Javascript 在不存在节点的上下文中引用该节点时发生DOM错误
我试图用DOM和javascript构建一个包含各种div的页面,如html中的:Javascript 在不存在节点的上下文中引用该节点时发生DOM错误,javascript,html,css,dom,Javascript,Html,Css,Dom,我试图用DOM和javascript构建一个包含各种div的页面,如html中的: <div id="divMain" class="divMain"> <div id="divT1" class="divT1"> onsite optimization </div> <div id="divR1" class="divR1"> <div id="divR11" class="divR11">img1</div>
<div id="divMain" class="divMain">
<div id="divT1" class="divT1"> onsite optimization </div>
<div id="divR1" class="divR1">
<div id="divR11" class="divR11">img1</div>
<a style="display:block" href="#">
<div id="divR12" class="divR12">btn1</div>
</a>
<div id="divR13" class="divR13">btn2</div>
</div>
</div>
但当我尝试此代码浏览器时,会显示一个警报:
读取响应时出错:NotFoundError:尝试在不存在节点的上下文中引用该节点
我认为是“oMRch.appendChild(oAR1ch);(最后一行)产生了错误,但为什么呢
提前谢谢
AM您可以使用DOM方法创建该结构,如下所示:
function createElement(tagName, id, className) {
var elm = document.createElement(tagName);
if (id) {
elm.id = id;
}
if (className) {
elm.className = className;
}
return elm;
}
var divMain = createElement('div', 'divMain', 'divMain');
var divT1 = createElement('div', 'divT1', 'divT1');
divT1.appendChild(document.createTextNode(" onsite optimization "));
divMain.appendChild(divT1);
var divR1 = createElement('div', 'divR1', 'divR1');
divMain.appendChild(divR1);
var divR11 = createElement('div', 'divR11', 'divR11');
divR11.appendChild(document.createTextNode('img1'));
divR1.appendChild(divR11);
var a = createElement('a');
a.style.display = "block";
a.href = "#";
divR1.appendChild(a);
var divR12 = createElement('div', 'divR12', 'divR12');
divR12.appendChild(document.createTextNode('btn1'));
a.appendChild(divR12);
var divR13 = createElement('div', 'divR13', 'divR13');
divR13.appendChild(document.createTextNode('btn2'));
divR1.appendChild(divR13);
document.body.appendChild(divMain);
或者您可以使用innerHTML
,它适用于所有浏览器:
var divMain = document.createElement('div');
divMain.className = divMain.id = "divMain";
divMain.innerHTML =
'<div id="divT1" class="divT1"> onsite optimization </div>' +
'<div id="divR1" class="divR1">' +
'<div id="divR11" class="divR11">img1</div>' +
'<a style="display:block" href="#">' +
'<div id="divR12" class="divR12">btn1</div>' +
'</a>' +
'<div id="divR13" class="divR13">btn2</div>' +
'</div>';
document.body.appendChild(divMain);
var divMain=document.createElement('div');
divMain.className=divMain.id=“divMain”;
divMain.innerHTML=
“现场优化”+
'' +
“img1”+
'' +
“btn2”+
'';
document.body.appendChild(divMain);
为什么要查找已有引用的元素?例如,您刚刚将oRt1
放在DOM中,那么为什么要document.getElementById(“divRt1”)呢
?您的oRt1
变量引用了它。您的代码使用了与您引用的结构不同的id
值,这使得它几乎无法遵循。例如,您的代码使用了id
“divRt1”
,它在示例结构中没有出现。
var divMain = document.createElement('div');
divMain.className = divMain.id = "divMain";
divMain.innerHTML =
'<div id="divT1" class="divT1"> onsite optimization </div>' +
'<div id="divR1" class="divR1">' +
'<div id="divR11" class="divR11">img1</div>' +
'<a style="display:block" href="#">' +
'<div id="divR12" class="divR12">btn1</div>' +
'</a>' +
'<div id="divR13" class="divR13">btn2</div>' +
'</div>';
document.body.appendChild(divMain);