Home > Backend Development > C++ > Why Am I Getting a 'Cannot Deserialize the Current JSON Object' Error During JSON Deserialization?

Why Am I Getting a 'Cannot Deserialize the Current JSON Object' Error During JSON Deserialization?

Barbara Streisand
Release: 2025-01-08 00:22:40
Original
453 people have browsed it

Why Am I Getting a

Troubleshooting "Cannot Deserialize JSON Object" Error

This common JSON deserialization error, "Cannot deserialize the current JSON object," stems from a discrepancy between the JSON data's structure and the expected data type in your deserialization code. The error message suggests a mismatch: the JSON is likely an object, but your code attempts to deserialize it as an array, or vice-versa.

Correcting the Deserialization

The problem arises from attempting to deserialize a JSON object into a list. The provided JSON is clearly an object, not an array. The solution is to adjust your deserialization to match the JSON structure. Instead of deserializing into List<RootObject>, deserialize directly into a RootObject instance.

Incorrect Code (Attempting to deserialize into a list):

<code class="language-csharp">List<RootObject> datalist = JsonConvert.DeserializeObject<List<RootObject>>(jsonString);</code>
Copy after login

Corrected Code (Deserializing into a single RootObject):

<code class="language-csharp">RootObject data = JsonConvert.DeserializeObject<RootObject>(jsonString);</code>
Copy after login

This corrected code accurately reflects the JSON's object structure, eliminating the deserialization error. Remember to ensure your RootObject class correctly maps to the properties within the JSON object. If the JSON contains a property holding an array of Datum objects, your RootObject class should have a corresponding property (e.g., List<Datum> data) to accommodate this.

The above is the detailed content of Why Am I Getting a 'Cannot Deserialize the Current JSON Object' Error During JSON Deserialization?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template