将字符串转换为 object 并调用 function

Question

I want to store an object as a string, and then convert ot back to an object and call a method of this object.

    user.delete() // this works
    self.user = JSON.stringify(user)
    const storeUser = JSON.parse(self.user)
    storeUser.delete() // Error: delete is not a function

Answer 1

As has been said in comments, functions/methods are not part of the JSON format so they are not present when you serialize to JSON. Same with class names and such. So, when you call JSON.stringify() , it does not include any information about what type of class it is or any of the methods associated with that object. When you then call JSON.parse() , all you will get back is instance data in a plain object. That's just one of the limitations of JSON.

Normally, you will not serialize code like methods. Instead, you will serialize what type of object it is, serialize the relevant instance data and then you want to reinstantiate the object, you will look at the data to see what type of object it is, call the constructor to make an object of that type and pass to the constructor the data you serialized from the prior object and you will have built a version of the constructor that takes exactly the data you are serializing so it can properly initialize the new object.

Answer 2

If JSON.parse and JSON.stringify would allow a copy of methods, your code would be insecure and would have risks of people running arbitrary code on your server/computer since normally JSON string comes from external sources.

You should allow your User's class to parse a json and create a new instance based on that json:

class User {
  constructor(someRandomProperty) {
    // ... code
  }

  static fromJson(userJson) {
    let props;
    try {
      props = JSON.parse(userJson);
    } catch (error) {
      throw new Error("Invalid Json User");
    }

    return new User(props.name, props.someRandomProperty);
  }

  delete() {
    // .... code
  }
}

And then you would copy the user like this:

User.fromJson(yourJsonStrinHere)