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javascript - jQuery 绑定(bind)/取消绑定(bind)不起作用

转载 作者:行者123 更新时间:2023-12-03 06:31:03 25 4
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我正在尝试创建一个简单的 slider 。这是一个示例,但 slider 下一个和上一个按钮无法正常工作。

// next
var next = $('.next').click(function() {
var storepos = $(".storepos").val();
$('.prev').bind('click');
$('.storepos').val($('.storepos').val() / 1 + 110);
$('.container').animate({
scrollLeft: $('.storepos').val()
}, 200);
});

//prev
$('.prev').click(function() {
var storepos = $(".storepos").val();
$('.next').bind('click');
$('.storepos').val($('.storepos').val() / 1 - 110);
$('.container').animate({
scrollLeft: $('.storepos').val()
}, 200);
});

//after scrollend right event
$('.container').bind('scroll', function() {
if ($('.container').scrollLeft() + $(this).innerWidth() >= $(this)[0].scrollWidth) {
$('.next').unbind('click');
}
});
//after scrollend left event
$('.container').bind('scroll', function() {
if ($('.container').scrollLeft() < 1) {
$('.prev').unbind('click');
}
});
.container {
overflow: hidden !important
}

.container::-webkit-scrollbar {
width: 0;
height: 0
}

.content {
width: 1600px
}

.items {
background: black;
color: white;
margin-left: 10px;
width: 100px;
height: 100px;
float: left;
text-align: center
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
<div class="content">
<div class="items">1</div>
<div class="items">2</div>
<div class="items">3</div>
<div class="items">4</div>
<div class="items">5</div>
<div class="items">6</div>
<div class="items">7</div>
<div class="items">8</div>
<div class="items">9</div>
<div class="items">10</div>
</div>
</div>
<a href="#" class="prev">Prev</a> / <a href="#" class="next">Next</a>

<input class="storeposx" value="" />
<input class="storepos" value="" />

fiddle

最佳答案

我看到两个错误。首先,前一个按钮从请求开始就处于事件状态,可以滚动到负值。其次,当到达两端时,您确实会解除事件的绑定(bind),但之后您不会将它们绑定(bind)回来。

我使用了两个变量来保存按钮状态。当我到达开始或结束位置时,我不会解除它们的绑定(bind),而是在单击时返回 false。

// next
var next = $('.next').click(function() {
if (!nextIsActive || $('.container').is(':animated')) return false;
var storepos = $(".storepos").val();
$('.prev').bind('click');
$('.storepos').val($('.storepos').val() / 1 + 110);
$('.container').animate({
scrollLeft: $('.storepos').val()
}, 200);
});

//prev
$('.prev').click(function() {
if (!prevIsActive || $('.container').is(':animated')) return false;
var storepos = $(".storepos").val();
$('.next').bind('click');
$('.storepos').val($('.storepos').val() / 1 - 110);
$('.container').animate({
scrollLeft: $('.storepos').val()
}, 200);
});

var nextIsActive=true;
var prevIsActive=false;
//after scrollend right event
$('.container').bind('scroll', function() {
if ($('.container').scrollLeft() + $(this).innerWidth() >= $(this)[0].scrollWidth) {
nextIsActive=false;
}else{
nextIsActive=true;
}
});
//after scrollend left event
$('.container').bind('scroll', function() {
if ($('.container').scrollLeft() < 1) {
prevIsActive=false;
}else{
prevIsActive=true;
}
});
.container{overflow:hidden !important}
.container::-webkit-scrollbar {
width:0;
height:0
}
.content {width:1600px}
.items { background:black;
color:white;
margin-left:10px;
width:100px;
height:100px;
float:left;
text-align:center
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
<div class="content">
<div class="items">1</div>
<div class="items">2</div>
<div class="items">3</div>
<div class="items">4</div>
<div class="items">5</div>
<div class="items">6</div>
<div class="items">7</div>
<div class="items">8</div>
<div class="items">9</div>
<div class="items">10</div>
</div>
</div>
<a href="#" class="prev">Prev</a> / <a href="#" class="next">Next</a>

<input class="storeposx" value="" />
<input class="storepos" value="" />

关于javascript - jQuery 绑定(bind)/取消绑定(bind)不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38451010/

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