameters serialization this to-json

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Contributed on Apr 15 2024

Colorful Copperhead

0 Answers Avg Quality 2/10

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[JsonPropertyName("xyz")] how to ignore this so serialization to json doesnt use these names, but insted use real names of parameters (vars)

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