Скачать или смотреть Solving the Decorator Field Issue in TypeScript Superclasses

Discover how to resolve the `decorator` field not working issue in TypeScript superclasses with an effective solution and detailed explanation.
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Understanding Decorator Fields in TypeScript Superclasses

When working with TypeScript and decorators, you may come across situations where properties from a superclass are not behaving as you expect. One common issue is when decorators applied to fields in your superclass did not seem to register correctly in a subclass. In this post, we will explore this problem in depth and provide a solution to ensure that decorator fields work seamlessly across inheritance.

The Problem: Decorator Fields Not Registering

Here, we will break down the core of the problem. Consider you have a base class Base, which uses a decorator JsonInclude to specify fields that should be included in JSON serialization. The specific issue arises when subclassing this base class:

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In this setup, you might expect that the name property from the Base class should also be included in the serialization output when creating an instance of Staff. However, it seems like the name field from Base isn't being added to the INCLUDED_FIELDS list as intended.

The Cause of the Issue

The root of the issue lies in how TypeScript handles decorators and inheritance:

Decorator Registration: When applying the JsonInclude decorator to name, the decorator registers it under the class name "Base". However, when toJSON() is called on an instance of Staff, the context for this.constructor.name returns "Staff". As a result, only the fields registered under "Staff" are retrieved, leading to the omission of the "name" field from Base.

Mapping Classes: The current mapping strategy in the decorator is based on class names. Since Base and Staff are treated differently due to their unique class names, their properties are not cross-referenced appropriately.

The Solution: Using a Unified Key Name for Classes

To address this issue, we need to modify our approach. We can develop a helper function to reliably fetch the "root" class name consistently for both parent and child classes. This will allow all fields, regardless of their specific class, to be serialized together.

Step 1: Implementing the getRootClassName Function

We will create a utility function to traverse the class prototype chain and find the topmost class name:

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Step 2: Refactoring the Decorator

Next, we will modify the JsonInclude decorator to use our new utility function. This ensures consistent behavior across class types:

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Step 3: Adjusting the toJSON Method

Lastly, we need to adjust the toJSON method in the Base class to use the root class name for accessing the INCLUDED_FIELDS map:

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Final Implementation

After these adjustments, our classes will look like this:

Base Class (base.ts)

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Staff Class (staff.ts)

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Conclusion

By implementing the getRootClassName function and adjusting our decorator mechanism, we can effectively address the issue of superclass fields not being included via decorators in TypeScript. This solution enables us to maintain a clear structure in our code while ensuring that properties from both parent and child classes are serialized as needed.

Feel free to implement this solution in your TypeScript projects, and experience seamless integration of decorators across class hierarchies.