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使用Java脚本删除动态元素

更新时间:2023-12-05 18:38:10

这是您要实现的目标吗?如果是这样,当您在 removeElement 函数中混合ID时。

Is this what you are trying to achieve? If so, when you mixed up ids in removeElement function.

//+Element
function addElement(div) {
  var ni = document.getElementById(div);
  var numi = document.getElementById('numVal');
  var num = (document.getElementById('numVal').value -1)+ 2;
  numi.value = num;
  var newdiv = document.createElement('div');
  newdiv.setAttribute('id',num);
  var unum="'"+div+"','"+num+"'";
  newdiv.innerHTML ='<input id="t" type="tel" placeholder="Tel.'+num+'" name="t'+num+1+'"><img id="del" onClick="removeElement('+unum+');" alt="del" src="images/del.gif" />';
  ni.appendChild(newdiv);
}
//-Element
function removeElement(divNum,div) {
  var d = document.getElementById(divNum);
  var olddiv = document.getElementById(div);
  d.removeChild(olddiv);
}

<form>
    <input type="hidden" value="0" id="numVal" />
    <div id="m">
        <div id="h125">
            <div class="fb">
                <input id="t" type="tel" placeholder="Tel.">
                <img id="add" Alt="add" src="images/add.gif" onclick="addElement('h125');" />
            </div>
        </div>
    </div>
    </form>