我是jQuery和堆栈溢出的新手,所以我会尝试具体,但请耐心等待。我正在尝试从头开始创建一个带有相关链接的文本滑块,使用模数迭代列表并重复。
这是我正在使用的代码:
ul#text { position: relative; margin-bottom: 40px; height: 40px; }
ul#text li { position: absolute; display: none; }
.active { font-weight: bold; }
<ul id="text">
<li id="textBody">Suffering is not a result of physical pain alone. It can be compounded by changes in one's life, and changes in the self. <em>We understand, and we can help.</em></li>
<li id="textFamily">Aggressive assessment of physical symptoms & pain in the body are key to support <em>the best possible quality of life</em>.</li>
<li id="textFunction">Chronic pain & illness may affect your role in your family. We work with you and your family as you confront those changes.</li>
<li id="textPsyche">Chronic pain and illness make even everyday activities challenging. We will help you maintain independence and physical function.</li>
<li id="textSuffering">Changes in the physical body mean changes in the self. We will provide support as you navigate those changes in the psyche.</li>
</ul>
<ul id="vivid_buttons">
<li><a href="#" id="buttonBody">BODY</a></li>
<li><a href="#" id="buttonFamily" class="active">FAMILY</a></li>
<li><a href="#" id="buttonFunction">FUNCTION</a></li>
<li><a href="#" id="buttonPsyche">PSYCHE</a></li>
<li><a href="#" id="buttonSuffering">SUFFERING</a></li>
</ul>
$(document).ready(function () {
function fadeAndMove() {
var nextLi = $("#text > li:nth-child(" + i % 5 + ")");
var nextA = $("#vivid_buttons > li:nth-child(" + i % 5 + ") > a");
nextLi.fadeIn(1000, function () {
$("#vivid_buttons > li > a").removeClass("active");
nextA.addClass("active");
setTimeout(function () {
nextLi.fadeOut(1000);
}, 4000);
});
}
for (i = 1; i < 7; i++) {
fadeAndMove($("#text"));
}
});
用简单的语言,我想淡化第一个列表中的一个句子,并突出显示底部列表中的相应链接。然后我希望它淡出并移动到下一个项目。
我认为我可以使用模数(%)和for循环来迭代并创建一个无限循环,但是当我把它放在它中时它就像它一次执行所有内容,而不是迭代(淡入和淡出)每个项目。
我知道这很令人困惑,但我很感激我能得到的任何帮助。
答案 0 :(得分:3)
摆脱for
循环,让setTimeout调用fadeAndMove()
函数,传递当前索引。
示例: http://jsfiddle.net/drWhE/
$(document).ready(function () {
// cache the LI elements
var $lis = $("#text > li");
var $aLis = $("#vivid_buttons > li");
function fadeAndMove( currentIndex ) {
var nextIndex = (currentIndex + 1) % 5;
var nextLi = $lis.eq( nextIndex );
nextLi.fadeIn(1000, function () {
$aLis.eq( currentIndex ).children('a').removeClass("active");
$aLis.eq( nextIndex ).children('a').addClass("active");
setTimeout(function () {
nextLi.fadeOut(1000, function() {
// Call fadeAndMove() passing nextIndex as the new currentIndex
fadeAndMove( nextIndex );
});
}, 4000);
});
}
// Get it started on index 0
fadeAndMove( 0 );
});
答案 1 :(得分:0)
一切都是动画,因为你的主循环一直在运行而你的淡出计时器等了四秒钟。
示意图,它是这样的(每行代表一秒):
li1.fadeIn
li2.fadeIn |
li3.fadeIn | |
li4.fadeIn | | | Timers
li5.fadeIn V | | | wait four
li1.fadeOut V | | | seconds
li2.fadeOut V | |
li3.fadeOut V |
li4.fadeOut V
li5.fadeOut
li1.fadeIn
li2.fadeIn
.
.
.
etc, etc, ad nauseam.
要解决此问题,您需要在淡出延迟函数中的当前项目之后立即将下一个调用链接到fadeAndMove()
:
nextLi.fadeIn(1000, function () {
$("#vivid_buttons > li > a").removeClass("active");
nextA.addClass("active");
setTimeout(function () {
nextLi.fadeOut(1000);
fadeAndMove(i + 1);
}, 4000);
});
(因为这是一个延迟调用,它不是递归的。无限循环不会粉碎堆栈。)
答案 2 :(得分:0)
这是获得您所需要的一种方式:
var $sentence_set = $('ul#text > li');
var $link_set = $('ul#vivid_buttons > li');
var highlight = function(which) {
var el = $sentence_set.eq(which - 1);
var position = el.prevAll('li').length;
$link_set.removeClass('active').eq(position).addClass('active');
$sentence_set.eq(position).siblings().fadeOut().end().fadeIn();
}
var loopcount = 0;
var repeater = setInterval(function() {
var theList = $('#text > li');
highlight(++loopcount % $sentence_set.length);
}, 4000);
这是the fiddle。
而且......橙色的酒吧告诉我帕特里克dw打败了我类似的东西。好吧,无论如何它就在这里。