我在页面上显示许多图片。我想创建一个动画,在屏幕中间居中显示用户click
的图片。然后,在任何click
事件之后,我想显示下一张图片,直到用户离开此演示模式。
注意:我将jquery scrollTo脚本添加到我的代码中。
以下是我迄今所做的工作:http://jsfiddle.net/qnQSP/3/
HTML
<div id="galleries">
<div id="pictures-content" class="1"><img src="http://www.sb-designs.co.uk/ckfinder/userfiles/images/free%20web%20hosting.jpg"></div>
<div id="pictures-content" class="2"><img src="http://www.mastercreations.net/wp-content/uploads/2012/10/5.jpg"></div>
<div id="pictures-content" class="3"><img src="http://www.sb-designs.co.uk/ckfinder/userfiles/images/free%20web%20hosting.jpg"></div>
<div id="pictures-content" class="4"><img src="http://www.webdesign4essex.co.uk/images/essex_website_design.jpg"></div>
<div id="pictures-content" class="5"><img src="http://www.mastercreations.net/wp-content/uploads/2012/10/5.jpg"></div>
</div>
Jquery
var next;
var element_already_focus = 0;
var oldcurrent = "";
$("#galleries #pictures-content").unbind("click");
$("#galleries #pictures-content").bind("click", function(event) {
// Count the number of pictures
var max_number_of_pictures= $('#galleries #pictures-content').length;
// console.log("max_number_of_pictures: "+max_number_of_pictures);
// Get the binding element class number
var picture_number = $(this).attr('class');
// console.log("picture_number: "+picture_number);
// Save the element inside the current variable
current=$(this);
// Do a loop to go to the top picture when
if(picture_number==max_number_of_pictures)
{
next = $("#galleries .1");
} else
{
next = $(this).next();
}
// Do a loop to go to the bottom picture
if(picture_number==1){
previous = $("#galleries ."+max_number_of_pictures);
} else { previous = $(this).prev();}
// console.log("current: "+$(this).attr("class"));
if(oldcurrent != "")
{
console.log("old: "+oldcurrent.attr("class"));
// Doing some test
class_test = parseInt(oldcurrent.attr("class"));
class_test = parseInt(class_test)+parseInt(1);
console.log("old: "+class_test);
}
console.log("previous: "+previous.attr("class"));
console.log("current: "+current.attr("class"));
console.log("next: "+next.attr("class"));
if(oldcurrent == "")
{
$("body").scrollTo(current, 800);
next = 0;
console.log("What we do:"+next);
}
else{
// If the element that we are binding have the same class as the old binding element (previous)
// We are redirecting the user to the next picture
console.log("oldcurrent "+oldcurrent.attr("class"));
console.log("current: "+current.attr("class"));
console.log("next: "+next.attr("class"));
// if((oldcurrent.attr("class"))==($(this).attr("class"))||(class_test)==(next.attr("class"))) {
if((oldcurrent.attr("class"))==(current.attr("class"))||(class_test)==(next.attr("class"))) {
$("body").scrollTo(next, 800);
next = 1;
console.log("What we do:"+next);
}else{
$("body").scrollTo(current, 800);
next = 0;
console.log("What we do:"+next);
}
}
oldcurrent=current;
});
我的问题:我必须在图片上点击两次才能转到下一张图片。如果我删除了这个条件,当我离开演示模式时,我就无法对另一张照片进行对焦。
我认为我必须在用户处于演示模式时(已经点击图片)定义状态,并在他离开演示模式时删除此状态。
有人有任何想法或解决方案吗?
答案 0 :(得分:1)
单击时只验证图像是否在渲染视图中。如果是,那么您可以假设点击是滚动到下一个图像。
您可以使用$(yourImage).position().top
了解图片的起始位置。并且$(yourImage).height()
知道它的高度。
var imageTop = $(yourImage).position().top,
imageHeight = $(yourImage).height(),
windowTop = $(yourImage).height(),
windowHeight = $(window).height();
if(imageTop>windowTop && top+height < windowTop+windowHeight){
Scroll to the next one
}
或者,如果您希望图像始终位于顶部:
if(imageTop==windowTop){
Scroll to the next one
}
+1用于链接小提琴