CODEWITHSUNDEEP
jsonangulartypescript
swati
swati

Reputation: 1267

Angular 4 - Map JSON to model and vice versa

I need to call [POST] a REST service from Angular 4 app. The rest service expects a JSON in the body of the request.

The JSON which I am constructing in Angular, based on some user inputs looks something like

{
    "val-type": "prop",
    "val-name": "prop1",
    "op-ter": "DIFF",
    "value": "Unknown",
    "match-twice": false,
    "case-sensitive": false
}

In my code, I am creating this json as 

let jsonSubpart = {
      "val-type": userInput.valtype,
      "val-name": userInput.valname,
      "op-ter": userInput.opter,
      "value": userInput.val,
      "match-twice": false,
      "case-sensitive": false
    }

I was hoping I can create a model for this structure, to ensure that the json structure is adhered to. So, I went ahead and create a model.ts file as below

export class SubPart {
    public valType: string;
    public valName: string;
    public opTer: string;
    public value: string;
    public matchTwice: boolean = false;
    public caseSensitive: boolean = false;

    constructor(valType: string, valName: string, opTer: string, value: string,
        matchTwice: boolean, caseSensitive: boolean) {
        this.valType=valType;
        this.valName= valName;
        this.opTer=opTer;
        this.value = value;
        this.matchTwice=matchTwice;
        this.caseSensitive = caseSensitive;
    }
}

My idea was to then use this model, when constructing the json -

import { Subpart} from './subpart.model.ts';

let jsonSubpart: Subpart = {
      "val-type": userInput.valtype,
      "val-name": userInput.valname,
      "op-ter": userInput.opter,
      "value": userInput.val,
      "match-twice": false,
      "case-sensitive": false
    }

However, this won't work as the field names in json and the class do not match. So, angular wouldn't know that val-type is the same as valType. I can't keep the variable names in .ts file as val-type as it's not a valid variable name due to a '-'.

Wanted to hear from the experts what's the best approach in such a scenario? Should I just construct the json without worrying about the class, or is there another way to get this kind of strong typing?

Upvotes: 3

Views: 5296

Answers (1)

jitender
jitender

Reputation: 10429

You can use Json typescript decorater to serialize your model while posting.

For example, declare class as usual:

export class SubPart {
    @JsonProperty('val-type')
    public valType: string;
}

Fill the model

  let jsonSubpart: Subpart = {
     valType: userInput.valtype
  }

and while posting the model

 var json = serialize(jsonSubpart);

Upvotes: 3

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