所以,标题总结得很好。我正在敲打这个 我知道空错误通常会由于时间错误而发生,因为我要求它在加载之前在文档中找到某些东西。但是,在我的JS代码中,我能够无问题地获取ElementsByTagName。此外,js函数的整个级联从window.onload开始,因此文档准备就绪,尽可能接近我所知道的。
在.html文件中证明:
<section id="main">
<header>
<a href="#">Enter Site</a>
</header>
<nav class="links">
<ul>
<li><a href="#">Galleries</a></li>
<li><a href="#">Video</a></li>
<li><a href="#">Albums</a></li>
<li><a href="#">Pricing</a></li>
<li><a href="#">Testimonials</a></li>
<li><a href="#">Wedding Planner</a></li>
<li><a href="#">Shopping Cart</a></li>
</ul>
</nav>
<address>
New Accents Photography ♦
14 Mountain View Road ♦
Ashland, OR 97521 ♦
(541) 555-6002
</address>
<article>
<p>New Accents is photography studio specializing in wedding
portraiture; serving the
needs of Southern Oregon and Northern California.
</p>
<p>We'll capture your wedding in whatever fashion you wish:
from romantic to zany and all the styles in between.
</p>
</article>
<footer>
<span class="left">New Accents Photography © 2015 (US)</span>
<span class="right">
<a href="#">About</a>
<a href="#">Terms</a>
<a href="#">Help</a>
</span>
</footer>
</section>
来自.js文件:
function displayThumbs(){
var figBox = document.createElement("figure");
figBox.setAttribute("id", "thumbnails");
for(var i = 0; i < allSheets.length; i++)
{
var sheetImg = document.createElement("img");
var imgName = allSheets[i].title + "_small.png";
sheetImg.setAttribute("src", imgName);
sheetImg.title = allSheets[i].title;
sheetImg.onclick = showSheet();
figBox.appendChild(sheetImg);
}
var thisMain = document.getElementById("main");
thisMain.appendChild(figBox);
}