Executing functions synchronously in javascript

Question

How to execute functions synchronously one after the another

   function test1(){
      setTimeout(function(){console.log("should print 1st");},1000);

   }
   function test(){
     test1();
     console.log("should print 2nd");
   }
     function aftertest(){
     var dfd = $.Deferred();

         dfd.done(test()).done(tester());
         //dfd.done(test,tester);   

         //$.when(test()).then(tester());

       console.log("should print 4th");
   }
    function tester(){
       console.log("should print 3rd");
    }
    aftertest();

Here is BIN of what I have tried so far.

remove setTimeout and change it to function test1(){ console.log("should print 1st"); } — Debabrata Mohanta
– Debabrata Mohanta, Commented May 18, 2016 at 7:45
please explain better what you need. You mix synchronous and asynchronous here. Either you want one or the other, — Regis Portalez
– Regis Portalez, Commented May 18, 2016 at 7:50

Avinash · Accepted Answer · 2016-05-18 08:02:12Z

3

Use .then :

function test1(){
      var defer = $.Deferred();
      setTimeout(function(){console.log("should print 1st");defer.resolve();},1000);
     return defer;
   }
   function test(){
     var defer = $.Deferred();
     test1().then(function() {
       console.log("should print 2nd");
       defer.resolve();
     });
     return defer;
   }
     function aftertest(){
     var dfd = $.Deferred();

         test().then(tester).then(function () {
           console.log("should print 4th");    
         });
         //dfd.done(test,tester);   

         //$.when(test()).then(tester());


   }
    function tester(){
       console.log("should print 3rd");
      return $.when();
    }
    aftertest();

edited May 18, 2016 at 8:02

answered May 18, 2016 at 7:53

Avinash

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