JavaScript/jQuery 中的叠加层和图像

发布于 2024-12-22 20:37:33 字数 515 浏览 2 评论 0原文

我试图在 X=100 & 处显示 16 x 16 像素图像Y=100 但我卡住了。该图像将显示在页面上任何内容的顶部。更改顶部和左侧属性不会执行任何操作。

无论我做什么，图像似乎都显示在 pos 0,0 处。

shimDiv.style.position = 'absolute';
    shimDiv.style.top = 100;
    shimDiv.style.left = 100;
    shimDiv.style.width = "1000px"; // todo: change me!
    shimDiv.style.height = "1000px"; // todo: change me!
    shimDiv.style.zIndex = 3;
    document.body.appendChild(shimDiv);
    $("#shim").append('<img id="icon" src="../Images/info.png" alt="todo:" />');

原文

I'm trying to show 16 by 16 pixel image at X=100 & Y=100 but am stuck. The image is to be displayed over the top of whatever is on the page. Altering the top and left properties don't do anything.

Whatever I do, the image seems to show up at pos 0,0.

shimDiv.style.position = 'absolute';
    shimDiv.style.top = 100;
    shimDiv.style.left = 100;
    shimDiv.style.width = "1000px"; // todo: change me!
    shimDiv.style.height = "1000px"; // todo: change me!
    shimDiv.style.zIndex = 3;
    document.body.appendChild(shimDiv);
    $("#shim").append('<img id="icon" src="../Images/info.png" alt="todo:" />');

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蝶舞 2024-12-29 20:37:33

这是因为您没有指定 top 和 left 位置的单位：

shimDiv.style.top = '100px';
shimDiv.style.left = '100px';

That's because you've not specified the units for the top and left positions:

shimDiv.style.top = '100px';
shimDiv.style.left = '100px';

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鸢与 2024-12-29 20:37:33

嗯，也许这过于简单化了，但这应该可以满足您的要求：

$('body').append('<img id="icon" src="../Images/info.png" style="position: absolute; top: 100px; left: 100px; width: 1000px; height: 1000px; z-index: 1000;" />');

在您的代码示例中似乎缺少父 div 元素的创建代码，但上面的代码应该是必需的。

希望有帮助...

Hmm, maybe this is an oversimplification, but this should accomplish what you are asking:

$('body').append('<img id="icon" src="../Images/info.png" style="position: absolute; top: 100px; left: 100px; width: 1000px; height: 1000px; z-index: 1000;" />');

It seems in your code sample that the creation code for the parent div element is missing, but the above code should be all that is necessary.

Hope that helps...

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