Question

Sure there’s something tiny wrong with my syntax… Here is stripped down version of my code…

function example1() {
    var thumbWidth = $(this).find('img').width();
    //some more code
};

function example2() {
  $('div').each(function(){
    example1();
    thumbTotal = thumbWidth + xx //etc.;
  });
};

$(document).ready(function(){
  example2();
});

My problem is that example1 is not being executed in the code above. I’ve tried inserting the example1 code manually and this works fine. I want to be able to call example1 elsewhere so it’s important that it’s isolated within a separate function.

Many thanks.

Answer 1

Based on you edited post you’ve got at least two ways to solve the situation:

1st: Define a return value for example1()

function example1() {
    var thumbWidth = $(this).find('img').width();
    //some more code
 };
 function example2() {
  $('div').each(function() {
      thumbTotal = example1() + xx //etc.;
  });
 };
 $(document).ready(function(){
   example2();
 });

2nd: Define an appropiate scope for the variable ‘thumbWidth’

    var thumbWidth;
    function example1() {
        thumbWidth = $(this).find('img').width();
        //some more code
     };
     function example2() {
      $('div').each(function() {
         //or here var thumbWidth;
         example1();     
         thumbTotal = thumbWidth + xx //etc.;
      });
     };

 $(document).ready(function(){
   example2();
 });