CODEWITHSUNDEEP
node.jspromise
Muzammil
Muzammil

Reputation: 467

nodejs - call functions in order after termination of the previous using Promise

I am trying to call functions call2() and then call3() in this order, so that call3() is called only when call2() terminates. I am using Promise to achieve that.

But call3() is called before call2() terminates. Here is my code:

function call2() {
    return new Promise(function (resolve, reject) {
        setTimeout(function () {
            console.log("calling 2");
            resolve(true);
        }, 3000);
    });
}

function call3() {
    console.log("calling 3");    
}

call2().then(call3());

I am obviously doing something wrong, or could not understand how to use promise. Any help please?

Upvotes: 0

Views: 42

Answers (2)

Alexander Antyneskul
Alexander Antyneskul

Reputation: 11

You have to wrap your call3 into the function: 

call2().then(()=>call3());

Upvotes: 0

Damian
Damian

Reputation: 2852

In then(call3()) you are calling the call3 function rather than passing it as a callback, change to this:

call2().then(call3);


function call2() {
   console.log('Start...');
    return new Promise(function (resolve, reject) {
        setTimeout(function () {
            console.log("calling 2");
            resolve(true);
        }, 3000);
    });
}

function call3() {
    console.log("calling 3");    
}

call2().then(call3);

Upvotes: 3

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