Question

我正在尝试创建一个脚本，用于在mouseover事件上更改元素的重复背景图像。不幸的是它无法正常工作。我已经找到了几种可能的方法来使用JavaScript，但它们都没有对我有用。我该如何解决这个问题？

以下代码无法正常运行：

    while (document.getElementById("content_" + modid + "_" + i) != null) {
        document.getElementById("content_" + modid + "_" + i).style.display = "none";
        document.getElementById("menu_" + modid + "_" + i).style.backgroundImage = "url(psycho_normal.jpg)";
        document.getElementById("menu_" + modid + "_" + i).style.backgroundPosition = "top left";
        document.getElementById("menu_" + modid + "_" + i).style.backgroundRepeat = "repeat-x";
        i++;
    }
    document.getElementById("menu_" + modid + "_" + ind).style.backgroundImage = "url(phycho_hover.jpg)";
    document.getElementById("menu_" + modid + "_" + ind).style.backgroundPosition = "top left";
    document.getElementById("menu_" + modid + "_" + ind).style.backgroundRepeat = "repeat-x";

但是如果我尝试使用backgroundColor属性，它可以正常工作：

    while (document.getElementById("content_" + modid + "_" + i) != null) {
        document.getElementById("content_" + modid + "_" + i).style.display = "none";
        document.getElementById("menu_" + modid + "_" + i).style.backgroundColor = "#000000";
        i++;
    }
    document.getElementById("menu_" + modid + "_" + ind).style.backgroundColor = "#ff0000";

Answer 1

写一个CSS类并在你的JavaScript中调用它

document.getElementById("menu_" + modid + "_" + i).className = "yourcssclass"

看看会发生什么。

Answer 2

此代码适用于我。也许你的代码中有错误？尝试在浏览器中启用JavaScript控制台，看看是否记录了任何内容。

<div id="menu_a_0" onmouseover="doit(0);" style="width:200px;height: 200px;">aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa</div>
<div id="menu_a_1" onmouseover="doit(1);" style="width:200px;height: 200px;">aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa</div>
<div id="menu_a_2" onmouseover="doit(2);" style="width:200px;height: 200px;">aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa</div>

<div id="content_a_0"></div>
<div id="content_a_1"></div>
<div id="content_a_2"></div>

<script>
    function doit(ind) {
        modid = "a";
        i = 0;

        while (document.getElementById("content_" + modid + "_" + i) != null) {
            document.getElementById("content_" + modid + "_" + i).style.display = "none";
            document.getElementById("menu_" + modid + "_" + i).style.backgroundImage = "url(psycho_normal.jpg)";
            document.getElementById("menu_" + modid + "_" + i).style.backgroundPosition = "top left";
            document.getElementById("menu_" + modid + "_" + i).style.backgroundRepeat = "repeat-x";
            i++;
        }
        document.getElementById("content_" + modid + "_" + ind).style.display = "block";
        document.getElementById("menu_" + modid + "_" + ind).style.backgroundImage = "url(phycho_hover.jpg)";
        document.getElementById("menu_" + modid + "_" + ind).style.backgroundPosition = "top left";
        document.getElementById("menu_" + modid + "_" + ind).style.backgroundRepeat = "repeat-x";

        return true;
    }
</script>

Answer 3

Homour me，

如果您尝试使用简单标记显示图像会发生什么？你看到了吗？

即

<img src="phycho_hover.jpg" />

另外，您对getElementById的多次调用无助于您的可读性或性能尝试类似这样的事情：

var objElem = document.getElementById("content_" + modid + "_" + i); 
while (objElem  != null) {
    objElem.style.display = "none";   
    objElem.style.backgroundImage = "url('psycho_normal.jpg')";
    objElem.style.backgroundPosition = "top left";
    objElem.style.backgroundRepeat = "repeat-x";
    i++;
    objElem = document.getElementById("content_" + modid + "_" + i); 
}

//same idea with these:
document.getElementById("menu_" + modid + "_" + ind).style.backgroundImage = "url('phycho_hover.jpg')";
document.getElementById("menu_" + modid + "_" + ind).style.backgroundPosition = "top left";
document.getElementById("menu_" + modid + "_" + ind).style.backgroundRepeat = "repeat-x";

使用JavaScript更改背景重复图像

3 个答案: