How to chain queries to mongoDB database (mongoose) based on condition

Question

Source Code

let placeForSearch="hampi"
let filteredHotelFacilities=req.body.filter //["Free Wifi"]

let hotels = await Hotel.find({placeForSearch})
      .where("facilities")
      .all(filteredHotelFacilities)
      .in(filteredHotelFacilities)
      .select(selectedProperties)
      .skip(pageNumber * pageSize)  
      .limit(pageSize);

The above code finds if any elements of req.body.filter is in the facilities array in the database and the above code outputs the collection given below.

  {
    facilities: [ 'Garden', 'Free Wifi' ],
    _id: 60f066e61e6971412cf5e727,
    placeForSearch: 'hampi'
  },
  {
    facilities: [ 'Free Wifi' ],
    _id: 60f180141121791e2884d6f0,
    placeForSearch: 'hampi'
  }

Database

  {
    facilities: [ 'Garden', 'Free Wifi' ],
    _id: 60f066e61e6971412cf5e727,
    placeForSearch: 'hampi'
  },
  {
    facilities: [ 'Garden', 'AC' ],
    _id: 60f066e61e6971412cf5e727,
    placeForSearch: 'hampi'
  },
  {
    facilities: [ 'Free Wifi' ],
    _id: 60f180141121791e2884d6f0,
    placeForSearch: 'hampi'
  }

Facilities array in the database can have any of the below-given elements or all of them.

  [
    "Free Wifi",
    "Garden",
    "Water park",
    "Spa and wellness centre",
    "Terrace",
    "Fitness centre",
    "Restaurant",
    "Room service",
    "Bar",
    "Hot tub/jacuzzi",
    "Swimming pool",
    "AC"
  ];

So how can I get all collections in the database if the user does not apply any filter?(req.body.filter=[])

In other words how can i remove .all(filteredHotelFacilities).in(filteredHotelFacilities) based on condition.

Nithin K Joy · Accepted Answer · 2021-07-19 09:51:52Z

1

You should return a promise(by removing await) and chain query based on condition and at last query use await keyword.

let hotels = Hotel.find({placeForSearch})  //remove await keyword

if (filteredHotelFacilities.length > 0) {
   hotels=hotels.where("facilities")       //remove await keyword
                .all(filteredHotelFacilities)
                .in(filteredHotelFacilities);
     }

hotels = await hotels                     // apply await keyword here
   .select(selectedProperties)
   .skip(pageNumber * pageSize)
   .limit(pageSize);

answered Jul 19, 2021 at 9:51

Nithin K Joy

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