我已经创建了幻灯片/轮播,但是我想根据屏幕尺寸显示偶数个图像。
因此,如果屏幕可以容纳5张图像,则滑块将显示5张等。问题是如果屏幕仅适合4个,我想删除第13、14和15个图像,以便一次在屏幕上显示4个图像,而最后不显示三个图像。
在此示例中:
var slides_div = document.getElementById("container");
for (var k=11; k<slides.length; k++) {
slides_div.removeChild(slides_div.childNodes[k]);
}
该位并没有删除任何元素,但是在我的计算机上却删除了图像5和6。在我的版本中,页面全部都是内联js / css,文件中没有其他内容。
对此将提供任何帮助或帮助。
var doc_width = $(document).outerWidth(true);
var slides = document.getElementsByClassName("slides");
if(doc_width > 1310){
no_slides = 5;
}else if(doc_width > 1060){
no_slides = 4;
var slides_div = document.getElementById("container");
for (var k=11; k<slides.length; k++) {
slides_div.removeChild(slides_div.childNodes[k]);
}
}else if(doc_width > 810){
no_slides = 3;
}else if(doc_width > 560){
no_slides = 2;
}else if(doc_width > 310){
no_slides = 1;
}
var first_slide = 0;
var last_slide = no_slides-1;
showDivs(first_slide,last_slide);
function prev_slides() {
showDivs(first_slide -= no_slides, last_slide -= no_slides);
}
function next_slides() {
showDivs(first_slide += no_slides, last_slide += no_slides);
}
function showDivs(f,l) {
if (l > slides.length) {
first_slide = 0;
last_slide = no_slides-1;
}
if (l < 1) {
first_slide = slides.length-no_slides;
last_slide = slides.length-1;
}
for (var i=0; i<slides.length; i++) {
$(slides[i]).css({"opacity": 0, "right":"+=250", "display":"none"});
}
for (var j=first_slide; j<=last_slide; j++){
$(slides[j]).css({"opacity": 1, "right":"0", "display":"inline"});
}
}
.container{
width:100%;
text-align:center;
}
.arrow{
vertical-align:middle;
width:30px;
}
.slides{
position:relative;
width:250px;
opacity:0;
right:-200px;
display:none;
vertical-align:middle;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<div class="container" id="container">
<img class="arrow" src="arrow_left.jpg" onclick="prev_slides()">
<img class="slides" src="https://via.placeholder.com/250x100/111111?text=1">
<img class="slides" src="https://via.placeholder.com/250x100/222222?text=2">
<img class="slides" src="https://via.placeholder.com/250x100/333333?text=3">
<img class="slides" src="https://via.placeholder.com/250x100/444444?text=4">
<img class="slides" src="https://via.placeholder.com/250x100/555555?text=5">
<img class="slides" src="https://via.placeholder.com/250x100/666666?text=6">
<img class="slides" src="https://via.placeholder.com/250x100/777777?text=7">
<img class="slides" src="https://via.placeholder.com/250x100/888888?text=8">
<img class="slides" src="https://via.placeholder.com/250x100/999999?text=9">
<img class="slides" src="https://via.placeholder.com/250x100/aaaaaa?text=10">
<img class="slides" src="https://via.placeholder.com/250x100/bbbbbb?text=11">
<img class="slides" src="https://via.placeholder.com/250x100/cccccc?text=12">
<img class="slides" src="https://via.placeholder.com/250x100/dddddd?text=13">
<img class="slides" src="https://via.placeholder.com/250x100/eeeeee?text=14">
<img class="slides" src="https://via.placeholder.com/250x100/ffffff?text=15">
<img class="arrow" src="arrow_right.jpg" onclick="next_slides()">
</div>
答案 0 :(得分:0)
由于某种原因,它在图像之间拾取了空对象。
然后在循环中,删除每个项目后对事物的索引将有所不同,因此必须向后循环。
将其更改为以下内容,并按照我想要的方式工作:
<div>Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nam ultricies.</div>
<div class="one-line">Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nam ultricies.</div>
答案 1 :(得分:-1)
您可以使用CSS媒体查询来隐藏最后3个HTML元素。
@media only screen and (max-width: 300px) {
.hideme {
display:none;
}
}
<div>1</div>
<div>2</div>
<div>3</div>
<div>4</div>
<div class="hideme">5</div>
<div class="hideme">6</div>
<div class="hideme">7</div>