AngularJS form. Optional properties in JSON

Question

I have a search form with several fields. If user types nothing into field, it still will be sent to the server with value "null".
However, there are about 25 optional (advanced) fields in this form, and if user types nothing into it, JSON shouldn't contain this field at all.
So, my question is: is there some way (pattern maybe) to avoid those 25 "if () else"? What is the best way to form that final JSON?
Thanks!

UPD: Thank you, @RIYAJKHAN for your answer! I've managed to solve it by myself using LoDash:

   private compactObject(obj) {
      let compactObj = this.removeEmptyProperties(_.cloneDeep(obj));
      _.forEach(compactObj, (value, key) => {
        if (_.isObject(value)) {
          compactObj[key] = this.compactObject(this.removeEmptyProperties(compactObj[key]));
        }
      });
      return this.removeEmptyProperties(compactObj);
    };

    private removeEmptyProperties(obj) {
      let trimedObj = _.cloneDeep(obj);
      _.forEach(trimedObj, (value, key) => {
        if (!value || _.isEmpty(trimedObj[key])) {
          delete trimedObj[key];
        }
      });
      return trimedObj;
    }

what's wrong with just sending them as null to the server? – toskv Nov 27 '16 at 14:46 — toskv, Nov 27 '16 at 14:46

score 2 · Answer 1 · answered Nov 27 '16 at 14:50

<form name="form" ng-submit="submit(form)">

    <input type="text" name="test1" ng-model="test1">
    <input type="text" name="test2" ng-model="test2">
    <input type="text" name="test3" ng-model="test3">

<input type="button" name="submit"> 
</form>

JavaScript :

$scope.submit = function(form){

    for(var prop in form){
        if(!!!form[prop]){
                delete form[prop];
        }
    }//form will contain form prop having only valid values
}

AngularJS form. Optional properties in JSON

1 Answers1