Nsryjdtyk

I tried apply chain of conditionals inside `filter. How to join result from previous conditionals and pass result the next:

this.fltRooms = this.rooms.filter(room => {

      let _rule;

      if (formValues["auditoryNumber"]) {

        _rule = formValues["auditoryNumber"] == room.title;

      }



      if (formValues["floor"]) {

        _rule = formValues["floor"] == room.floor; // Should be AND

      }



      if (formValues["corpus"]) {

        _rule = formValues["corpus"] == room.building; // Should be AND

      }



      return _rule;

    });

So, logic is, if exist formValues["auditoryNumber"] and formValues["floor"] and formValues["corpus"] then filter by all three conditionals with AND between.

Basically you want to say if the filter is not set that it is true. If the filter is set than check out if the value matches.

this.fltRooms = this.rooms.filter(room => 

  (!formValues["auditoryNumber"] || formValues["auditoryNumber"] == room.title) &&

  (!formValues["floor"] || formValues["floor"] == room.floor) &&

  (!formValues["corpus"] || formValues["corpus"] == room.building)

)

Other way with ifs would be using the boolean in each check

this.fltRooms = this.rooms.filter(room => {

  let _rule = true

  if (formValues["auditoryNumber"]) {

    _rule = formValues["auditoryNumber"] == room.title;

  }



  if (_rule && formValues["floor"]) {

    _rule = formValues["floor"] == room.floor; // Should be AND

  }



  if (_rule && formValues["corpus"]) {

    _rule = formValues["corpus"] == room.building; // Should be AND

  }



  return _rule;

});

And what if either doesn't exist?

As you describe it, it should be quite straightforward. Something like

this.fltRooms = this.rooms.filter(room =>

    formValues["auditoryNumber"] == room.title

    && formValues["floor"] == room.floor

    && formValues["corpus"] == room.building);

Alternative to @bipll correct answer, using a defined array of keys to compare

this.filtRooms = this.rooms.filter(x => [

    'auditoryNumber',

    'floor',

    'corpus',

 ].every(y => x[y] === formValues[y]));

If keys are different on both objects to compare

this.filtRooms = this.rooms.filter(x => [

    [

      'auditoryNumber',

      'title',

    ],

    [

      'floor',

      'floor',

    ],

    [

      'corpus',

      'building',

    ],

  ].every(([

    key1,

    key2,

  ]) => x[key1] === void 0 || x[key1] === formValues[key2]));

this.filterRooms = this.rooms.filter(meetsSearchParams);



    function meetsSearchParams(room){

      return  room.auditoryNumber == formValues["auditoryNumber"] &&

              room.floor == formValues["floor"] &&

              room.corpus == formValues["corpus"];

    }

For the sake of keeping this as readable as possible.

If you're going to need to extend it, make a map array between the two structures

var fields = [

  {f: "auditoryNumber", p: "title"},

  {f: "floor", p: "floor"},

  {f: "corpus", p: "building"}

];



this.fltRooms = this.rooms.filter(room => 

  fields.every(field => !formValues[field.f] || formValues[field.f] == room[field.p]);

)