Javascript 如何检查其他divs类属性?
我正在尝试制作简单的图片 到现在为止,我已经成功地添加/删除了存储在数组中的div元素上的活动类 我的问题是,每当我在特定的div中添加或删除类(选中该div)时,其他类都会保持不变,直到函数出现在同一个div中 我不知道如何在函数indexPlus触发时检查其他divs类。我在考虑for循环中的一些内容:Javascript 如何检查其他divs类属性?,javascript,html,slider,Javascript,Html,Slider,我正在尝试制作简单的图片 到现在为止,我已经成功地添加/删除了存储在数组中的div元素上的活动类 我的问题是,每当我在特定的div中添加或删除类(选中该div)时,其他类都会保持不变,直到函数出现在同一个div中 我不知道如何在函数indexPlus触发时检查其他divs类。我在考虑for循环中的一些内容: for (var i = 0; i < pictures.length; i++) { //here should be the code that ch
for (var i = 0; i < pictures.length; i++) {
//here should be the code that checks other "inactive (divs that have class
//"picture active" but do not have the the same index as the selected
// picture)" divs
}
目前,我的代码工作正常,但不是我所希望的那样。我连续两天在谷歌上搜索,试图让这一切顺利进行。也许我被错误的方向迷住了,没有用一种合乎逻辑的方式思考,我是JS的一个彻头彻尾的傻瓜。请随意更正我的代码和/或使其更高效。对不起,我的英语不是我的第一语言。谢谢你的时间和回答
以下是我的滑块HTML代码:
<div class="slider">
<div class="kontrola back">
<i class="fa fa-angle-left"></i>
</div>
<div class="picture" id="0" style="background-image: urlsomeUrl)">
</div>
<div class="picture" id="1" style="background-image: url(someUrl)">
</div>
<div class="picture" id="2" style="background-image: url(someUrl)">
</div>
<div class="picture" id="3" style="background-image: url(someUrl)">
</div>
<div class="picture" id="4" style="background-image: url(someUrl)">
</div>
<div class="kontrola next">
<i class="fa fa-angle-right"></i>
</div>
</div>
和我的JavaScript代码:
var back = document.getElementsByClassName('back')[0];
var next = document.getElementsByClassName('next')[0];
var picturesArray = []; //creating array of pictures
//populating array by getting ids
for (var i = 0; i < 5; i++) {
picturesArray[i] = document.getElementById(i);
}
//setting the first picture
var index = 0;
var firstPicture = document.getElementById(index);
var class2 = firstPicture.className;
if(class2 == class2) {
firstPicture.className += " active";
} else {
firstPicture.className -= " active";
}
//function that executes on click of next button
function indexPlus() {
if (index >= picturesArray.length - 1) { //this resets the index
index = 0; //this resets the index
var activePicture = document.getElementById(index);
var class1 = activePicture.className;
if(class1 == "picture active") { //this checks class of the active picture
activePicture.classList.remove("active"); //this removes active class
} else {
activePicture.className += " active"; //this adds active class
}
} else {
index++;
var activePicture = document.getElementById(index);
var class1 = activePicture.className;
if(class1 == "picture active") {
activePicture.classList.remove("active");
} else {
activePicture.className += " active";
}
}
//before mentioned possible for loop
for (var i = 0; i < picturesArray.length; i++) {
//code...
}
}
//function that executes on click of back button
function indexMinus() {
}
//event listeners
back.addEventListener('click', indexMinus);
next.addEventListener('click', indexPlus);
我通过创建另一个名为inactivePictures的数组解决了我的问题,该数组包含除活动索引以外的所有内容的图片数组
function indexPlus() {
//added code
var inactivePictures = [];
var notIndex = [0,1,2,3,4];
var notIndex1 = notIndex.splice(index, 1);
for (var i = 0; i < picturesArray.length; i++) {
inactivePictures[i] = document.getElementById(notIndex1);
}
if (index >= picturesArray.length - 1) { //this resets the index
index = 0; //this resets the index
var activePicture = document.getElementById(index);
var class1 = activePicture.className;
if(class1 == "picture active") { //this checks class of the active picture
activePicture.classList.remove("active"); //this removes active class
} else {
activePicture.className += " active"; //this adds active class
}
} else {
index++;
var activePicture = document.getElementById(index);
var class1 = activePicture.className;
if(class1 == "picture active") {
activePicture.classList.remove("active");
} else {
activePicture.className += " active";
}
}
//added code
for (var i = 0; i < picturesArray.length; i++) {
if(activePicture.className == inactivePictures[i].className) {
inactivePictures[i].classList.remove("active");
}
}
}
我的问题现在解决了,如果有任何进一步的建议,请随时发表评论 如果您使用的是jQuery,则可以使用.hasClass、.addClass和removClass方法。@Prashanth抱歉,我不能使用jQuery。谢谢你的回答。你能做一把小提琴吗?@John谢谢你的回答!虽然它并没有直接解决我的问题,但它让我以更先进的方式思考。