我想要将五个图像作为图库进行迭代。因此,当用户单击“前进按钮”时,数组中的下一个图像将显示在div中。并且当点击“后退按钮”时。对不起,如果这是愚蠢的编码,但我在这里很新。
$(document).ready(function () {
// when thumbnail is clicked
$("img#thumb").click(function () {
var imgpath = $(this).attr("src");
$("img#cover").attr("src", imgpath);
});
$(function () {
$("img#thumb").bind('click', function () {
var img = $(this).clone(); //this clones the img, so you do not need to re-load that using src path
$("img#cover").hide().html(img).fadeIn(500); //remove .hide() segment if you just are hiding by css rules
});
});
//when the nextimage link is clicked
var imgs = ["images/exteriors/abandonned/img0.jpg",
"images/exteriors/abandonned/img1.jpg",
"images/exteriors/abandonned/img2.jpg"];
$("img#nextimage").click(function () {
$.each(imgs, function () {
$("img#cover").attr("src", imgs); // I want to iterate each image and display when it is clicked
});
});
});
HTML:
<div id="thumbs"> <!--gallery thumbs-->
<img id="thumb1" src="images/exteriors/abandonned/img0.jpg" width="100px" height="80px" class="" /><br>
<img id="thumb2" src="images/exteriors/abandonned/img1.jpg" width="100px" height="80px" class="" /><br>
<img id="thumb3" src="images/exteriors/abandonned/img2.jpg" width="100px" height="80px" class="" /><br>
<img id="thumb" src="images/exteriors/abandonned/img3.jpg" width="100px" height="80px" class="" /><br>
<img id="thumb" src="images/exteriors/abandonned/img3.jpg" width="100px" height="80px" class="" /><br>
</div>
答案 0 :(得分:0)
每次点击下一个或上一个按钮时,您无需迭代图像;相反,使用变量将索引存储到imgs数组中,让您的单击处理程序更新该索引,然后使用数组中该索引处的URL更新标记的src属性。这样的事情,也许是:
var imgindex = 0;
$('img#nextimage').click(function() {
imgindex++; // increment the index into array 'imgs'
if (imgindex > (imgs.length - 1)) { imgindex = 0; }; // if we just walked off the far end of the array, reset the index to zero (loop around to the 1st image)
$('img#cover').attr('src', imgs[imgindex]); // update the img src
});
$('img#previmage').click(function() {
imgindex--; // decrement the array index
if (imgindex < 0) { imgindex = (images.length - 1); }; // if we just walked off the near end of the array, loop around to the last image
$('img#cover').attr('src', imgs[imgindex]);
});