Question

I am developing a wordpress theme. I am working on the theme options page right now. I added farbtastic (4 fields) and the problem is that every time I click on the input, the color picker appears on the other 3 fields too. Anybody knows how to fix this? Thank you!

<div> <br />
  <label for="<?php echo $colorPicker['ID']; ?>"><?php _e($colorPicker['label']); ?></label>
  <input type="text" class="color-picker" id="<?php echo $colorPicker['ID']; ?>" value="<?php echo get_option($colorPicker['ID']); ?>" name="<?php echo $colorPicker['ID']; ?>" />
  <div id="<?php echo $colorPicker['ID']; ?>_color" class="fabox"></div>            </div>          
<?php endforeach; ?>            
<p><input type="submit" name="update_options" value="Update Options" class="button-primary" /></p>
</form>
</div>

<script type="text/javascript">
jQuery(document).ready(function($) {
    var colorPickers = $('.color-picker');
    console.log(colorPickers);
    for (e in colorPickers) {
        if (colorPickers[e].id != undefined) {
            var colorPickerID = colorPickers[e].id;
            $('#' + colorPickerID + '_color').farbtastic('#' + colorPickerID);
        }
    }

    $('.fabox').hide();


    $('.color-picker').click(function() {
        $('.fabox').fadeIn();
    });

    $(document).mousedown(function() {
        $('.fabox').each(function() {
            var display = $(this).css('display');
            if (display == 'block') $(this).fadeOut();
        });
    });
});​
</script>

HTML OUTPUT:

<form method="POST" action="">  

                        <div>

            <br />

            <label for="color_1"><strong>Post Title</strong></label>

            <input type="text" class="color-picker" id="color_1" value="#273990" name="color_1" />

            <div id="color_1_color" class="fabox"></div>

            </div>

                        <div>

            <br />

            <label for="color_2"><strong>Paragraph Text</strong></label>

            <input type="text" class="color-picker" id="color_2" value="#840000" name="color_2" />

            <div id="color_2_color" class="fabox"></div>

            </div>

                        <div>

            <br />

            <label for="color_3"><strong>Example</strong></label>

            <input type="text" class="color-picker" id="color_3" value="#4377df" name="color_3" />

            <div id="color_3_color" class="fabox"></div>

            </div>

                        <div>

            <br />

            <label for="color_4"><strong>And Another Example</strong></label>

            <input type="text" class="color-picker" id="color_4" value="#3c8400" name="color_4" />

            <div id="color_4_color" class="fabox"></div>

            </div>

                        <p><input type="submit" name="update_options" value="Update Options" class="button-primary" /></p>

        </form>

    </div>

Answer 1

You’re referencing too broad of an element with your jQuery selector. Essentially your code says every time you click anything with the color-picker class, show anything with the fabox class.

You should make your reference more specific to the currently clicked .color-picker.

I recommend replacing this:

$('.fabox').fadeIn();

With this:

$(this).parent().find('.fabox').fadeIn();

So you are only referencing the .fabox that is connected to the .color-picker you just clicked.

EDIT: As gillesc noted, it would actually be quicker to use:

$(this).next().fadeIn();

So long as the the .fabox always follows the .color-picker.

If the .fabox was inside the same container, but not the very next element you could use:

$(this).next('.fabox').fadeIn();