2 回答
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TA贡献1860条经验 获得超9个赞
发现孩子直接使用Element.querySelector()带有属性选择器,并将其删除:
function removeChild(dataID) {
const child = document.querySelector(`[data-id='${dataID}']`);
if(child) child.remove();
};
removeChild(2);
<div class="parent-div">
<div class="child-item" data-id="1"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="3"> SomeContent </div>
</div>
如果可以有多个具有相同属性的项目,请使用Document.querySelectorAll(),然后使用 forEach 迭代结果,并删除项目:
function removeChild(dataID) {
const children = document.querySelectorAll(`[data-id='${dataID}']`);
children.forEach(el => el.remove());
};
removeChild(2);
<div class="parent-div">
<div class="child-item" data-id="1"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="2"> SomeContent </div>
<div class="child-item" data-id="3"> SomeContent </div>
</div>
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