How to Properly Deserialize JSON to a C# POCO Class Using JsonConvert.DeserializeObject?

Deserialize JSON to C# POCO class using JsonConvert.DeserializeObject

Question:

When trying to use JsonConvert.DeserializeObject to deserialize JSON data to a C# POCO class, an exception is thrown due to incorrect deserialization of a specific property.

Answer:

In order to use JsonConvert.DeserializeObject to correctly deserialize JSON to a POCO class, it is important to ensure that the property names in the class match the property names in the JSON data. Additionally, the following steps may be required:

1. Use the JsonProperty attribute: Use the JsonProperty attribute to specify the JSON attribute name corresponding to a specific class attribute. This is especially useful when the JSON property name is different from the class property name.
2. Consider attribute types: Verify that the data type of each attribute in the POCO class is consistent with the expected type in the JSON data. For example, if the JSON data contains an array, the corresponding property in the POCO class should be declared as a list.

Example:

Consider the following example where the Accounts property in the User class needs to be deserialized from a JSON object:

public class User
{
    [JsonProperty("accounts")]
    public Account Accounts { get; set; }

    // 其他User属性此处省略，以简洁起见
}

public class Account
{
    public string github { get; set; }
}

By using the JsonProperty attribute and declaring the Accounts property as a specific Account object, deserialization will correctly handle the JSON object for that property.

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