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;
}
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";