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Learn how to effectively convert a JSON string into Java objects using the Jackson library by following our step-by-step guide.
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This video is based on the question https://stackoverflow.com/q/68983736/ asked by the user 'Shiv' ( https://stackoverflow.com/u/13763219/ ) and on the answer https://stackoverflow.com/a/68983974/ provided by the user 'Tushar Kesarwani' ( https://stackoverflow.com/u/1221085/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

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How to Serialize a JSON String to a Java Object Using Jackson

If you're working with Java and JSON, a common task you'll encounter is deserializing a JSON string into a corresponding Java object. This operation can often be tricky, especially if the structure of your JSON does not map directly to your classes. In this post, we'll address how to properly deserialize a JSON string using the Jackson library by solving a common issue with JSON structure.

Understanding the Problem

You might find yourself facing an error similar to the following:

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This usually occurs when the JSON structure you're trying to deserialize does not match the structure defined in your Java classes. For example, consider the following JSON string:

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Now, when using the ObjectMapper to convert this JSON string to a Java object, you may find that the structure of the defined classes does not align with the JSON, particularly in how the "alert" properties are defined.

The Solution

To resolve the issue, you'll need to create a wrapper class for the alert property to make it align with the JSON structure. Let's walk through the changes you need to make step-by-step.

Step 1: Create an AlertWrapper Class

First, you need to create a new class called AlertWrapper to encapsulate the alert property.

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Step 2: Modify the AlertConfig Class

Next, you’ll need to update your AlertConfig class to use AlertWrapper for each of its properties:

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Step 3: Ensure JobConfig Class Remains the Same

The JobConfig class will still reference AlertConfig as is:

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Putting It All Together

After making these changes, your complete structure will look similar to this:

Alert.java

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AlertWrapper.java

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AlertConfig.java

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JobConfig.java

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Conclusion

By wrapping your alert property in a new AlertWrapper class, you ensure that your JSON structure aligns correctly with your Java class definitions. This adjustment eliminates the UnrecognizedPropertyException error and allows you to successfully deserialize the JSON string into Java objects.

Feel free to reach out if you run into any issues or need further clarification on any points made in this guide. Happy coding!