How to combine two json response and send it as a whole?

Question

I am using zomato API which sends only 20 names in one api call and I want atleast 60 responses so I thought of calling the same api three times and combining all the responses together.

app.get('/locations/:query', async (req, res) => {
  const query = req.params.query;
  const data = await zomato.cities({ q: query,count: 1 })
  const cityId= await (data[0].id);
  const restaurants1 = await zomato.search({ entity_id: cityId, entity_type: 'city', start:0, count:20, sort:'rating', order:'desc' })
  const restaurants2 = await zomato.search({ entity_id: cityId, entity_type: 'city', start:20, count:40, sort:'rating', order:'desc' })
  const restaurants =  Object.assign(restaurants1,...restaurants2);
  res.send(restaurants);
  
})

As of now I have only tried till 40 but even this does not work. If another constant restaurant3 is there which has start 40 and end 60, how do I merge these three and send them back?

Answer 1

You shouldn't be spreading restaurants2 in your Object.assign() . Use one of the following:

const restaurants = Object.assign(restaurants1, restaurants2);

// or

const restaurants = { ...restaurants1, ...restaurants2 };

Answer 2

I haven't worked with zomato-api before but according to their API-documentation , you should be able to do this:

app.get('/locations/:query', async (req, res) => {
  const query = req.params.query;
  const data = await zomato.cities({ q: query,count: 1 })
  const cityId= data[0].id;
  const result = [];
  const nrOfRequests = 2; // change this accordingly to increase the nr of requests
  let currCount = 0;  
  const nrOfEntries = 20;
  for(let i=0; i < nrOfRequests ; i++) {
    const response = await zomato.search({ entity_id: cityId, entity_type: 'city', start:currCount, count:nrOfEntries, sort:'rating', order:'desc' });
    result.push(...response.restaurants);
    currCount += nrOfEntries;
  }      
  res.send(result);      
});

Basically you should be able to loop for the desired amount of requests by updating the start parameter for each iteration and then storing the resulting restaurants in your result.