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How Can I Deserialize JSON with Dynamic Property Names as Identifiers into a Strongly Typed C# Object?

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2025-01-30 00:01:09933browse

How Can I Deserialize JSON with Dynamic Property Names as Identifiers into a Strongly Typed C# Object?

JSON created a strong type C# object with the attribute name as the identifier

Problem description

When trying to create a strong type C# object from JSON data, the author encountered the problem of processing the attribute name in the JSON object. These attribute names are the identifier of the corresponding session. The existing code attempts to use the layered class structure to serialize the JSON file, but the attribute name (session logo) is not recognized as an object attribute.

Solution

In order to solve this problem, the author needs to modify the root object in the C# class as

. This change will allow the JSON object attribute name to be mapped as a dictionary key.

code example Dictionary<string, SessionPerformanceStats>

Or, if the attribute name in the JSON object is always numbered, the dictionary key can be defined as long and upstormic:

<code class="language-csharp">var dictionary = JsonConvert.DeserializeObject<Dictionary<string, SessionPerformanceStats>>(theJsonResponse);</code>

Additional description

The goals of these modifications are to enable the correctly serialization of the JSON object into a dictionary. This method allows the attribute name in the JSON object as a dictionary key to effectively map the JSON object structure to the C# dictionary. This solves the problem that the session identifier is not recognized as an object attribute.
<code class="language-csharp">var dictionary = JsonConvert.DeserializeObject<Dictionary<long, SessionPerformanceStats>>(theJsonResponse);</code>

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