How to wait for all async tasks to finish in Node.js?

Question

My program needs to run many async tasks.

And one task must run when all async tasks complete.

How can I make this function to wait all async functions?

let urls=[....]

for(var i=0; i<urls.length; i++){
    doRequest(urls[i])
}
finish()



function doRequest(url){
    request({
      uri: url,
    }, function(error, response, body) {
        //do some tasks
    });
}

function finish(){
    console.log('finish')
}

Answer 1

Promise are exactly what you need. They're native in JS nowadays, no need for extra libraries.

function finish () {
    console.log("finish");
}

function doRequest (uri) {
    // wrap code in promise
    return new Promise((resolve, reject) => {
        request(
            { uri },
            (error, response, body) => {
                if (error) {
                    // error case, similar to "throw"
                    reject(error);
                }
                else {
                    // success case, similar to "return"
                    resolve({ response, body });
                }
            }
        );
    });
}

let urls=[...];

// Basic way to do it:
Promise.all(urls.map(doRequest)).then(finish);

// async/await notation:
// you must be in an "async" environement to use "await"
async function wrapper () {
    await Promise.all(urls.map(doRequest));
    finish();
}
// async function return a promise transparently
wrapper();

Answer 2

The easiest solution that doesn't require rewriting everything using promises, would be to check the number of completed requests at the end of doRequest function, and call finish if all requests have been completed.

let urls=[....]
let completed = 0

for(var i=0; i<urls.length; i++){
    doRequest(urls[i])
}

function doRequest(url){
    request({
        uri: url,
    }, function(error, response, body) {
        //do some tasks

        completed++
        if (completed === urls.length) {
            finish()
        }
    });
}

function finish(){
    console.log('finish')
}