Angular2 - Parse JSON into an Object

Question

Example: I have a Entity-Class named "Person"

constructor(name:string,surname:string,birthdate:string) {
    this.name = name;
    this.surname = surname;
    this.birthdate = birthdate;
}

And in a "Manager"-Class I get a string that looks like a JSON:

{
    "name" : "testName",
    "surname" : "testSurrname",
    "birthdate" : "JJJJ:MM:DD hh:mm:ss"
}

So how to parse the JSON into a "Person"

personData : Person;
jsonData : JSON;
public toPerson(data: string): Person {
    this.jsonData = JSON.parse(data);
    .?
    .?
    .?
    personData = new Person(....);
    return personData;
}

Where is this string coming from and why is it broken JSON? Assuming you can sort that out, the best approach is to rewrite your constructor so it looks like constructor(data) { Object.assign(this, data); }, then invoke it with new Person(this.jsondata). — user663031
– user663031, Commented Mar 7, 2017 at 16:57

El houcine bougarfaoui · Accepted Answer · 2018-12-20 08:25:12Z

3

public toPerson(data: string): Person {
    let jsonData = JSON.parse(data);

    personData = new Person(jsonData.name, jsonData.surname, jsonData.birthdate);
    return personData;
}

edited Dec 20, 2018 at 8:25

user8280225

answered Mar 7, 2017 at 16:59

El houcine bougarfaoui

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7 Comments

ALSTRA Over a year ago

at this.jsonData. i can only choose between this.jsonData.stringify and this.jsonData.parse...

El houcine bougarfaoui Over a year ago

the data is already stringified why you want to use stringify ?

ALSTRA Over a year ago

could you join the chat, it would be faster :) chat.stackoverflow.com/rooms/137472/…

ssube Over a year ago

this. on each of the parameters to new Person is unnecessary, it should just be jsonData.*

El houcine bougarfaoui Over a year ago

thnx @ssube I forgot the this cuz I updated the code many times :)

|

Serginho · Accepted Answer · 2018-12-20 08:15:59Z

2

One more elegant solution is to use JSON.parse reviver:

public static fromJSON(json: any): Person {
    if (typeof json === 'string') {
        return JSON.parse(json, Person.reviver);
    } else if (json !== undefined && json !== null) {
        let person = Object.create(Person.prototype);
        return Object.assign(person, json);
    } else {
        return json;
    }
}

public static reviver(key: string, value: any): any {
    return key === '' ? Person.fromJSON(value) : value;
}

edited Dec 20, 2018 at 8:15

user8280225

answered Jun 16, 2017 at 0:03

Serginho

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