php - AJAX 关闭图像在 Chrome 中运行时无法在 Firefox 上运行

Question

关闭图像 close2.jpg,close1.jpg 在 Firefox 中不起作用，而在 Google Chrome 和 IE 中运行。URL 是 www.srfads.com --> 主页--> 在印度查找广告--> 点击更改

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我的 js 文件是

enter code here
function show(id,x,y)
{
    var element = document.getElementById(id);
    //var exit_element = document.getElementById(exit_id);
    var width = window.innerWidth ? window.innerWidth : document.documentElement.clientWidth;
    if(width == 0)
    {
        width = 1280;
    }
    var height = window.innerHeight ? window.innerHeight : document.documentElement.clientHeight;
    if(height == 0)
    {
        height = 786;
    }
    element.style.position = "absolute";
    element.style.display = "block";

    element.style.left = (document.documentElement.scrollLeft + (width - element.clientWidth)/2 + x) + 'px';
    element.style.top = (document.documentElement.scrollTop + (height - element.clientHeight)/2 + y) + 'px';

    var xmlhttpobject = new XMLHttpRequest();
    xmlhttpobject.open("GET","sct.php",true);
    xmlhttpobject.send();
    xmlhttpobject.onreadystatechange = function ()
    {
        if(xmlhttpobject.readyState == 4 && xmlhttpobject.status == 200)
        {
            document.getElementById('shct').innerHTML = xmlhttpobject.responseText;
        }
    }
}

我的 php 页面是

改变

                        <div id="showcity" style="display:none;border:8px solid rgb(120,120,120); border-radius:8px;">
                            <table style="background:white;">
                            <tr>
                                <td style="text-align: left;padding: 10px; width:300px; height:20px;border-bottom:1px solid;">Select a City</td>
                                <td style="text-align: right;padding: 10px; width:300px; height:20px;border-bottom:1px solid;">
                                <div id="clsdiv1" style="display: block;"><img src="img\close2.jpg" id="close1" onmouseover="show1()" /></div>
                                <div id="clsdiv2" style="display: none;"><img src="img\close1.jpg" id="close2" onmouseout="show2()" onclick="popupexit('showcity')" /></div>
                                </td>
                            </tr>
                            <tr>
                                <td colspan="2" style="height:200px;">
                                <div id="shct"></div>
                                </td>
                            </tr>
                            </table>
                        </div>

Answer 1

将移动这条线：

xmlhttpobject.send();

在此代码下方：

xmlhttpobject.onreadystatechange = function ()
{
    if(xmlhttpobject.readyState == 4 && xmlhttpobject.status == 200)
    {
        document.getElementById('shct').innerHTML = xmlhttpobject.responseText;
    }
}

帮助？