我想使用jQuery自动滚动基于鼠标位置的div。
如果你看到这个fiddle here,你可以看到一些在可滚动的div中水平排序的图像:
<div id="parent">
<div id="propertyThumbnails">
<img src="http://www.millport.org/wp-content/uploads/2013/05/Flower-festival.jpg" />
<img src="http://www.millport.org/wp-content/uploads/2013/05/Flower-festival.jpg" />
<img src="http://www.millport.org/wp-content/uploads/2013/05/Flower-festival.jpg" />
<img src="http://www.millport.org/wp-content/uploads/2013/05/Flower-festival.jpg" />
<img src="http://www.millport.org/wp-content/uploads/2013/05/Flower-festival.jpg" />
</div>
</div>
CSS:
#parent {
height: 300px;
width: 100%;
background: #ddd;
}
#propertyThumbnails {
background: #666;
height: 80px;
white-space: nowrap;
overflow: scroll;
}
#propertyThumbnails img {
width: 125px;
height: 80px;
display: inline-block;
margin: 3px;
margin-right: 0;
opacity: 0.6;
}
我发现您可以使用$("#container").scrollLeft(position)
来设置滚动条的位置,但我想根据父级的鼠标位置来设置它。因此,当鼠标完全位于右侧时,将显示最右侧的图像,当鼠标完全左侧时,将显示最左侧的图像。
我该怎么做?
答案 0 :(得分:7)
实现您需要的方式略有不同:
jQuery(function($) {
$(window).load(function() {
var $gal = $("#propertyThumbnails"),
galW = $gal.outerWidth(true),
galSW = $gal[0].scrollWidth,
wDiff = (galSW / galW) - 1, // widths difference ratio
mPadd = 60, // Mousemove Padding
damp = 20, // Mousemove response softness
mX = 0, // Real mouse position
mX2 = 0, // Modified mouse position
posX = 0,
mmAA = galW - (mPadd * 2), // The mousemove available area
mmAAr = (galW / mmAA); // get available mousemove fidderence ratio
$gal.mousemove(function(e) {
mX = e.pageX - $(this).offset().left;
mX2 = Math.min(Math.max(0, mX - mPadd), mmAA) * mmAAr;
});
setInterval(function() {
posX += (mX2 - posX) / damp; // zeno's paradox equation "catching delay"
$gal.scrollLeft(posX * wDiff);
}, 10);
});
});
#parent {
position: relative;
margin: 0 auto;
width: 60%;
height: 260px;
}
#propertyThumbnails {
position: relative;
overflow: hidden;
background: #444;
width: 100%;
height: 262px;
white-space: nowrap;
}
#propertyThumbnails img {
vertical-align: middle;
height: 100%;
display: inline;
margin-left: -4px;
}
<div id="parent">
<div id="propertyThumbnails">
<img src="//placehold.it/600x400/0bf" />
<img src="//placehold.it/600x400/f0b" />
<img src="//placehold.it/600x400/0fb" />
<img src="//placehold.it/600x400/b0f" />
<img src="//placehold.it/600x400/bf0" />
</div>
</div>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
其中mPadd
是区域(在PX中,在左右边界区域),没有任何敏感性以防止用户沮丧:)
答案 1 :(得分:2)
至少应该让你朝着正确的方向前进。
var parent = $('#parent');
var img = $('img:first-child');
parent.on('mousemove', function(e) {
mouseX = e.pageX
img.css('margin-left',-mouseX/parent.width()*100);
});