Serialising and deserialising TypeScript types in local storage

Question

Serialising and deserialising TypeScript types in local storage

I'm currently working on a Typescript application where I store objects using local storage for development purposes. However, I've run into some trouble with deserialization.

Specifically, I have an object called meeting of type MeetingModel:

export interface MeetingModel {
    date: moment.Moment; // using the library momentjs
}

In order to store this object in local storage, I use JSON.stringify(meeting).

I assumed that the stringify call would utilize moment.toJson(), which returns an iso string. This results in the value being stored as:

{"date":"2016-12-26T15:03:54.586Z"}

.

However, when I retrieve this object and attempt to parse it, I encounter an issue:

const stored = window.localStorage.getItem("meeting");
const meeting: MeetingModel = JSON.parse(stored);

The problem is that meeting.date now contains a string instead of a Moment object!

Firstly, I find myself questioning why TypeScript allows me to assign a string value instead of a Moment and why the compiler doesn't catch this error?

Secondly, I'm looking for a solution to convert plain JSON objects (aka strings) back into Typescript types.

While I can create a factory function to handle this conversion, it will become cumbersome as my object database grows.

Is there a more efficient way to handle storing objects in local storage while still maintaining their original data types?

Thank you

json typescript local-storage

Answer 1

Answer №1

1) TypeScript offers optional typing, allowing developers to work around the strictness of its type system. The use of the any type enables dynamic typing, which can be useful when used correctly but also risky if not handled carefully.

For example, the following code will compile:

var x: string = <any> 1;

In this scenario, the number 1 is casted as any, indicating to TypeScript that the developer knows how to handle it. Even though assigning any to a string may lead to errors during runtime, TypeScript allows it due to its design philosophy of types existing only at compile time.

When it comes to serializing and deserializing data, using JSON.stringify and JSON.parse may not always suffice. Type information can be lost during these processes, leading to potential discrepancies between in-memory storage and transfer/storage requirements.

It is recommended to define interfaces for objects to be saved, create a .toJson method on model objects for serialization, and utilize factory functions or constructors for deserialization. This approach provides more control over data storage and helps prevent deserialization issues.

Best of luck with your coding endeavors!

Answer 2

1) TypeScript offers optional typing, allowing developers to work around the strictness of its type system. The use of the any type enables dynamic typing, which can be useful when used correctly but also risky if not handled carefully.

For example, the following code will compile:

var x: string = <any> 1;

In this scenario, the number 1 is casted as any, indicating to TypeScript that the developer knows how to handle it. Even though assigning any to a string may lead to errors during runtime, TypeScript allows it due to its design philosophy of types existing only at compile time.

When it comes to serializing and deserializing data, using JSON.stringify and JSON.parse may not always suffice. Type information can be lost during these processes, leading to potential discrepancies between in-memory storage and transfer/storage requirements.

It is recommended to define interfaces for objects to be saved, create a .toJson method on model objects for serialization, and utilize factory functions or constructors for deserialization. This approach provides more control over data storage and helps prevent deserialization issues.

Best of luck with your coding endeavors!

Answer 3

Answer №2

If you want to transform the string back into a moment, consider utilizing the reviver functionality of JSON.parse:

JSON.parse(input, (key, value) => {
  if (key == "date") {
    return parseStringAsMoment(value);
  } else {
    return value;
});

However, keep in mind that browser support for reviver may vary from basic JSON.parse.

Answer 4

If you want to transform the string back into a moment, consider utilizing the reviver functionality of JSON.parse:

JSON.parse(input, (key, value) => {
  if (key == "date") {
    return parseStringAsMoment(value);
  } else {
    return value;
});

However, keep in mind that browser support for reviver may vary from basic JSON.parse.

Serialising and deserialising TypeScript types in local storage

Answer №1

Answer №2

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