当我到达滚动图像的末尾时,我正在尝试使我的javascript / jquery滑块按钮停用(当图像一直向右移动时,必须停用MoveRight按钮并且只保留MoveLeft按钮激活,同样为移动LeftButton),任何人都可以帮忙吗?我不确定我是否正在使用 .attr()和removeAttr()正确。我在下面粘贴了我的代码。
<script type="text/javascript">
$(document).ready(function(){
//Check width of Gallery div
var galleryWidth = $("#Gallery").innerWidth();
//Check width of GalleryItem
var NoListed = $("ul.GalleryItem li").length;
var galleryItemWidth = 1881;
$('.MoveRight')
$('.GalleryItem').css('width', galleryItemWidth);
//Check width of Gallery div on resize
$(window).resize(function(){
var galleryWidth = $("#Gallery").innerWidth();
});
$('.MoveLeft').click(function() {
$(".GalleryItem2").animate({"left": "-=350px"}, "slow");
$(".GalleryItem3").animate({"left": "-=280px"}, "slow");
$(".GalleryItem").animate({
left: '-=230',
}, "slow", function() {
position = $(".GalleryItem").position();
galleryItemLeft = position.left;
if(galleryItemLeft <= galleryWidth - galleryItemWidth) {
$('.MoveLeft').removeAttr('disabled');}
else{
$('.MoveLeft').attr('disabled','disabled');
$('.MoveRight').attr('disabled','disabled');
}
});
});
$('.MoveRight').click(function() {
$(".GalleryItem2").animate({"left": "+=350px"}, "slow");
$(".GalleryItem3").animate({"left": "+=280px"}, "slow");
$(".GalleryItem").animate({
left: '+=230',
}, "slow", function() {
position = $(".GalleryItem").position();
galleryItemLeft = position.left;
if(galleryItemLeft >= "0") {
$('.MoveLeft').removeAttr('disabled');}
else{
$('.MoveLeft').attr('disabled','disabled');
$('.MoveRight').attr('disabled','disabled');
}
});
});
});
</script>
答案 0 :(得分:0)
Okey,首先删除这个逗号;
left: '-=230',
-------------^
并在此处添加px
left: '-=230px'
------------^
然后尝试添加此内容;
if(galleryItemLeft <= 0) {
return false;
}
在$('.MoveLeft').click(function() {
之后
和正确的一样