JavaScript的“这个”不工作的HTML元素

问题描述 投票:0回答:1

我想我敢通过img访问数据库php元素显示在DIV的信息。我关联onmouseover event每个元素并通过this和一些参数作为参数,但我鼠标移到图片,DIV显示在浏览器的左上角和不显示图像的旁边。

我也试图与删除this,结果是相同的旧的。

为图像元素代码

         $row = mysqli_fetch_array($result);

          $firstname = $row['firstname'];

          $lastname = $row['lastname'];

          $image =    $row['image'];

          $phone =    $row['phone'];

          $email =    $row['email'];

     $realtorData = $firstname.'|'.$lastname.'|'.$phone.'|'.$email.'|';
echo "<img src='/../../Realtors/$image'onmouseover='showRealtorInfo(this,\"".$realtorData."\" );' onmouseout='hideRealtorInfo();'>";

JavaScript代码

  function showRealtorInfo(element, realtorInfo)
{
    var realtorArray = realtorInfo.split('|');

    var firstname = realtorArray[0];
    var lastname  = realtorArray[1];
    var phone     = realtorArray[2];
    var email     = realtorArray[3];

    var realtorInfoDiv = document.getElementById('realtorinfo');

    var myHTML  = "<p><b>" + firstname + " " + lastname + "</b><br /><br />";
        myHTML += "Phone: " + phone + "<br />";
        myHTML += "Email: " + email + "<br />";

    realtorInfoDiv.innerHTML = myHTML;

    x = element.offsetLeft;
    y = element.offsetTop;

    //alert(x);

    realtorInfoDiv.style.left = y + 100;
    realtorInfoDiv.style.top = x + 550;



    realtorInfoDiv.style.visibility = 'visible';


}

为div元素CSS代码

#realtorinfo{

position: absolute;
left: 10px;
top: 10px;
width: 200px;
height: 150px;
padding: 5px;
background-color: yellow;
visibility: hidden;
float: left;

} 
javascript php html css mysql
1个回答
0
投票

您还没有在后续LOC提到“PX”。

realtorInfoDiv.style.left = y + 100;
realtorInfoDiv.style.top = x + 550;

这应该工作:

realtorInfoDiv.style.left = y + 100 + "px";
realtorInfoDiv.style.top = x + 550 + "px";
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