I am struggling to get the img location from my database and execute it to use as a background image using css any ideas here is the code
Image location is set in data base in table users and named layout
thanks
Code:
<?php
session_start();
include "includes/db_connect.php";
include "includes/functions.php";
logincheck();
$username=$_SESSION['username'];
$viewuser=$_GET['viewuser'];
$fetch=mysql_fetch_object(mysql_query("SELECT * FROM users WHERE username='$username'"));
$select = mysql_query("SELECT * FROM users WHERE online > '$timenow' ORDER by id");
$num = mysql_num_rows($select);
$test=mysql_fetch_object(mysql_query("SELECT * FROM users WHERE username='$viewuser'"));
?>
<head>
<SCRIPT LANGUAGE="javascript">
function click() {
if (event.button==2) {
alert('Sorry, this function is disabled.')
}
}
document.onMouseDown=click
</SCRIPT>
<style type="text/css">
body {
background-color: #151515;
margin-left: 0px;
margin-top: 0px;
margin-right: 0px;
margin-bottom: 0px;
padding: 0;
SCROLLBAR-FACE-COLOR: #2b0303; SCROLLBAR-HIGHLIGHT-COLOR: #4b0505; SCROLLBAR-SHADOW-COLOR: #140101; SCROLLBAR-3DLIGHT-COLOR: 140101; SCROLLBAR-ARROW-COLOR: #FFFFFF; SCROLLBAR-TRACK-COLOR: #222222; SCROLLBAR-DARKSHADOW-COLOR: #210303;
}
body,td,th {
font-family: verdana;
font-size: 11px;
color: black;
}
a:link {
text-decoration: none;
}
a:active {
text-decoration: none;
}
a:hover {
text-decoration: none;
}
a:visited {
text-decoration: none;
}
.gradient {
font-family: Arial;
font-weight: Bold;
font-size: 12px;
font-style: normal;
line-height: 23px;
color: #FFFFFF;
background-image: url('<? echo"$test->layout"; ?>');
background-repeat: repeat-x;
cursor: pointer;
text-align: center;
}
a {
width: 140px;
display: block;
padding: 3px;
padding-right: 5px;
padding-left: 5px;
border-top: 1px solid black;
background-color: #333333;
color: #FFFFFF;
}
a:hover {
color: #FFFFFF;
background: #CD0000;
}
</style>
<!--[if IE]>
<style type="text/css">
a {
width: 150px;
display: block;
padding: 3px;
padding-right: 5px;
padding-left: 5px;
border-top: 1px solid black;
background-color: #333333;
color: #FFFFFF;
}
</style>
<![endif]-->
<script language="javascript" type="text/javascript">
function popitup(url) {
newwindow=window.open(url,'name','height=520,width=640');
if (window.focus) {newwindow.focus()}
return false;
}
function hide(id){
var hide = document.all? document.all[id + "_block"] : document.getElementById? document.getElementById(id + "_block") : "";
if(hide.style.display == 'none'){ hide.style.display = ''; }
else{ hide.style.display = 'none'; }
}
</script>
<table width="140" border="0" cellspacing="0" cellpadding="0" bgcolor="#151515">
<?php if ($fetch->userlevel == "2"){
echo "<tr>
<td class=gradient>Staff</td>
</tr>
</tbody>
</table>
Image location is set in data base in table users and named layout
thanks
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