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I have a very simple webpage that contains 4 color images. When I hover over any of the images the original (xxxx1.jpg) is replaced by a black and white version (xxxxx2.jpg) of the same image, and then it reverts back to the original image after hover.

I'm declaring the same variable for both handlerIn and handlerOut:

var src = $(this).attr("src")

Can I declare the var just once somehow? I just can't seem to figure this out myself.

$(document).ready(function(){
    $( "img" ).hover(
        function() {
            var src = $(this).attr("src").replace("1.jpg", "2.jpg");
            $(this).attr("src", src);
        }, 
        function() {
            var src = $(this).attr("src").replace("2.jpg", "1.jpg");
            $(this).attr("src", src);
        }
    );
});
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  • 4
    \$\begingroup\$ Use CSS background-image to set the image on element, on :hover change the background image. \$\endgroup\$
    – Tushar
    Dec 19, 2016 at 5:49
  • \$\begingroup\$ @Tushar I'll definitely do that for the practice, but this is an assignment for school that needs to be done entirely in jquery, there is no css for this page. \$\endgroup\$
    – Krista P
    Dec 19, 2016 at 6:01
  • \$\begingroup\$ @nnnnnn Thanks, this works. I'm literally on day 2 of jquery so I don't quite understand how this is working but I'll start picking it apart to figure it out:) \$\endgroup\$
    – Krista P
    Dec 19, 2016 at 6:24
  • \$\begingroup\$ I've expanded a little on that comment in an answer (so have deleted the comment). \$\endgroup\$
    – nnnnnn
    Dec 19, 2016 at 6:32
  • \$\begingroup\$ You can use the same approach, with jQuery hover() and CSS classes and toggleClass in callbacks. \$\endgroup\$
    – Tushar
    Dec 19, 2016 at 6:47

1 Answer 1

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The first thing I would do to make the code shorter is to take advantage of the fact that many jQuery methods like .attr() accept a function as the second argument, allowing you to replace this:

var src = $(this).attr("src").replace("1.jpg", "2.jpg");
$(this).attr("src", src);

...with this:

$(this).attr("src", function(i, v) { return v.replace("1.jpg", "2.jpg"); });

(and similar in the other function.) What this does is call the function in the second argument with the current value of the attribute as the v argument, so you don't need a separate call to .attr() to find out what the current value is, and you don't need the src variable.

Then I'd change the .hover() code to pass a single function to run on mouseenter and mouseleave, and use the event object to check whether it is mouseenter or mouseleave:

$(document).ready(function(){
  $("img").hover(function(e) {
    $(this).attr("src", function(i, v) {
      return v.replace(/[12]\.jpg/, (e.type === "mouseenter" ? "2" : "1") + ".jpg");
    });
  });
});

This uses a regular expression /[12]\.jpg/ with .replace() to find a string ending with "1.jpg" or "2.jpg", and replace it with a value depending on the event.type.

Demo: https://jsfiddle.net/g5ncjjxo/

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  • \$\begingroup\$ @Tushar - I didn't mention CSS in my answer because of the comments above about having to do it without CSS. But how would you use CSS classes to change images with different sources? \$\endgroup\$
    – nnnnnn
    Dec 19, 2016 at 7:17
  • \$\begingroup\$ Fair Enough. Using CSS, instead of using <img /> use other element and add class to it. The CSS Class will have background-image: 1.jpg and some other positioning styles. Another class will have background-image: 2.jpg which can be used to change the background image. In hover() callbacks $(this).toggleClass('originalImage hoverImage') \$\endgroup\$
    – Tushar
    Dec 19, 2016 at 7:22
  • \$\begingroup\$ @Tushar - I meant, how would you apply that to multiple different elements in the page, each of which has its own pair of images, without having to create classes for each element? \$\endgroup\$
    – nnnnnn
    Dec 19, 2016 at 9:27

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