CODEWITHSUNDEEP
jqueryimageswapslidetoggle
ja408
ja408

Reputation: 808

jquery slidetoggle image swap

I'm trying to get slideToggle to swap images based on if a div is expanded or collapsed. I have it working one way, where it will change an image from plus to minus, but it only works on one div and I have multiple and it only works once. It doesn't toggle.

Here is the jQuery:

$(".content").toggle(1);
$(".expand").click(function () {
    $(this).next(".content").slideToggle(400);

    if ($(".image1").attr('src', "../nathan/plus.png")) {
        $(".image1").attr(
            'src', 
            $(".image1").attr('src').replace('plus', 'minus')
        );
    } else {
        $(".image1").attr(
            'src', 
            $(".image1").attr('src').replace('minus', 'plus')
        );
    }
});
});

The IndexOf() works, but if I was to duplicate the divs they all change when one is clicked. Here is the HTML:

<div class="list">
<h2 class="expand">Title1<img class="image1" src="../nathan/plus.png"></h2>
<div class="content">
</div>
</div>

<div class="list">
<h2 class="expand">Title2<img class="image1" src="../nathan/plus.png"></h2>
<div class="content">
</div>
</div>

Upvotes: 3

Views: 2856

Answers (1)

wirey00
wirey00

Reputation: 33661

You are using the setter method to check your `src'

if ($(".image1").attr('src', "../nathan/plus.png"))

instead use

if ($(".image1").attr('src') == "../nathan/plus.png"))

Or even better just check if the source contains the image using indexOf()

if ($(".image1").attr('src').indexOf("plus.png") > -1)

EDIT:

You can probably do it like this to prevent all from changing - this is the context the to look for the image

$(".expand").click(function() {
    $(this).next(".content").slideToggle(400);
    var $img = $(".image1", this);
    if ($img.attr('src').indexOf("plus.png") > -1) {
        $img.attr('src', $img.attr('src').replace('plus', 'minus'));
    } else {
        $img.attr('src', $img.attr('src').replace('minus', 'plus'));
    }
});​

so

$(".image1", this);

would be the same as

$(this).find('.image1');

Upvotes: 2

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