Question

Possible Duplicate:
Variable in JavaScript callback functions always gets last value in loop?

I am not able to pass the value of k to the callback function of fadeOut function.
my loop is as given below.

 for(var k=0;k<image1.length;k++)
 {

    $(img[k]).fadeOut(200,function(k) {
        alert(k);
        $(this).attr('src', image2[k]);
        $(this).fadeIn(200);
    });
  }

Answer 1

The jQuery fadeOut function takes a callback function with no arguments. From the jQuery documentation, “The callback is not sent any arguments”. If you want to capture the value of k, do something like the following:

for(var k=0;k<image1.length;k++) {
    (function(k) {
        $(img[k]).fadeOut(200,function() {
            alert(k);
            $(this).attr('src', image2[k]);
            $(this).fadeIn(200);
        });
    })(k);
}