How to ignore an entity field in TypeScript when sending a JSON to a Web Service?

Question

I have this class:

export class TblColabAdmin {
    snomatrcompl: string;
    nflativo: number;
    ativo: boolean;
}

The attribute ativo doesn't exist in my web service entity, so I would like to avoid that it was added in JSON.

In Java, for example, we have @JsonIgnore annotation. Does exist something similar in TypeScript?

nope, you need to create a new object and set only the properties you want to send. — toskv
– toskv, Commented Dec 12, 2016 at 13:40
Did you look at adding a 'toJSON' method? You could tr something like toJSON() { const obj = Object.assign({}, this); delete obj.ativo; return obj; }. Or toJSON() { return Object.assign({}, this, {ativo: undefined}); }. — user663031
– user663031, Commented Dec 12, 2016 at 15:31
Using a custom replacer as suggested in this answer will help, I believe. stackoverflow.com/a/41685627/2685234 — ramtech
– ramtech, Commented Jun 14, 2017 at 7:10
Found one more approach of having toJSON function in each type.. stackoverflow.com/a/35138341/2685234 — ramtech
– ramtech, Commented Jun 14, 2017 at 7:22

Nitzan Tomer · Accepted Answer · 2016-12-12 15:29:43Z

12

You can create a JsonIgnore decorator so that it will work like with java:

const IGNORE_FIELDS = new Map<string, string[]>();
function JsonIgnore(cls: any, name: string) {
    let clsName = cls.constructor.name;
    let list: string[];

    if (IGNORE_FIELDS.has(clsName)) {
        list = IGNORE_FIELDS.get(clsName);
    } else {
        list = [];
        IGNORE_FIELDS.set(clsName, list);
    }

    list.push(name);
}

class Base {
    toJson(): { [name: string]: any } {
        let json = {};
        let ignore = IGNORE_FIELDS.get(this.constructor.name);

        Object.getOwnPropertyNames(this).filter(name => ignore.indexOf(name) < 0).forEach(name => {
            json[name] = this[name];
        });

        return json;
    }
}

class TblColabAdmin extends Base {
    snomatrcompl: string;
    nflativo: number;

    @JsonIgnore
    ativo: boolean;

    constructor(snomatrcompl: string, nflativo: number, ativo: boolean) {
        super();

        this.snomatrcompl = snomatrcompl;
        this.nflativo = nflativo;
        this.ativo = ativo;
    }
}

let obj = new TblColabAdmin("str", 43, true).toJson();
console.log(obj); // Object {snomatrcompl: "few", nflativo: 43}

(code in playground)

It's quite a lot of work if you're only doing it once, but if it's a common issue in your code then this approach should work well.

edited Dec 12, 2016 at 15:29

answered Dec 12, 2016 at 14:05

Nitzan Tomer

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4 Comments

Gabriel Cerutti Over a year ago

This works well if you are not using the JSON.stringify, in other case a good approach is using a replacer as suggested in this post stackoverflow.com/questions/41685082/…

Luke Kroon Over a year ago

I use this in my project and extend all my classes from the Base class. Just make sure to check let ignore before you filter because if you do not have a @JsonIgnore in your class it will be undefined and json will be {}. So if(!ignore) json = this;

Gourav Garg Over a year ago

@LukeKroon I updated this code for a property which is ref of another class having same kind of functionality

Nitzan Tomer Over a year ago

@GouravGarg your edit changes my reply, which is old and i prefer to keep it as is. you're welcome to write a new answer with your improved code.

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