CODEWITHSUNDEEP
javascriptnode.jspromise
Krisalay
Krisalay

Reputation: 75

Return a data from callback

    var myVar;    
    myModule.myFunction().then(function(data){
       myVar = data;     
    });
    console.log(myVar);

I have a module named myModule which exports a function myFunction with some promise. Now when I assign the data to myVar, and print it, it shows undefined. What will be the proper solution to achieve this result?

Upvotes: 0

Views: 74

Answers (3)

synthet1c
synthet1c

Reputation: 6282

The code you have posted is not synchronous, so it is working correctly.

This is the flow of your code to give you a better understanding of what the code is actually doing.

// myVar is declared undefined
var myVar;
// you call console.log with undefined
console.log(myVar);
// sometime later .then is called
myModule.myFunction().then(function(data){
   // later you set the global variable to the data
   myVar = data;
   // at this point you can log the value of myVar 
});

In terms of returning something from an ajax call you can't do it synchronously, but you can chain another .then after it's returned.

myModule.myFunction().then(function(data){
   // do something with the data
   return data; 
})
.then(function(data){
   // do something else with the data
})

You can keep chaining then as long as you want, and there are more useful methods you can use for error handling, or calling promises in parallel or synchronously

Upvotes: 1

James Monger
James Monger

Reputation: 10665

Because then is asynchronous, console.log will be called before then, so at the time of the log, it is undefined.

This will give you the result you're after:

var myVar;    
myModule.myFunction().then(function(data){
    myVar = data; 
    console.log(myVar);  
});

Upvotes: 0

petur
petur

Reputation: 1386

You may have to reconsider your structure, but to achieve what you're asking for is to have the console.log inside the promise.

 var myVar;    
    myModule.myFunction().then(function(data){
       myVar = data;  
       console.log(myVar);   
    });

Upvotes: 1

Related Questions