如何复制 div onclick 事件?

时间:2023-02-11
本文介绍了如何复制 div onclick 事件?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我希望在单击按钮时复制一个 div.我虽然是这样的;但它不起作用.谁能帮帮我?

I want a div to be duplicated when a button is clicked. I though something like this; but it's not working. Can anyone help me?

HTML

<div id="duplicater"> 
duplicate EVERYTHING INSIDE THIS DIV
</div>

JAVASCRIPT

function duplicate()
{
var div = duplicate("div");
    div.id = "duplicater";
div.appendChild(duplicate("duplicater"));
}

推荐答案

你正在创建一个无限递归!

You are creating an infinite recursion!

function duplicate()
{
    var div = duplicate("div");

函数一遍又一遍地调用自己.使用 cloneNode():

The function is calling itself over and over again. Use cloneNode():

HTML:

<div id="duplicater0"> 
duplicate EVERYTHING INSIDE THIS DIV
</div>

JavaScript:

JavaScript:

var i = 0;

function duplicate() {
    var original = document.getElementById('duplicater' + i);
    var clone = original.cloneNode(true); // "deep" clone
   clone.id = "duplicater" + ++i; // there can only be one element with an ID
    clone.onclick = duplicate; // event handlers are not cloned
    original.parentNode.appendChild(clone);
}

工作演示

或者没有 ID:

function duplicate() {
    var clone = this.cloneNode(true); // "deep" clone
    clone.id = ""; // there can only be one element with an ID
    clone.onclick = duplicate; // event handlers are not cloned
    this.parentNode.appendChild(clone);
}

更新:

如果你想在按钮点击时克隆 div,你可以使用稍微不同的版本:

If you want to clone the div on button click, you can use a slightly different version:

HTML:

<button id="button" onclick="duplicate()">Click me</button>
<div id="duplicater"> 
    duplicate EVERYTHING INSIDE THIS DIV
</div>

JavaScript:

JavaScript:

var i = 0;
var original = document.getElementById('duplicater');

function duplicate() {
    var clone = original.cloneNode(true); // "deep" clone
    clone.id = "duplicater" + ++i;
    // or clone.id = ""; if the divs don't need an ID
    original.parentNode.appendChild(clone);
}

如果你不在表单中,你应该使用 <button> 而不是 <input type="button">.

If you are not in a form, you should use <button> instead of <input type="button">.

工作演示 2

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