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

Question

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?

Damian · Accepted Answer · 2017-01-20 15:58:15Z

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);

Alexander Antyneskul · Accepted Answer · 2017-01-20 16:03:58Z

0

You have to wrap your call3 into the function:

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

answered Jan 20, 2017 at 16:03

Alexander Antyneskul

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