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rated 0 times [  38] [ 5]  / answers: 1 / hits: 46118  / 9 Years ago, mon, september 28, 2015, 12:00:00

In my Node.js code I need to make 2 or 3 API calls, and each will return some data. After all API calls are complete, I want to collect all the data into a single JSON object to send to the frontend.


I know how to do this using the API callbacks (the next call will happen in the previous call's callback) but this would be slow:


//1st request
request('http://www.example.com', function (err1, res1, body) {

//2nd request
request('http://www.example2.com', function (err2, res2, body2) {

//combine data and do something with it

});

});

I know you could also do something similar and neater with promises, but I think the same concept applies where the next call won't execute until the current one has finished.


Is there a way to call all functions at the same time, but for my final block of code to wait for all API calls to complete and supply data before executing?


More From » node.js

 Answers
6

Promises give you Promise.all() (this is true for native promises as well as library ones like bluebird's).



Update: Since Node 8, you can use util.promisify() like you would with Bluebird's .promisify()



var requestAsync = util.promisify(request); // const util = require('util')
var urls = ['url1', 'url2'];
Promise.all(urls.map(requestAsync)).then(allData => {
// All data available here in the order of the elements in the array
});


So what you can do (native):



function requestAsync(url) {
return new Promise(function(resolve, reject) {
request(url, function(err, res, body) {
if (err) { return reject(err); }
return resolve([res, body]);
});
});
}
Promise.all([requestAsync('url1'), requestAsync('url2')])
.then(function(allData) {
// All data available here in the order it was called.
});



If you have bluebird, this is even simpler:



var requestAsync = Promise.promisify(request);
var urls = ['url1', 'url2'];
Promise.all(urls.map(requestAsync)).then(allData => {
// All data available here in the order of the elements in the array
});

[#64909] Thursday, September 24, 2015, 9 Years  [reply] [flag answer]
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maxinec

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